Calculus and Its Applications, 10th Edition

Citation preview

10

th

EDITION

Calculus and its applications Marvin L. Bittinger Indiana University Purdue University Indianapolis

David J. Ellenbogen Community College of Vermont

Scott A. Surgent Arizona State University

Addison-Wesley Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

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Library of Congress Cataloging-in-Publication Data Bittinger, Marvin L. Calculus and its applications. — 10th ed./Marvin L. Bittinger, David J. Ellenbogen, Scott A. Surgent. p. cm. Includes index. 1. Calculus—Textbooks. I. Ellenbogen, David. II. Surgent, Scott A. III. Title. QA303.2.B466 2012 515—dc22 2010017892 Copyright © 2012 Pearson Education, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. For information on obtaining permission for use of material in this work, please submit a written request to Pearson Education, Inc., Rights and Contracts Department, 501 Boylston Street, Boston, MA, 02116. 1 2 3 4 5 6 7 8 9 10—CRK—13 12 11 10 © 2012, 2008, 2004 Pearson Education, Inc.

ISBN-10: 0-321-69433-3 ISBN-13: 978-0-321-69433-1

To: Elaine, Victoria, and Beth

Contents CHAPTER 2

Preface vii Prerequisite Skills Diagnostic Test xix

Applications of Differentiation

197

2.1 CHAPTER R

Functions, Graphs, and Models R.1 Graphs and Equations 2 R.2 Functions and Models 13 R.3 Finding Domain and Range 24 R.4 Slope and Linear Functions 33 R.5 Nonlinear Functions and Models 50 R.6 Mathematical Modeling and Curve Fitting Chapter Summary 79 Chapter Review Exercises 85 Chapter Test 88 Extended Technology Application Average Price of a Movie Ticket 90

1

68

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs 198 2.2 Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs 216 2.3 Graph Sketching: Asymptotes and Rational Functions 234 2.4 Using Derivatives to Find Absolute Maximum and Minimum Values 250 2.5 Maximum–Minimum Problems; Business and Economics Applications 262 2.6 Marginals and Differentials 277 2.7 Implicit Differentiation and Related Rates 288 Chapter Summary 295 Chapter Review Exercises 301 Chapter Test 303 Extended Technology Application Maximum Sustainable Harvest 305

CHAPTER 1

Differentiation 1.1

Limits: A Numerical and Graphical Approach 94 1.2 Algebraic Limits and Continuity 109 1.3 Average Rates of Change 121 1.4 Differentiation Using Limits of Difference Quotients 132 1.5 Differentiation Techniques: The Power and Sum–Difference Rules 144 1.6 Differentiation Techniques: The Product and Quotient Rules 158 1.7 The Chain Rule 166 1.8 Higher-Order Derivatives 177 Chapter Summary 185 Chapter Review Exercises 190 Chapter Test 192 Extended Technology Application Path of a Baseball: The Tale of the Tape 194

iv

93

CHAPTER 3

Exponential and Logarithmic Functions 3.1 Exponential Functions 308 3.2 Logarithmic Functions 322 3.3 Applications: Uninhibited and Limited Growth Models 337 3.4 Applications: Decay 353 3.5 The Derivatives of ax and loga x 366 3.6 An Economics Application: Elasticity of Demand 371 Chapter Summary 378 Chapter Review Exercises 382 Chapter Test 384 Extended Technology Application The Business of Motion Picture Revenue and DVD Release 385

307

v

CONTENTS

CHAPTER 4

Integration

389

4.1 4.2 4.3 4.4 4.5 4.6

Antidifferentiation 390 Antiderivatives as Areas 399 Area and Definite Integrals 410 Properties of Definite Integrals 425 Integration Techniques: Substitution 436 Integration Techniques: Integration by Parts 445 4.7 Integration Techniques: Tables 454 Chapter Summary 459 Chapter Review Exercises 466 Chapter Test 468 Extended Technology Application Business: Distribution of Wealth 469

Cumulative Review 602 Appendix A: Review of Basic Algebra Appendix B: Regression and Microsoft Excel 616 Appendix C: MathPrint Operating System for TI-84 and TI-84 Plus Silver Edition

Table A:

Areas for a Standard Normal Distribution

Answers A-1 Index of Applications Index I-4

605

618

621 I-1

CHAPTER 5

Applications of Integration An Economics Application: Consumer Surplus and Producer Surplus 5.2 Applications of Integrating Growth and Decay Models 480 5.3 Improper Integrals 492 5.4 Probability 498 5.5 Probability: Expected Value; The Normal Distribution 508 5.6 Volume 522 5.7 Differential Equations 526 Chapter Summary 535 Chapter Review Exercises 541 Chapter Test 543 Extended Technology Application Curve Fitting and Volumes of Containers 545

473

5.1

474

CHAPTER 6

Functions of Several Variables 547 6.1 6.2 6.3 6.4

Functions of Several Variables 548 Partial Derivatives 556 Maximum–Minimum Problems 565 An Application: The Least-Squares Technique 572 6.5 Constrained Optimization 579 6.6 Double Integrals 588 Chapter Summary 594 Chapter Review Exercises 598 Chapter Test 599 Extended Technology Application Minimizing Employees’ Travel Time in a Building 600

SUPPLEMENTARY CHAPTERS

Available to qualified instructors through the Pearson Instructor Resource Center (www.pearsonhighered.com/ irc) and to students at the downloadable student resources site (www.pearsonhighered.com/mathstatsresources) or within MyMathLab

A. Sequences and Series

online

A.1 Infinite Sequences A.2 Infinite Series A.3 The Ratio Test and Power Series A.4 Taylor Series and Taylor Polynomials Summary and Review Test

B. Differential Equations

online

B.1 Further Background B.2 First-Order Linear Differential Equations B.3 Graphical Analysis B.4 Numerical Analysis: Euler’s Method Summary and Review Test

C. Trigonometric Functions C.1 Introduction to Trigonometry C.2 Derivatives of the Trigonometric Functions C.3 Integration of the Trigonometric Functions C.4 Inverse Trigonometric Functions Summary and Review Test

online

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vi

Preface Calculus and Its Applications is the most student-oriented applied calculus text on the market, and this tenth edition continues to improve on that approach. The authors believe that appealing to students’ intuition and speaking in a direct, down-to-earth manner make this text accessible to any student possessing the prerequisite math skills. By presenting more topics in a conceptual and often visual manner and adding student self-assessment and teaching aids, this revision addresses students’ needs better than ever before. However, it is not enough for a text to be accessible—it must also provide students with motivation to learn. Tapping into areas of student interest, the authors provide an abundant supply of examples and exercises rich in real-world data from business, economics, environmental studies, health care, and the life sciences. New examples cover applications ranging from the distribution of wealth to the growth of membership in Facebook. Found in every chapter, realistic applications draw students into the discipline and help them to generalize the material and apply it to new and novel situations. To further spark student interest, hundreds of meticulously drawn graphs and illustrations appear throughout the text, making it a favorite among students who are visual learners. Appropriate for a one-term course, this text is an introduction to applied calculus. A course in intermediate algebra is a prerequisite, although Appendix A: Review of Basic Algebra, together with Chapter R, provides a sufficient foundation to unify the diverse backgrounds of most students. For schools offering a two-term course, additional chapters are available online; see the Contents.

Our Approach Intuitive Presentation Although the word “intuitive” has many meanings and interpretations, its use here means “experience based, without proof.” Throughout the text, when a concept is discussed, its presentation is designed so that the students’ learning process is based on their earlier mathematical experience. This is illustrated by the following situations. ■ Before the formal definition of continuity is presented, an informal explanation is given, complete with graphs that make use of student intuition into ways in which a function could be discontinuous (see pp. 113–114). ■ The definition of derivative, in Chapter 1, is presented in the context of a discussion of average rates of change (see p. 135). This presentation is more accessible and realistic than the strictly geometric idea of slope. ■ When maximization problems involving volume are introduced (see p. 264), a function is derived that is to be maximized. Instead of forging ahead with the standard calculus solution, the student is first asked to stop, make a table of function values, graph the function, and then estimate the maximum value. This experience provides students with more insight into the problem. They recognize that not only do different dimensions yield different volumes, but also that the dimensions yielding the maximum volume may be conjectured or estimated as a result of the calculations. ■ Relative maxima and minima (Sections 2.1 and 2.2) and absolute maxima and minima (Section 2.4) are covered in separate sections in Chapter 2, so that students gradually build up an understanding of these topics as they consider graphing using calculus concepts (see pp. 198–234 and 250–262). ■ The explanation underlying the definition of the number e is presented in Chapter 3 both graphically and through a discussion of continuously compounded interest (see pp. 345–347).

vii

viii

P R E FA C E

Strong Algebra Review One of the most critical factors underlying success in this course is a strong foundation in algebra skills. We recognize that students start this course with varying degrees of skills, so we have included multiple opportunities to help students target their weak areas and remediate or refresh the needed skills. New! ■ Prerequisite Skills Diagnostic Test (Part A). This portion of the diagnostic test assesses skills refreshed in Appendix A: Review of Basic Algebra. Answers to the questions reference specific examples within the appendix. ■ Appendix A: Review of Basic Algebra. This 11-page appendix provides examples on topics such as exponents, equations, and inequalities and applied problems. It ends with an exercise set, for which answers are provided at the back of the book so that students can check their understanding. New! ■ Prerequisite Skills Diagnostic Test (Part B). This portion of the diagnostic test assesses skills that are reviewed in Chapter R, and the answers reference specific sections in that chapter. Some instructors may choose to cover these topics thoroughly in class, making this assessment less critical. Other instructors may use all or portions of this test to determine whether there is a need to spend time remediating before moving on with Chapter 1. ■ Chapter R. This chapter covers basic concepts related to functions, graphing, and modeling. It is an optional chapter based on the prerequisite skills students have. New! ■ “Getting Ready for Calculus” in MyMathLab. This optional chapter within MyMathLab provides students with the opportunity to self-remediate in an online environment. Assessment and guidance are provided.

Applications Relevant and factual applications drawn from a broad spectrum of fields are integrated throughout the text as applied examples and exercises and are also featured in separate application sections. These have been updated and expanded in this edition to include even more applications using real data. In addition, each chapter opener in this edition includes an application that serves as a preview of what students will learn in the chapter. The applications in the exercise sets are grouped under headings that identify them as reflecting real-life situations: Business and Economics, Life and Physical Sciences, Social Sciences, and General Interest. This organization allows the instructor to gear the assigned exercises to a particular student and also allows the student to know whether a particular exercise applies to his or her major. Furthermore, the Index of Applications at the back of the book provides students and instructors with a comprehensive list of the many different fields considered throughout the text.

Approach to Technology This edition continues to emphasize mathematical modeling, utilizing the advantages of technology as appropriate. Though the use of technology is optional with this text, its use meshes well with the text’s more intuitive approach to applied calculus. For example, the use of the graphing calculator in modeling, as an optional topic, is introduced in Section R.6 and then reinforced many times throughout the text. Technology Connections Technology Connections are included throughout the text to illustrate the use of technology. Whenever appropriate, art that simulates graphs or tables generated by a graphing calculator is included as well. This edition also includes discussion of the iPhone applications Graphicus, iPlot, and Grafly New! to take advantage of technology to which many students have access. There are four types of Technology Connections for students and instructors to use for exploring key ideas. ■ Lesson/ Teaching. These provide students with an example, followed by exercises to work within the lesson. ■ Checking. These tell the students how to verify a solution within an example by using a graphing calculator. ■ Exploratory/Investigation. These provide questions to guide students through an investigation.

ix

P R E FA C E ■

Technology Connection Exercises. Most exercise sets contain technology-based exercises identified with either a icon or the heading “Technology Connection.” These exercises also appear in the Chapter Review Exercises and the Chapter Tests. The Printable Test Forms include technology-based exercises as well.

Use of Art and Color y

f (x) = x2

1

−1

g(x) =

One of the hallmarks of this text is the pervasive use of color as a pedagogical tool. Color is used in a methodical and precise manner so that it enhances the readability of the text for students and instructors. When two curves are graphed using the same set of axes, one is usually red and the other blue with the red graph being the curve of major importance. This is exemplified in the graphs from Chapter R (pp. 54 and 82) at the left. Note that the equation labels are the same color as the curve. When the instructions say “Graph,” the dots match the color of the curve. The following figure from Chapter 1 (p. 134) shows the use of colors to distinguish between secant and tangent lines. Throughout the text, blue is used for secant lines and red for tangent lines.

x

1 −1

x3

Slope m 1

y1 = x3, y2 = 3x + 1 y1 y2

8

Q1

Slope m 2 Secant lines

Q2

Slope m 3 –3

3

Q3

Slope m 4 T

Q4

–5

Slope m = instantaneous rate of change at P

Tangent line

P

We next find second coordinates by substituting the critical values in the original function: f1 - 32 = 1- 323 + 31- 322 - 91- 32 - 13 = 14; f112 = 1123 + 31122 - 9112 - 13 = - 18. Are the points 1 - 3, 142 and 11, - 182 relative extrema? Let’s look at the second derivative. We use the Second-Derivative Test with the critical values - 3 and 1: f –1 - 32 = 61- 32 + 6 = - 12 6 0;

Relative maximum

f –112 = 6112 + 6 = 12 7 0.

Relative minimum

Thus, f1 - 32 = 14 is a relative maximum and f112 = - 18 is a relative minimum. We plot both 1- 3, 142 and 11, - 182, including short arcs at each point to indicate the graph’s concavity. Then, by calculating and plotting a few more points, we can make a sketch, as shown below. y

y

40

40

Relative maximum (−3, 14)

30

Relative maximum

20

(−3, 14)

−4

−3

−2

−1

1

2

3

4

x

−5

−4

−3

−2

(3, 14)

1

2

3

(0, −13) (−5, −18)

(1, −18) Relative minimum

4

x

(2, −11)

(−1, −2)

−10 −20

20

(−2, 9) 10

(−4, 7)

10 −5

f

30

In the text from Chapter 2 (p. 219) shown at the left, the color red denotes substitution in equations and blue highlights the corresponding outputs, including maximum and minimum values. The specific use of color is carried out in the figure that follows. Note that when dots are used for emphasis other than just merely plotting, they are black. Beginning with the discussion of integration in Chapter 4, the color amber is used to highlight areas in graphs. The figure to the left below, from Chapter 4 (p. 427), illustrates the use of blue and red for the curves and labels and amber for the area.

−20

(1, −18) Relative minimum

y

z

Absolute maximum

y = f(x) y = g(x)

Relative maximum

A

a

b

Graph of f

x y

x

Relative minimum

New! In Chapter 6, all of the three-dimensional art has been redrawn for this edition, making it even easier for students to visualize the complex graphs presented in this chapter, like the one above (p. 565).

x

P R E FA C E

Pedagogy of Calculus and Its Applications, Tenth Edition Chapter Openers New!

Exponential and Logarithmic Functions

Each newly designed chapter opener provides a “Chapter Snapshot” that gives students a preview of the topics in the chapter and an application that whets their appetite for the chapter material and provides an intuitive introduction to a key calculus topic. (See pp. 197, 307, and 389.)

3

Chapter Snapshot What You’ll Learn

Why It’s Important

3.1 Exponential Functions 3.2 Logarithmic Functions 3.3 Applications: Uninhibited and Limited Growth Models 3.4 Applications: Decay 3.5 The Derivatives of ax and loga x 3.6 An Economics Application: Elasticity of Demand

In this chapter, we consider two types of functions that are closely related: exponential functions and logarithmic functions. After learning to find derivatives of such functions, we will study applications in the areas of population growth and decay, continuously compounded interest, spread of disease, and carbon dating.

Section Objectives As each new section begins, its objectives are stated in the margin. These can be spotted easily by the student, and they provide the answer to the typical question “What should I be able to do after completing this section?” (See pp. 322, 399, and 480.)

Where It’s Used

p

3.1

COMIC BOOK VALUE BY YEAR V(t)

MILLION-DOLLAR COMIC BOOK

This problem appears as Example 7 in Section 3.3.

OBJECTIVES Value of comic book

A 1939 comic book with the first appearance of the “Caped Crusader,” Batman, sold at auction in Dallas in 2010 for $1.075 million. The comic book originally cost 10¢. What will the value of the comic book be in 2020? After what time will the value of the comic book be $30 million?

$1.075 million

• Graph exponential functions.

g

Exponential Functions Graphs of Exponential Functions Consider the following graph. The rapid rise of the graph indicates that it approximates an exponential function. We now consider such functions and many of their applications.

• Differentiate exponential functions.

V(t) = 0.10e 0.228t

WORLD POPULATION GROWTH 8 billion* 2020 6 billion 1998

$0.10

5 billion 1987 81

Number of years since 1939

4 billion 1974

t

307

3 billion 1960 2 billion 1927 1 billion 1804 1800 *Projected

1900

2000

2100

Year

(Source: U.S. Census Bureau.)

Let’s review definitions of expressions of the form a x, where x is a rational number. l

Technology Connections

TECHNOLOGY CONNECTION

The text allows the instructor to incorporate graphing calculators, spreadsheets, and smart phone applications into classes. All use of technology is clearly labeled so that it can be included or omitted as desired. (See pp. 54–56 and 209–212.)

Check the results of Example 3 by entering each function as y1 and letting y2 = nDeriv1y1, x, x2. Then enter the derivatives from Example 3 as y3 and use graphs or a table to compare y2 and y3. Using iPlot, graph f 1x2 = 3e x in red. Turn on Derivate. Now suppose the derivative was mistakenly found to be f ¿1x2 = 3e x - 1. Graph this incorrect function in some other color. What happens? Explain. Then describe a procedure for checking the results of Example 3 using iPlot. Using Graphicus, graph g1x2 = x 2e x as a first function. Then touch + and choose Add derivative. Suppose the derivative was mistakenly found to be g¿1x2 = 2xe x. Graph this incorrect function. What happens? Explain. Then describe a procedure for checking the results of Example 3 using Graphicus.

Exploratory

Quick Check Exercises New!

Finding Derivatives of Functions Involving e We can use Theorem 1 in combination with other theorems derived earlier to differentiate a variety of functions.

I EXAMPLE 3 Find dy>dx: a) y = 3e x; b) y = x 2e x; c) y =

ex . x3

Solution a)

b)

c)

d d 13e x2 = 3 e x dx dx = 3e x

Recall that

d 3c # f 1x24 = c # f ¿1x2. dx

d 2 x 1x e 2 = x 2 # e x + e x # 2x dx = e x1x 2 + 2x2, or xe x1x + 22 x

d e x a b = dx x 3 = =

3

#e

x

- e x6

x

# 3x

x 2e x1x - 32 x6 e x1x - 32 x

Using the Product Rule Factoring

2

4

Using the Quotient Rule

Factoring

Simplifying

Suppose that we have a more complicated function in the exponent, as in h1x2 = e x

2

- 5x

.

This is a composition of functions. For such a function, we have h1x2 = g1 f1x22 = e f1x2, where g1x2 = e x and f1x2 = x 2 - 5x. Now g¿1x2 = e x. Then by the Chain Rule (Section 1.7), we have h¿1x2 = g¿1 f1x22 # f ¿1x2 = e f 1x2 # f ¿1x2.

For the case above, f1x2 = x 2 - 5x, so f ¿1x2 = 2x - 5. Then h¿1x2 = g¿1 f1x22 # f ¿1x2

O

Giving students the opportunity to check their understanding of a new concept or skill is vital to their learning and their confidence. In this edition, Quick Check exercises follow and mirror selected examples in the text, allowing students to both practice and assess the skills they are learning. Instructors may include these as part of a lecture as a means of gauging skills and gaining immediate feedback. Answers to the Quick Check exercises are provided at the end of each section following the exercise set. (See pp. 236, 331, and 412.)

In Section 3.5, we will develop a formula for the derivative of the more general exponential function given by y = a x.

= e f 1x2 # f ¿1x2 2 = e x - 5x12x - 52.

Quick Check 3 Differentiate: a) y = 6e x; b) y = x 3e x; ex c) y = 2 . x

The next theorem, which we have proven using the Chain Rule, allows us to find derivatives of functions like the one above.

O Quick Check 3

P R E FA C E Section Summary • The exponential function f1x2 = e x, where e L 2.71828, has the derivative f ¿1x2 = e x. That is, the slope of a tangent line to the graph of y = e x is the same as the function value at x.

• The graph of f 1x2 = e x is an increasing function with no critical values, no maximum or minimum values, and no points of inflection. The graph is concave up, with lim f1x2 = q

lim f1x2 = 0.

and

x: q

x: -q

• Calculus is rich in applications of exponential functions.

EXERCISE SET

3.1

Graph. 1. y = 4 x

2. y = 5x

3. y = 10.252x

4. y = 10.22x

5. f 1x2 =

A 32 B x

7. g1x2 =

A 23 B x x

6. f 1x2 =

A 43 B x

8. g1x2 =

A 34 B x

Differentiate. 12. f 1x2 = e x

13. g1x2 = e 3x

14. g1x2 = e 2x

48. y = e x + x 3 - xe x

49. y = 1 - e -x

50. y = 1 - e -3x

APPLICATIONS

- 7x

54. g1x2 = 15x 2 - 8x2e x

- 4x

2

81. U.S. exports. U.S. exports of goods are increasing exponentially. The value of the exports, t years after 2009, can be approximated by V1t2 = 1.6e 0.046t, where t = 0 corresponds to 2009 and V is in billions of dollars. (Source: U.S. Commerce Department.) a) Estimate the value of U.S. exports in 2009 and 2020. b) What is the doubling time for the value of U.S. exports?

84. Marginal cost. A company’s total cost, in millions of dollars, is given by C1t2 = 200 - 40e -t, where t is the time in years since the start-up date. C(t) 200

50 –2

2

4

6

10 t

8

Find each of the following. a) The marginal cost C¿1t2 b) C¿102 c) C¿152 (Round to the nearest thousand.) d) Find lim C1t2 and lim C¿1t2. Why do you think t: q t: q the company’s costs tend to level off as time passes? 85. Marginal demand. At a price of x dollars, the demand, in thousands of units, for a certain music player is given by the demand function

A1t2 = 2.43e 0.18t. (Source: Nutrition Business Journal, 2004.) a) Estimate the amount that Americans spent on organic food and beverages in 2009. b) Estimate the rate at which spending on organic food and beverages was growing in 2006. 83. Marginal cost. A company’s total cost, in millions of dollars, is given by C1t2 = 100 - 50e -t, where t is the time in years since the start-up date. C(t) C(t) = 100 – 50e –t

80

86. Marginal supply. At a price of x dollars, the supply function for the music player in Exercise 85 is given by q = 75e 0.004x, where q is in thousands of units. a) How many music players will be supplied at a price of $250? Round to the nearest thousand. b) Graph the supply function for 0 … x … 400. c) Find the marginal supply, q¿1x2. d) Interpret the meaning of the derivative.

C1t2 = 10t 2e -t.

40 20 5

10

15

a) Find the concentration after 0 hr, 1 hr, 2 hr, 3 hr, and 10 hr. b) Sketch a graph of the function for 0 … t … 10. c) Find the rate of change of the concentration, C¿1t2. d) Find the maximum value of the concentration and the time at which it occurs. e) Interpret the meaning of the derivative.

t

20

Find each of the following. a) The marginal cost, C¿1t2 b) C¿102

Social Sciences

TECHNOLOGY CONNECTION

88. Ebbinghaus learning model. Suppose that you are given the task of learning 100% of a block of knowledge. Human nature is such that we retain only a percentage P of knowledge t weeks after we have learned it. The Ebbinghaus learning model asserts that P is given by P1t2 = Q + 1100 - Q2e -kt, where Q is the percentage that we would never forget and k is a constant that depends on the knowledge learned. Suppose that Q = 40 and k = 0.7. a) Find the percentage retained after 0 weeks, 1 week, 2 weeks, 6 weeks, and 10 weeks. b) Find lim P1t2. t: q

c) Sketch a graph of P. d) Find the rate of change of P with respect to time t. e) Interpret the meaning of the derivative.

Use a graphing calculator (or iPlot or Graphicus) to graph each function in Exercises 107 and 108, and find all relative extrema. 107. f 1x2 = x 2e -x

108. f 1x2 = e -x

2

For each of the functions in Exercises 109–112, graph f, f ¿, and f –. 109. f 1x2 = e x

110. f 1x2 = e -x

111. f 1x2 = 2e 0.3x

112. f 1x2 = 1000e -0.08x

113. Graph 1 x f 1x2 = a1 + b . x Use the TABLE feature and very large values of x to confirm that e is approached as a limit.

Answers to Quick Checks

SYNTHESIS Differentiate. 2

89. y = 1e 3x + 125

93. y =

a) How many music players will be bought at a price of $250? Round to the nearest thousand. b) Graph the demand function for 0 … x … 400. c) Find the marginal demand, q¿1x2. d) Interpret the meaning of the derivative.

87. Medication concentration. The concentration C, in parts per million, of a medication in the body t hours after ingestion is given by the function

60

91. y =

q = 240e -0.003x.

Life and Physical Sciences

100

90. y = 1e x - 224

e 3t - e 7t e

1 1 1 1. 1, 3, 9, 27, , , 3 9 27

3

ex 1 - ex

94. y =

95. f 1x2 = e 2x + 2e x 97. f 1x2 = e x>2 # 2x - 1 e x - e -x 99. f1x2 = x e + e -x

1 96. f 1x2 = x + e 1>x e

8

6

6 4

98. f 1x2 =

xe

-x

1 + x2

100. f 1x2 = e

f(x)  3 x

2 −2

−1

1

2

1 x g(x)  () 3

2 −2

x

−1

1

(b) x 2e x 1x + 32; (c)

3. (a) 6e x;

4. (a) - 4e -4x; 5. (a)

(b) e x

3

+ 8x

x3

13x 2 + 82;

(c)

(b)

y

t:0

1>1t - 12

, we have e = lim g1t2. Find g10.52, t:1

g10.92, g10.992, g10.9992, and g10.99982. 103. Find the maximum value of f 1x2 = x 2e -x over 30, 44. 104. Find the minimum value of f 1x2 = xe over 3 - 2, 04. x

105. A student made the following error on a test: d x e = xe x - 1. dx Identify the error and explain how to correct it. 106. Describe the differences in the graphs of f 1x2 = 3x and g1x2 = x 3.

2 −1

+5

6

6

−2

2

2x 2 + 5

8

4

4

f 10.52, f 10.22, f 10.12, and f 10.0012.

x

xe 2x

y

8

ex

101. For f 1t2 = 11 + t21>t, we have e = lim f 1t2. Find f 112,

2

e 1x - 22 x

Exercises 101 and 102 each give an expression for e. Find the function values that are approximations for e. Round to five decimal places.

102. For g1t2 = t

y

8

4

ex x2 + 1

f(x)  2ex 1

2

x

No critical values Decreasing on 1- q , q2 No inflection points Concave up on 1- q , q2 (c) y

2 −2

−1

g(x)  2e x 1

2

x

No critical values Increasing on 1 - q, q2 No inflection points Concave up on 1- q , q2

1

0.8 0.6

h(x)  1  ex

0.4 0.2 2

4

6

There are over 3500 exercises in this edition. All exercise sets are enhanced by the inclusion of real-world applications, detailed art pieces, and illustrative graphs.

8 10 x

No critical values Increasing on 1- q , q2 No inflection points Concave down on 1- q, q 2 6. (a) 20, 26.8, 47.1, 61.9, 74.6, 78.4 (b) (c) N¿1t2 = 7.2e -0.12t; after t days, the rate of change of number of phones produced per day is given by 7.2e -0.12t. (d) 80 phones produced per day

N(t) 80 70 60 50 40 30 20 10

N(t)  80  60e0.12t

10

20

30

Applications A section of applied problems is included in nearly every exercise set. The problems are grouped under headings that identify them as business and economics, life and physical sciences, social sciences, or general interest. Each problem is accompanied by a brief description of its subject matter (see pp. 155–157, 347–351, and 397–398). Thinking and Writing Exercises Identified by a , these exercises ask students to explain mathematical concepts in their own words, thereby strengthening their understanding (see pp. 143, 249, and 422). Synthesis Exercises Synthesis exercises are included in every exercise set, including the Chapter Review Exercises and Chapter Tests. They require students to go beyond the immediate objectives of the section or chapter and are designed to both challenge students and make them think about what they are learning (see pp. 176, 276, and 364). Technology Connection Exercises These exercises appear in the Technology Connections (see pp. 29, 141, and 327) and in the exercise sets (see pp. 120, 249, and 425). They allow students to solve problems or check solutions using a graphing calculator or smart phone. Concept Reinforcement Exercises As always, each chapter closes with a set of Chapter Review Exercises, which includes 8 to 14 Concept Reinforcement exercises at the beginning. The exercises are confidence builders for students who have completed their study of the chapter. Presented in matching, true/false, or fill-in-the-blank format, these exercises can also be used in class as oral exercises. Like all review exercises, each concept reinforcement exercise is accompanied by a bracketed section reference to indicate where discussion of the concept appears in the chapter. (See pp. 301, 382, and 466.)

1 1 1 2. 1, , , , 3, 9, 27 3 9 27

y

92. y = 2e 3t + t

4t

Variety of Exercises

C(t) = 200 – 40e –t

150 100

82. Organic food. More Americans are buying organic fruit and vegetables and products made with organic ingredients. The amount A1t2, in billions of dollars, spent on organic food and beverages t years after 1995 can be approximated by

To assist students in identifying the key topics for each section, a Section Summary now precedes every exercise set. Key concepts and definitions are presented in bulleted list format to help focus students’ attention on the most important ideas presented in the section. (See pp. 106, 246, and 360.)

Graph each function. Then determine critical values, inflection points, intervals over which the function is increasing or decreasing and the concavity

c) C¿142 (Round to the nearest thousand.) d) Find lim C1t2 and lim C¿1t2. Why do you think t: q t: q the company’s costs tend to level off as time passes?

Business and Economics

Section Summary New!

52. y = 1 - e -mx

53. g1x2 = 14x 2 + 3x2e x

2

10. f 1x2 = 11.22

11. f 1x2 = e -x

46. y = 2e x + 1

47. y = xe -2x + e -x + x 3

51. y = 1 - e -kx x

9. f 1x2 = 12.52

45. y = 2e x - 1

xi

40

t

P R E FA C E

Redesigned Chapter Summary New!

CHAPTER 5 SUMMARY

We introduced chapter summaries in the Ninth Edition, and they were well received by students and instructors. To make the summaries even more user-friendly in this Tenth Edition, we have reformatted them in a tabular style that makes it even easier for students to distill key ideas. Each chapter summary presents a section-bysection list of key definitions, concepts, and theorems, with examples for further clarification. (See pp. 185, 295, and 378.)

EXAMPLES

Price

Consumer surplus

Q

p ⫽D(x) Total expenditure

If p = S1x2 is a supply function, then the producer surplus at a point 1Q, P2 is S1x2 dx.

Units

Price S(x)

Consumer surplus P

D(x)

Producer surplus Q

Let p = 12 - 1.5x be a demand function and p = 4 + 0.5x be a supply function. The two curves intersect at (4, 6), the equilibrium point. At this point, the consumer surplus is 4

L0

112 - 1.5x2 dx - 142162 = 36 - 24 = $12,

Chapter Reviews and Tests

and the producer surplus is 4

142162 -

L0

14 + 0.5x2 dx = 24 - 20 = $4.

At the end of each chapter are review exercises and a test. The Chapter Review Exercises, which include bracketed references to the sections in which the related course content first appears, provide comprehensive coverage of each chapter’s material (see pp. 190–192). The Chapter Test includes synthesis and technology questions (see pp. 192–193). There is also a Cumulative Review at the end of the text that can serve as a practice final examination. The answers, including section references, to the chapter tests and the Cumulative Review are at the back of the book. Six additional forms of each of the chapter tests and the final examination, with answer keys and ready for classroom use, appear in the Printable Test Forms. y

Consumer surplus ⫽ $12

$12 11 10 9 8 7 6 5 4

0

Producer surplus ⫽ $4

p ⫽ 4 ⫹ 0.5x (x E, p E) ⫽ (4, 6) p ⫽ 12 ⫺ 1.5x

1

2

3

4

5

x

CHAPTER 4 REVIEW EXERCISES

CHAPTER 4 TEST

These review exercises are for test preparation. They can also be used as a practice test. Answers are at the back of the book. The blue bracketed section references tell you what part(s) of the chapter to restudy if your answer is incorrect.

REVIEW EXERCISES

Classify each statement as either true or false. 1. Riemann sums are a way of approximating the area under a curve by using rectangles. [4.2] b

2. If a and b are both negative, then

La

negative. [4.3]

f 1x2 dx is

3. For any continuous function f defined over 3 - 1, 74, it follows that 2

7

f1x2 dx +

f1x2 dx =

f1x2 dx . [4.4]

5. 6. 7. 8. 9. 10.

1 L 2x L

200

0

100

b) - x-1 + C

2x L 1 + x2 1 L x2

2. 4.

L L

23x dx

Evaluate. [4.1] L

e) 2x1>2 + C

dx

ae x +

L 11 + x 222

dx

f) ln 11 + x22 + C

L

e -0.5x dx

13e x + 22 dx

L-1

C¿1x2 = - 0 .2x + 500, C102 = 0 .

where W(t) is the speed, in words per minute, at time t. How many words are translated during the second minute (from t = 1 to t = 2)? 24. A robot leaving a spacecraft has velocity given by v1t2 = - 0 .4t 2 + 2t, where v(t) is in kilometers per hour and t is the number of hours since the robot left the spacecraft. Find the total distance traveled during the first 3 hr.

12x + 3x 22 dx e2

10.

dx Le x

5

t Time (in days)

dx L x17 - x2

W1t2 = - 6t 2 + 12t + 90, t in 30, 44,

e -2x dx

L0

x 3 ln x 4 dx

Find the total cost of producing 100 air conditioners.

t Time (in hours)

1

9.

L

23. Social science: learning curve. A translator’s speed over 4-min interval is given by

2

8.

18.

L

19.

22. Business: cost from marginal cost. An air conditioning company determines that the marginal cost, in dollars, for the xth air conditioner is given by

Evaluate.

In each case, give an interpretation of the shaded region. [4.2, 4.3]

17.

21. Find the area of the region in the first quadrant bounded by y = x and y = x 5 .

4 ; 31 , 34 x

6. y =

xe 5x dx

2 x dx

20. Find the average value of y = 4t 3 + 2t over 3- 1, 24 .

1000x 5 dx

7. Give an interpretation of the shaded area.

15. y = 4 - x 2 ; 3 - 2, 14

t Time (in minutes)

L

L

Evaluate using Table 1.

5 x

1 + x 3>8 b dx 1assume x 7 02 x

1 a3t + 5t + b dt 1assume t 7 02 14. t L

17.

4

3.

5. y = x - x 2 ; 30, 14 13.

3

Find the area under the curve over the indicated interval.

16. y = x 2 + 2x + 1; 30, 34 1 d) - 11 + 2x2-1 + C 2

dx

2x

c) - 11 + x22-1 + C

2

Evaluate.

50 100 150 200 250 300 x Number of units produced

20x 4 dx

1

18.

Find the area under the curve over the indicated interval. [4.3]

1 dx, x 7 0 Lx

16.

5

2

11 + 2x2-2 dx

14.

t 31t 4 + 329 dt

Evaluate using integration by parts. Do not use Table 1.

10

C′

300

4

L

dx L x + 12 L

15

400

a C¿1x i2 ¢x , with ¢x = 50. [4.2]

12.

13. 15.

20

500

i=1

a) ln x + C

dx

y 25

Approximate the total cost of producing 200 car stereos by computing the sum

L-1 L2 L-1 4. Every integral can be evaluated using integration by parts. [4.6]

Column B

y 600

0

7

Match each integral in column A with the corresponding antiderivative in column B. [4.1, 4.5] Column A

5

125 - x 22 dx L0 by computing the area of each rectangle and adding.

C¿1x2 = 0.004x 2 - 2x + 500 .

CONCEPT REINFORCEMENT

Evaluate using substitution. Assume u 7 0 when ln u appears. Do not use Table 1.

1. Approximate

11. Business: total cost. The marginal cost, in dollars, of producing the xth car stereo is given by

x2 , 11. g1x2 dx , where g1x2 = b 6 - x, L0

for x … 2 , for x 7 2

b

12. Decide whether

La

f1x2 dx is positive, negative, or zero.

y f a

b x

Integrate using any method. Assume u>0 when ln u appears. 25.

6 dx L 5 + 7x

26.

L

x 5e x dx

Extended Technology Applications

Extended Technology Application

Extended Technology Applications at the end of each chapter use real applications and real data. They require a step-by-step analysis that encourages group work. More challenging in nature, the exercises in these features involve the use of regression to create models on a graphing calculator.

Business: Distribution of Wealth

(1, 1) Line of equality

(0, 0)

x Percentage of population

Most equitable distribution

y

(0, 0)

(1, 1)

f(x) x Percentage of population

y

(0, 0)

(1, 1)

f(x) x Percentage of population

y

(1, 1)

Percentage of wealth

y

“x% of the population owns x% of the wealth,” and this is modeled by the function f1x2 = x, where 0 … x … 1. This is an example of a Lorenz function that is often called the line of equality. In many societies, the distribution of wealth is not equitable. For example, the Lorenz function f1x2 = x 3 would represent a society in which a large percentage of the population owns a small percentage of the wealth. For example, in this society, we observe that f10.72 = 0.73 = 0.343, meaning that 70% of the population owns just 34.3% of the wealth, with the implication that the other 30% owns the remaining 65.7% of the wealth. In the graphs below, we see the line of equality in the left-most graph, and increasingly inequitable distributions as we move to the right.

Percentage of wealth

Lorenz Functions and the Gini Coefficient The distribution of wealth within a population is of great interest to many economists and sociologists. Let y = f 1x2 represent the percentage of wealth owned by x percent of the population, with x and y expressed as decimals between 0 and 1. The assumptions are that 0% of the population owns 0% of the wealth and that 100% of the population owns 100% of the wealth. With these requirements in place, the Lorenz function is defined to be any continuous, increasing and concave upward function connecting the points 10, 02 and 11, 12, which represent the two extremes. The function is named for economist Max Otto Lorenz (1880–1962), who developed these concepts as a graduate student in 1905–1906. If the collective wealth of a society is equitably distributed among its population, we would observe that

Percentage of wealth

The equilibrium point 1x E, PE2 is the point at which the supply and demand curves intersect.

Units

Percentage of wealth

L0

Total receipts

Units

Q

QP -

p ⫽S(x)

(Q, P)

(Q, P)

D1x2 dx - QP.

Running speed (in miles per hour)

L0

Producer surplus

Price

Sales on the tth day

If p = D1x2 is a demand function, then the consumer surplus at a point 1Q, P2 is

Marginal cost (in dollars)

KEY TERMS AND CONCEPTS SECTION 5.1

Keyboard speed (in words per minute)

xii

f(x)

(0, 0)

x Percentage of population

Less equitable distribution

P R E FA C E

xiii

New and Revised Content In response to faculty and student feedback, we have made many changes to the text’s content for this edition. New examples and exercises have been added throughout each chapter, as well as new problems to each chapter’s review exercises and chapter test. Data have been updated wherever achievable, so that problems use the most up-to-date information possible. Following is an overview of the major changes in each chapter. Chapter R Chapter R contains numerous updated problems involving real-world data. We have continued to stress the use of regression for modeling throughout the text. New to this edition is the introduction in Section R.5 of two apps for the iPhone: iPlot and Graphicus. These are, in effect, graphing calculator apps. Though they do not perform all of the tasks that graphing calculators like the TI-83 Plus and the TI-84 Plus do, they are accessibly priced and visually appealing. (See pp. 55–56.) Chapter 1 Chapter 1 contains 10 new examples designed to reinforce the main concepts and applications of limits, continuity, derivatives, and the Chain Rule. Some of these examples serve as a bridge between concepts. In Section 1.5, we have added an expanded demonstration of the Power Rule of differentiation. Though not a complete proof, it ties together skills developed in earlier sections to obtain the derivative of a positive-integer power. To see a general demonstration of this fact may help convince some students that the derivative form is not a “lucky accident.” In a new example in Section 1.5, the derivative is used as a means to demonstrate behaviors of a function. This material is developed more fully in Chapter 2, but it is valuable to introduce an easy example early, so that students have some familiarity with the derivative as an analytical tool, as opposed to a formula to be memorized. In Section 1.6, more detail is shown for the steps in the Product and Quotient Rules. Finally, in Section 1.8, a new example continues the discussion from Section 1.7 in which we “hint” at the change in value of a derivative and the concept of concavity, although that specific term is not introduced until Chapter 2. Chapter 2 Sections 2.1 and 2.2 are refreshed by adding clarification for key themes and a discussion of optimization from both algebraic and calculus viewpoints. A new example in Section 2.2 ties together the discussion in earlier examples. In Section 2.3, a new example asks the student to “build” a function based on some given facts about its behavior. This serves as a gauge as to whether the student understands the concepts as opposed to memorizing steps. Section 2.6 has significant new material on using differentials as a means for approximation in real-world settings. Chapter 3 This chapter also has many new applications and updates of data in examples and exercises. New applications include those focusing on the exponential growth in the value of the Forever Stamp, of Facebook membership, of costs of attending a 4-year college or university, of the number of subscribers to Sirius XM radio, of net sales of Green Mountain Coffee Roasters, and of the value of antique Batman and Superman comic books. There are also new examples on exponential decay of the number of farms in the United States, of the number of cases of tuberculosis, and of the magnitude of earthquakes in Haiti and Chile. Chapter 4 This chapter’s presentation of integration has been significantly rearranged. The chapter starts with general antidifferentiation in Section 4.1. We feel this is a good way to segue from differential to integral calculus. Students at this stage may not yet know “why” they need to understand antiderivatives, but they can at least draw upon their skills of differentiation to learn the process of antidifferentiation. At the end of Section 4.1, a new Technology Connection introduces area under a curve. Although area under a curve is not formally discussed until Section 4.2, we feel that walking the students through the process may allow them to make the connection that antidifferentiation has something to do with area. When they start Section 4.2, they will have some basic skills of antidifferentiation and some idea of its significance. In Section 4.2, we concentrate on the geometry behind integration: Riemann sums and the development of the definite integral.

xiv

P R E FA C E

Many basic examples are presented in order to show various cases where area under a curve “makes sense.” Finally, in Section 4.3, we bring the two processes together with the Fundamental Theorem of Calculus. New examples throughout the remainder of Chapter 4 show some of the concepts in a different light. For example, in Section 4.5, we include a new example that extends the usual u-du method of substitution. This concept can be applied to integration by parts (Section 4.6), to show students that sometimes there may be more than one way to find an antiderivative. Many of these concepts are further discussed in the Synthesis sections of the exercise sets. Finally, the new Extended Technology Application for Chapter 4 shows how Lorenz functions and Gini coefficients are used to analyze distribution of wealth (or resources) in a society. Chapter 5 Chapter 5 begins with a discussion of consumer and producer surplus, which has been rewritten and the graphs rerendered to illustrate some of the concepts more clearly. Section 5.2 has been entirely rewritten. Reviewers made several suggestions that improved the clarity of this section. (We especially want to thank Bruce Thomas of Kennesaw State University for his extensive help.) Section 5.5 includes a significant amount of new material on percentiles, including three new examples, and Section 5.6 contains a new example illustrating the use of volumes by rotation. Finally, a brief discussion of the solution of general first-order linear differential equations is included in the Synthesis section of Exercise Set 5.7. Chapter 6 Many new examples have been added to Chapter 6. One in Section 6.1 shows how tables are used in real life to express a multivariable concept (payments on an amortized loan). A more formal discussion and an example on domains of a two-variable function are presented later. Section 6.4 has a new Technology Connection discussing a method of finding solutions to two-variable linear systems using matrices. Although systems of equations are not covered formally in this text, the need to solve such a system is central to the topic of regression, covered in Section 6.4. The method presented in the Technology Connection allows the student to better understand this aspect of the long process of regression more quickly. In Section 6.5, a more formal discussion of constrained optimization on a closed and bounded region allows us to include the Extreme-Value Theorem and extend the ideas of path constraints. Section 6.6 now has an extra example illustrating the use of a double integral. Finally, another smart phone app, Grafly, is introduced in Chapter 6. Accessibly priced, it can be used to create visually appealing graphs of functions of two variables.

New! Appendixes This edition includes two new appendixes. Appendix B: Regression and Microsoft Excel shows how regression can be done with Excel (2007 and later versions) far more robustly than with the TI calculators. Appendix C: MathPrint Operating System for TI-84 and TI-84 Plus Silver Edition shows how students can transition to the new operating system for TI-84 calculators.

More Applications and Exercises For most instructors, the ultimate goal is for students to be able to apply what they learn in this course to everyday scenarios. This ability motivates learning and brings student understanding to a higher level. ■ Over 300 applications have been added or updated. ■ Data in applications has been updated whenever achievable. ■ Over 630 exercises are new or updated. ■ The number of business and finance applications has been increased by over 10%. Section 5.2 contains numerous new problems on present and future value, accumulated future value, and accumulated present value.

New! Annotated Instructor’s Edition An Annotated Instructor’s Edition has been added to the long list of instructor resources. Located in the margins in the AIE are Teaching Tips, which are ideal for new or less experienced instructors. In addition, answers to exercises are provided on the same page, making it easier than ever to check student work.

P R E FA C E

xv

Supplements STUDENT SUPPLEMENTS

INSTRUCTOR SUPPLEMENTS

New!

Student’s Solutions Manual

Annotated Instructor’s Edition

(ISBN: 0-321-74495-0 | 978-0-321-74495-1)

(ISBN: 0-321-72511-5 | 978-0-321-72511-0)

• Provides detailed solutions to all odd-numbered exercises, with the exception of the Thinking and Writing exercises

• Includes numerous Teaching Tips • Includes all of the answers, usually on the same page as the exercises, for quick reference

Graphing Calculator Manual

Online Instructor’s Solutions Manual (downloadable) • Provides complete solutions to all text exercises • Available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhighered. com/irc, and MyMathLab

(ISBN: 0-321-74496-9 | 978-0-321-74496-8)

• Provides instructions and keystroke operations for the TI-83/84 Plus, and TI-84 Plus with new operating system, featuring MathPrintTM. • Includes worked-out examples taken directly from the text • Topic order corresponds with that of the text Video Lectures on DVD-ROM with optional captioning (ISBN: 0-321-74498-5 | 978-0-321-74498-2)

• Complete set of digitized videos for student use at home or on campus • Ideal for distance learning or supplemental instruction Supplementary Chapters • Three chapters: Sequences and Series, Differential Equations, and Trigonometric Functions • Include many applications and optional technology material • Available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhighered.com/ irc, and to students at the downloadable student resources site, www.pearsonhighered.com/ mathstatsresources, or within MyMathLab

All of the student supplements listed above are included in MyMathLab.

Printable Test Forms (downloadable) • Contains six alternative tests per chapter • Contains six comprehensive final exams • Includes answer keys • Available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhighered.com/ irc, and MyMathLab TestGen® • Enables instructors to build, edit, and print, and administer tests using a computerized bank of questions developed to cover all the objectives of the text • Algorithmically based, allowing instructors to create multiple but equivalent versions of the same question or test with the click of a button • Allows instructors to modify test bank questions or add new questions • Available for download from www.pearsoned.com/ testgen PowerPoint Lecture Presentation • Classroom presentation software oriented specifically to the text’s topic sequence • Available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhighered.com/ irc, and MyMathLab

Media Supplements MyMathLab® Online Course (access code required) MyMathLab® is a text-specific, easily customizable online course that integrates interactive multimedia instruction with textbook content. MyMathLab gives an instructor the tools to deliver all or a portion of the course online, whether students are in a lab setting or working from home. ■ Interactive homework exercises, correlated to the textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are free-response and provide guided solutions, sample problems, and tutorial learning aids for extra help. ■ Personalized homework assignments can be designed to meet the needs of individual students. MyMathLab tailors the assignment for each student based on his or her test or quiz scores. Each student receives a homework assignment that contains only the problems he or she still needs to master.

xvi

P R E FA C E ■

■

New!

■

■

■

Personalized Study Plan, generated when students complete a test or quiz or homework, indicates which topics have been mastered and gives links to tutorial exercises for topics not yet mastered. The instructor can customize the Study Plan so that the topics available match course content. Multimedia learning aids, such as video lectures and podcasts, animations and interactive figures, and a complete multimedia textbook, help students independently improve their understanding and performance. These multimedia learning aids can be assigned as homework to help students grasp the concepts. Interactive figures are included within MyMathLab as both teaching and learning tools. These figures, which use the static figures in the text as a starting point, were created by Charles Stevens of Skagit Valley College. They can be used by instructors during lectures to illustrate some of the more difficult and visually challenging calculus topics. Used in this manner, the figures engage students more fully and save time otherwise spent rendering figures by hand. Instructors may also choose to assign the questions that accompany the figures, which leads students to discover key concepts. The interactive figures are also available to students, who may explore them on their own as a way to better visualize the concepts being presented. ■ Homework and Test Manager allows instructors to assign homework, quizzes, and tests that are automatically graded. Just the right mix of questions can be selected from the MyMathLab exercise bank, instructor-created custom exercises, and/or TestGen® test items. ■ Gradebook, designed specifically for mathematics and statistics, automatically tracks students’ results, letting the instructor stay on top of student performance and providing control over how to calculate final grades. Instructors can also add offline (paper-and-pencil) grades to the gradebook. MathXL Exercise Builder allows instructors to create static and algorithmic exercises for online assignments. They can use the library of sample exercises as an easy starting point or edit any course-related exercise. Pearson Tutor Center (www.pearsontutorservices.com) access is automatically included with MyMathLab. The Tutor Center is staffed by qualified math instructors who provide textbookspecific tutoring for students via toll-free phone, fax, email, and interactive Web sessions.

Students do their assignments in the Flash®-based MathXL Player, which is compatible with almost any browser (Firefox®, SafariTM, or Internet Explorer®) on either common platform (Macintosh® or Windows®). MyMathLab is powered by CourseCompassTM, Pearson Education’s online teaching and learning environment, and by MathXL®, its online homework, tutorial, and assessment system. MyMathLab is available to qualified adopters. For more information, visit www.mymathlab.com or contact your Pearson representative.

MathXL® Online Course (access code required) MathXL® is a powerful online homework, tutorial, and assessment system that accompanies Pearson Education’s textbooks in mathematics or statistics. With MathXL, instructors can ■ create, edit, and assign online homework and tests using algorithmically generated exercises correlated at the objective level to the textbook; ■ create and assign their own online exercises and import TestGen tests for added flexibility; and ■ maintain records of all student work tracked in MathXL’s online gradebook. With MathXL, students can ■ ■ ■

take chapter tests in MathXL and receive personalized study plans and/or personalized homework assignments based on their test results; use the study plan and/or the homework to link directly to tutorial exercises for the objectives they need to study; and access supplemental animations and video clips directly from selected exercises.

MathXL is available to qualified adopters. For more information, visit www.mathxl.com, or contact your Pearson representative.

P R E FA C E

xvii

InterAct Math Tutorial Website: www.interactmath.com Get practice and tutorial help online! This interactive tutorial website provides algorithmically generated practice exercises that correlate directly to the exercises in the textbook. Students can retry an exercise as many times as they like, with new values each time for unlimited practice and mastery. Every exercise is accompanied by an interactive guided solution that provides helpful feedback in response to incorrect answers, and students can also view a worked-out sample problem that steps them through an exercise similar to the one they’re working on.

Acknowledgments As authors, we have taken many steps to ensure the accuracy of this text. Many devoted individuals comprised the team that was responsible for monitoring the revision and production process in a manner that makes this a work of which we can all be proud. We are thankful for our publishing team at Pearson, as well as all of the Pearson representatives who share our book with educators across the country. Many thanks to Michelle Christian, who was instrumental in getting Scott Surgent’s first book printed and in bringing him to the attention of the Pearson team. We would like to thank Jane Hoover for her many helpful suggestions, proofreading, and checking of art. Jane’s attention to detail and pleasant demeanor made our work as low in stress as humanly possible, given the demands of the production process. We also wish to thank Michelle Beecher Lanosga for her incredibly helpful data research. Her efforts make the real-world problems in this text as up-to-date as possible, given the production deadlines we faced. Geri Davis deserves credit for both the attractive design of the text and the coordination of the many illustrations, photos, and graphs. She is always a distinct pleasure to work with and sets the standard by which all other art editors are measured. We are very grateful for Mary Ann Teel’s contributions to this edition: her thoughtful comments while reviewing draft chapters, her careful reading of the exposition for accuracy and consistency, and her work on the testing manual. Many thanks to Lisa Grilli for her insightful review and for providing helpful teaching tips for the Annotated Instructor’s Edition. We greatly appreciate Dave Dubriske’s work on the solutions manuals and Steve Ouellette’s work on Appendix C and the Graphing Calculator Manual. Many thanks also to Lauri Semarne, Patricia Nelson, Deanna Raymond, and Doug Ewert for their careful checking of the manuscript and typeset pages. Finally, the following reviewers provided thoughtful and insightful comments that helped immeasurably in the revision of this text. Nilay Tanik Argon, The University of North Carolina at Chapel Hill Debra S. Carney, University of Denver Hugh Cornell, University of North Florida Rakissa Cribari, Ph.D., University of Colorado–Denver Jerry DeGroot, Purdue University North Central Samantha C. Fay, Jefferson College Burt K. Fischer, CPA Lewis A. Germann, Troy University Lisa Grilli, Northern Illinois University John R. Griggs, North Carolina State University Mary Beth Headlee, Manatee Community College Glenn Jablonski, Triton College Darin Kapanjie, Temple University Rebecca E. Lynn, Colorado State University Shahla Peterman, University of Missouri–St. Louis Mohammed Rajah, MiraCosta College Scott R. Sykes, University of West Georgia Charlie Snygg, Pikes Peak Community College Mary Ann Teel, University of North Texas Bruce Thomas, Kennesaw State University Ani P. Velo, University of San Diego Patrick Ward, Illinois Central College

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Prerequisite Skills Diagnostic Test To the Student and the Instructor Part A of this diagnostic test covers basic algebra concepts, such as properties of exponents, multiplying and factoring polynomials, equation solving, and applied problems. Part B covers topics, discussed in Chapter R, such as graphs, slope, equations of lines, and functions, most of which come from a course in intermediate or college algebra. This diagnostic test does not cover regression, though it is considered in Chapter R and used throughout the text. This test can be used to assess student needs for this course. Students who miss most of the questions in part A should study Appendix A before moving to Chapter R. Those who miss most of the questions in part B should study Chapter R. Students who miss just a few questions might study the related topics in either Appendix A or Chapter R before continuing with the calculus chapters.

Part A: Answers and locations of worked-out solutions appear on p. A-37.

Part B: Answers and locations of worked-out solutions appear on p. A-37.

Express each of the following without an exponent.

Graph.

1. 4

2. 1- 22

3

5

3.

A B

1 3 2

4. 1- 2x2

1

5. e

0

Express each of the following without a negative exponent. 6. x

-5

7.

A B

1 -2 4

8. t

-1

Multiply. Express each answer without a negative exponent. 9. x 5 # x 6

10. x -5 # x 6

11. 2x -3 # 5x -4

Divide. Express each answer without a negative exponent. 12.

a3

13.

a2

e3

Multiply. 16. 31x - 52

15. 12x 4y -5z 32-3

19. 12x - t22

17. 1x - 521x + 32

20. 13c + d213c - d2

18. 1a + b21a + b2

22. x 2 - 6xy + 9y 2

24. 6x 2 + 7x - 5

23. x 2 - 5x - 14

25. x 3 - 7x 2 - 4x + 28

Solve. 26.

- 56 x

+ 10 =

28. 4x 3 = x

1 2x

29.

+ 2

3. y = x 2 - 1

4. x = y 2

5. A function f is given by f 1x2 = 3x 2 - 2x + 8. Find each of the following: f 102, f 1- 52, and f 17a2. 6. A function f is given by f1x2 = x - x 2. Find and simplify f1x + h2 - f1x2 , for h Z 0. h for x Ú 0, 4, f1x2 = c 3 - x 2, for 0 6 x … 2, 2x - 6, for x 7 2.

8. Write interval notation for 5x ƒ - 4 6 x … 56. 9. Find the domain: f1x2 =

Factor. 21. 2xh + h2

2. 3x + 5y = 10

7. Graph the function f defined as follows:

e -4

Simplify. Express each answer without a negative exponent. 14. 1x -223

1. y = 2x + 1

27. 3x1x - 2215x + 42 = 0

2x 6 18 = 2 x x - 3 x - 3x

10. Find the slope and y-intercept of 2x - 4y - 7 = 0. 11. Find an equation of the line that has slope 3 and contains the point 1- 1, - 52.

12. Find the slope of the line containing the points 1- 2, 62 and 1- 4, 92. Graph. 13. f1x2 = x 2 - 2x - 3

14. f1x2 = x 3

1 x

16. f1x2 = ƒ x ƒ

30. 17 - 8x Ú 5x - 4

15. f1x2 =

31. After a 5% gain in weight, a grizzly bear weighs 693 lb. What was the bear’s original weight?

17. f1x2 = - 2x

32. Raggs, Ltd., a clothing firm, determines that its total revenue, in dollars, from the sale of x suits is given by 200x + 50. Determine the number of suits the firm must sell to ensure that its total revenue will be more than $70,050.

3 . 2x - 5

18. Suppose that $1000 is invested at 5%, compounded annually. How much is the investment worth at the end of 2 yr?

xix

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Functions, Graphs, and Models

R

Chapter Snapshot What You’ll Learn

Why It’s Important

R.1 R.2 R.3 R.4 R.5 R.6

This chapter introduces functions and covers their graphs, notation, and applications. Also presented are many topics that we will consider often throughout the text: supply and demand, total cost, total revenue, total profit, the concept of a mathematical model, and curve fitting. Skills in using a graphing calculator are also introduced in optional Technology Connections. Details on keystrokes are given in the Graphing Calculator Manual (GCM). Part A of the diagnostic test (p. xix), on basic algebra concepts, allows students to determine whether they need to review Appendix A (p. 605) before studying this chapter. Part B, on college algebra topics, assesses the need to study this chapter before moving on to the calculus chapters.

Graphs and Equations Functions and Models Finding Domain and Range Slope and Linear Functions Nonlinear Functions and Models Mathematical Modeling and Curve Fitting

Where It’s Used B IRTH R ATES

What is the average number of live births per 1000 women age 20? This problem appears as an example in a Technology Connection in Section R.6.

BIRTH RATES FOR WOMEN OF SELECTED AGES AGE, x

AVERAGE NUMBER OF LIVE BIRTHS PER 1000 WOMEN

16 18.5 22 27 32 37 42

34 86.5 111.1 113.9 84.5 35.4 6.8

OTH ER

886.9 3 − 3.22x2 + 101.2x − f(x) = 0.031x y 120 110 100 90 80 70 60 50 40 30 20 10 0

bies born Number of ba omen w 00 10 r pe

BIRTH RATES

BY AG E O F M

10

20

30

40

50

60

x

Women’s age

(Source: Centers for Disease Control and Prevention.)

1

2

CHAPTER R

•

Functions, Graphs, and Models

R.1

Graphs and Equations What Is Calculus? What is calculus? This is a common question at the start of a course like this. Lets consider a simplified answer for now. Consider a protein energy drink box, as shown below, at left. The following is a typical problem from an algebra course. Try to solve it. (If you need some algebra review, refer to Appendix A at the end of the book.)

OBJECTIVES • Graph equations. • Use graphs as mathematical models to make predictions. • Carry out calculations involving compound interest.

Algebra Problem The sum of the height, width, and length of a box is 207 mm. If the height is three times the width and the length is 7 mm more than the width, find the dimensions of the box.

The box has a width of 40 mm, a length of 47 mm, and a height of 120 mm.* The following is a calculus problem that a manufacturer of boxes might need to solve.

Calculus Problem

Height

A protein energy drink box is to hold 200 cm3 (6.75 fl oz) of protein energy drink. If the height of the box must be twice the width, what dimensions will minimize the surface area of the box? Length Width

Dimensions that assume that the height is twice the width Width, w

Height, h

Length, l

Total Surface Area, 2wh + 2lh + 2wl

5 cm 4 cm 3 cm 2 cm

10 cm 8 cm 6 cm 4 cm

4 cm 6.25 cm 11.1 cm 25 cm

220 cm2 214 cm2 236 cm2 316 cm2

⎫⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎬ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎭ We selected combinations for which h = 2w and the product w # h # l is 200.

Smallest

One way to solve this problem might be to choose several sets of dimensions for a 200-cm3 box that is twice as tall as it is wide, compute the resulting areas, and determine which is the least. If you have access to spreadsheet software, you might create a spreadsheet and expand the table at left. We let w = width, h = height, l = length, and A = surface area. Then the surface area A is given by A = 2wh + 2lh + 2wl, with h = 2w.

From the data in the table, we might conclude that the smallest surface area is 214 cm2. But how can we be certain that there are no other dimensions that yield a smaller area? We need the tools of calculus to answer this. We will study such maximum–minimum problems in more detail in Chapter 2. Other topics we will consider in calculus are the slope of a curve at a point, rates of change, area under a curve, accumulations of quantities, and some statistical applications.

*To find this, let w = width, h = height, and l = length. Then h = 3w and l = w + 7, so w + 3w + w + 7 = 207. This yields w = 40, and thus h = 120 and l = 47.

R.1

•

3

Graphs and Equations

Graphs The study of graphs is an essential aspect of calculus. A graph offers the opportunity to visualize relationships. For instance, the graph below shows how life expectancy has changed over time in the United States. One topic that we consider later in calculus is how a change on one axis affects the change on another. ESTIMATED LIFE EXPECTANCY OF U.S. NEWBORNS BY YEAR OF BIRTH, 1929–2010 Life expectancy (in years)

100 80 60 40 20 0

’29

’35

’41

’47

’53

’59

’65 ’71 ’77 Year of birth

’83

’89

’95

’01

’07

’13

(Source: U.S. National Center for Health Statistics.)

Ordered Pairs and Graphs Each point in a plane corresponds to an ordered pair of numbers. Note in the figure at the right that the point corresponding to the pair 12, 52 is different from the point corresponding to the pair 15, 22. This is why we call a pair like 12, 52 an ordered pair. The first number is called the first coordinate of the point, and the second number is called the second coordinate. Together these are the coordinates of the point. The vertical line is often called the y-axis, and the horizontal line is often called the x-axis.

y 6

(–3, 5)

5

(2, 5)

4

(4, 3)

3

(5, 2)

2 1 –6 –5 –4 –3 –2 –1 –1

(–4, – 2)

1

2

3

4

5

6

x

–2 –3 –4 –5

(3, –4)

–6

Graphs of Equations A solution of an equation in two variables is an ordered pair of numbers that, when substituted for the variables, forms a true sentence. If not directed otherwise, we usually take the variables in alphabetical order. For example, 1- 1, 22 is a solution of the equation 3x 2 + y = 5, because when we substitute - 1 for x and 2 for y, we get a true sentence: 3x 2 + y = 31- 122 + 2  3 + 2 5 =

5 5 5 5. TRUE

DEFINITION The graph of an equation is a drawing that represents all ordered pairs that are solutions of the equation.

4

CHAPTER R

•

Functions, Graphs, and Models

We obtain the graph of an equation by plotting enough ordered pairs (that are solutions) to see a pattern. The graph could be a line, a curve (or curves), or some other configuration.

■ EXAMPLE 1 Graph: y = 2x + 1. Solution We first find some ordered pairs that are solutions and arrange them in a table. To find an ordered pair, we can choose any number for x and then determine y. For example, if we choose - 2 for x and substitute in y = 2x + 1, we find that y = 21- 22 + 1 = - 4 + 1 = - 3. Thus, 1- 2, - 32 is a solution. We select both negative numbers and positive numbers, as well as 0, for x. If a number takes us off the graph paper, we usually omit the pair from the graph. y

y

1x, y2

-2

-3

-1

-1

0

1

10, 12

3

11, 32

5

12, 52

x

1 2

5

(2, 5)

1- 2, - 32

4

y = 2x + 1

1- 1, - 12

2

3

1 –4 –3 –2 –1

(1, 3) (0, 1) 1

2

3

4

x

(–1, –1) –1 –2

(–2, –3)

–3 –4

(1) Choose any x. (2) Compute y. (3) Form the pair 1x, y2. (4) Plot the points.

Quick Check 1 Graph: y = 3 - x.

After we plot the points, we look for a pattern in the graph. If we had enough points, they would suggest a solid line. We draw the line with a straightedge and label it y = 2x + 1.

O Now try Quick Check 1 ■ EXAMPLE 2 Graph: 3x + 5y = 10. Solution We could choose x-values, substitute, and solve for y-values, but we first solve for y to ease the calculations.* 3x + 5y = 10 3x + 5y - 3x = 10 - 3x 5y = 10 - 3x 1 5

# 5y

=

1 5

Subtracting 3x from both sides

# 110 - 3x2

y = 15 # 1102 = 2 - 35 x = -

3 5x

1 5

# 13x2

Simplifying Multiplying both sides by 15, or dividing both sides by 5 Using the distributive law Simplifying

+ 2

*Be sure to consult Appendix A, as needed, for a review of algebra.

O

R.1

•

5

Graphs and Equations

Next we use y = - 53 x + 2 to find three ordered pairs, choosing multiples of 5 for x to avoid fractions. y

x

1x, y2

y

0

(–5, 5)

5

10, 22

2

4

5

-1

15, - 12

-5

5

1- 5, 52

3

3x + 5y = 10

2

(0, 2)

1 –5 –4 –3 –2 –1

1

2

3

–1

4

5

x

(5, –1)

–2 –3

Quick Check 2

O

Graph: 3x - 5y = 10.

We plot the points, draw the line, and label the graph as shown.

O Now try Quick Check 2 Examples 1 and 2 show graphs of linear equations. Such graphs are considered in greater detail in Section R.4.

■ EXAMPLE 3 Graph: y = x 2 - 1. Solution y

x

y

1- 2, 32

0

1- 1, 02

(–1, 0)

0

-1

10, - 12

–3 –2 –1

1

0

11, 02

2

3

12, 32

-1

Graph: y = 2 - x 2.

4

3

-2

Quick Check 3

1x, y2 (–2, 3)

3

(2, 3)

2

y = x2 – 1

1

(1, 0) 1

2

3

4

x

–1 –2

(0, –1)

This time the pattern of the points is a curve called a parabola. We plot enough points to see a pattern and draw the graph.

O Now try Quick Check 3 ■ EXAMPLE 4 Graph: x = y 2. Solution In this case, x is expressed in terms of the variable y. Thus, we first choose numbers for y and then compute x.

O

6

CHAPTER R

•

Functions, Graphs, and Models y

y

1x, y2

4

-2

14, - 22

1

-1

11, - 12

x

(1) Choose any y. (2) Compute x. (3) Form the pair 1x, y2. (4) Plot the points.

0

0

10, 02

1

1

11, 12

4

2

14, 22

3 2

(4, 2)

x = y2

(1, 1)

1

(0, 0)

1

2

3

4

5

6

x

–1

–2 (1, –1)

(4, – 2)

–3

We plot these points, keeping in mind that x is still the first coordinate and y the second. We look for a pattern and complete the graph, usually connecting the points.

Quick Check 4

O Now try Quick Check 4

O

Graph: x = 1 + y 2.

TECHNOLOGY CONNECTION

Introduction to the Use of a Graphing Calculator: Windows and Graphs Viewing Windows In this first of the optional Technology Connections, we begin to create graphs using a graphing calculator. Most of the coverage will refer to a TI-84 Plus or TI-83 Plus graphing calculator but in a somewhat generic manner, discussing features common to most graphing calculators. Although some keystrokes will be listed, exact keystrokes can be found in the owner’s manual for your calculator or in the Graphing Calculator Manual (GCM) that accompanies this text. The viewing window is a feature common to all graphing calculators. This is the rectangular screen in which a graph appears. Windows are described by four numbers, [L, R, B, T], which represent the Left and Right endpoints of the x-axis and the Bottom and Top endpoints of the y-axis. A WINDOW feature can be used to set these dimensions. Below is a window setting of 3- 20, 20, - 5, 54 with axis scaling denoted as Xscl = 5 and Yscl = 1, which means that there are 5 units between tick marks extending from - 20 to 20 on the x-axis and 1 unit between tick marks extending from - 5 to 5 on the y-axis.

20

–20

Graphs Let’s use a graphing calculator to graph the equation y = x 3 - 5x + 1. The equation can be entered using the notation y=x^3–5x+1. We obtain the following graph in the standard viewing window. y = x3 – 5x + 1

10

10

–10

5 WINDOW Xmin = –20 Xmax = 20 Xscl = 5 Ymin = –5 Ymax = 5 Yscl = 1 Xres = 1

Scales should be chosen with care, since tick marks become blurred and indistinguishable when too many appear. On most graphing calculators, a setting of 3- 10, 10, - 10, 104, Xscl = 1, Yscl = 1, Xres = 1 is considered standard. Graphs are made up of black rectangular dots called pixels. The setting Xres allows users to set pixel resolution at 1 through 8 for graphs of equations. At Xres = 1, equations are evaluated and graphed at each pixel on the x-axis. At Xres = 8, equations are evaluated and graphed at every eighth pixel on the x-axis. The resolution is better for smaller Xres values than for larger values.

–10

It is often necessary to change viewing windows in order to best reveal the curvature of a graph. For example, –5

(continued)

R.1

each of the following is a graph of y = 3x 5 - 20x 3, but with a different viewing window. Which do you think best displays the curvature of the graph? y=

3x 5

–

•

Graphs and Equations

7

(See Example 2.) Its graph is shown below in the standard window. y = – 35 x + 2 10

20x 3

5000

–10 –5

10

5

–10 –5000

To graph an equation like x = y 2, we solve for y and get

y = 3x5 – 20x3

y = 2x or y = - 2x, which can be written as y = ; 2x.

500

We then graph the individual equations y1 = 2x and y2 = - 2x. –4

4

EXERCISES Graph each of the following equations. Select the standard window, 3- 10, 10, - 10, 104, with axis scaling Xscl = 1 and Yscl = 1.

–500

y = 3x5 – 20x3

80

–4

4

1. y = x + 3

2. y = x - 5

3. y = 2x - 1

4. y = 3x + 1

5. y =

- 23 x

+ 4

7. 2x - 3y = 18 9. y = x

–80

In general, choosing a window that best reveals a graph’s characteristics involves some trial and error and, in some cases, some knowledge about the shape of that graph. We will learn more about the shape of graphs as we continue through the text. To graph an equation like 3x + 5y = 10, most calculators require that the equation be solved for y. Thus, we must rewrite and enter the equation as y =

1- 3x + 102 5

,

or

3 y = a - bx + 2. 5

2

4

6. y = - 5 x + 3 8. 5y + 3x = 4

10. y = 1x + 422

11. y = 8 - x 2

12. y = 4 - 3x - x 2

13. y + 10 = 5x 2 - 3x

14. y - 2 = x 3

15. y = x 3 - 7x - 2

16. y = x 4 - 3x 2 + x

17. y = ƒ x ƒ (On most calculators, this is entered as

y = abs1x2.)

18. y = ƒ x - 5 ƒ 19. y = ƒ x ƒ - 5 20. y = 9 - ƒ x ƒ

Mathematical Models When a real-world situation can be described in mathematical language, the description is a mathematical model. For example, the natural numbers constitute a mathematical model for situations in which counting is essential. Situations in which algebra can be brought to bear often require the use of functions as models. See Example 5, which follows. Mathematical models are abstracted from real-world situations. The mathematical model may give results that allow us to predict what will happen in the real-world situation. If the predictions are inaccurate or the results of experimentation do not conform to the model, the model must be changed or discarded. Mathematical modeling is often an ongoing process. For example, finding a mathematical model that will provide an accurate prediction of population growth is not a simple task. Any population model that one might devise would need to be reshaped as further information is acquired.

CHAPTER R

•

Functions, Graphs, and Models

CREATING A MATHEMATICAL MODEL

■ EXAMPLE 5 The graph below shows participation by females in high school athletics from 2000 to 2009.

1. Recognize a real-world problem.

2. Collect data.

3. Analyze the data.

FEMALES IN HIGH SCHOOL ATHLETICS Number of female high school athletes (in millions)

8

y 3.0 2.8

2.68

2.95 2.86 2.87 2.91 2.78 2.80

3.02 3.06

2.6 2.4 2.2 2.0 2000 2001 2002 2003 2004 2005 2006 2007 2009 x Year

(Source: National Federation of State High School Associations.)

4. Construct a model.

Use the model N = 0.042t + 2.71, where t is the number of years after 2000 and N is the number of participants, in millions, to predict the number of female high school athletes in 2012.

Solution Since 2012 is 12 years after 2000, we substitute 12 for t: 5. Test and refine the model

N = 0.042t + 2.71 = 0.0421122 + 2.71 = 3.214. According to this model, in 2012, approximately 3.21 million females will participate in high school athletics.

6. Explain and predict.

As is the case with many kinds of models, the model in Example 5 is not perfect. For example, for t = 1, we get N = 2.752, a number slightly different from the 2.78 in the original data. But, for purposes of estimating, the model is adequate. The cubic model N = 0.001x 3 - 0.014x 2 + 0.087x + 2.69 also fits the data, at least in the short term: For t = 1, we get N L 2.76, close to the original data value. But for t = 12, N L 3.45, quite different from the prediction in Example 5. The difficulty with a cubic model here is that, eventually, its predictions get too high. For example, the model in Example 5 predicts that there will be 3.57 million female high school athletes in 2020, but the cubic model predicts 8.00 million. We always have to subject our models to careful scrutiny. One important model that is extremely precise involves compound interest. Suppose that we invest P dollars at interest rate i, expressed as a decimal and compounded annually. The amount A1 in the account at the end of the first year is given by A1 = P + Pi The original amount invested, P, is called the principal. = P11 + i2 = Pr, where, for convenience, we let r = 1 + i. Going into the second year, we have Pr dollars, so by the end of the second year, we will have the amount A2 given by A2 = A1 # r = 1Pr2r = Pr 2 = P11 + i22.

Going into the third year, we have Pr 2 dollars, so by the end of the third year, we will have the amount A3 given by A3 = A2 # r = 1Pr 22r = Pr 3 = P11 + i23.

R.1

•

Graphs and Equations

9

In general, we have the following theorem.

THEOREM 1 If an amount P is invested at interest rate i, expressed as a decimal and compounded annually, in t years it will grow to the amount A given by A = P11 + i2t.

■ EXAMPLE 6 Business: Compound Interest. Suppose that $1000 is invested in Fibonacci Investment Fund at 5%, compounded annually. How much is in the account at the end of 2 yr?

Solution We substitute 1000 for P, 0.05 for i, and 2 for t into the equation A = P11 + i2t and get A = = = = Quick Check 5 Business. Repeat Example 6 for an interest rate of 6%.

100011 + 0.0522 100011.0522 100011.10252 $1102.50.

Adding terms in parentheses Squaring Multiplying

There is $1102.50 in the account after 2 yr.

O Now try Quick Check 5 For interest that is compounded quarterly (four times per year), we can find a formula like the one above, as illustrated in the following diagram.

A = P(1 + i) t

The number of times interest is compounded goes from t to 4t.

Each time interest is compounded, the i rate used is −. 4

(

)

i 4t

A = P 1 + 4−

“Compounded quarterly” means that the interest is divided by 4 and compounded four times per year. In general, the following theorem applies.

THEOREM 2 If a principal P is invested at interest rate i, expressed as a decimal and compounded n times a year, in t years it will grow to an amount A given by A = Pa1 +

i nt b . n

■ EXAMPLE 7 Business: Compound Interest. Suppose that $1000 is invested in Wellington Investment Fund at 5%, compounded quarterly. How much is in the account at the end of 3 yr?

O

10

CHAPTER R

•

Functions, Graphs, and Models

Solution We use the equation A = P11 + i>n2nt, substituting 1000 for P, 0.05 for i, 4 for n (compounding quarterly), and 3 for t. Then we get

#

A = 1000a1 + = = = = L

Quick Check 6

O

Business: Compound Interest. Repeat Example 7 for an interest rate of 6%.

0.05 4 3 b 4

100011 + 0.0125212 100011.0125212 100011.1607545182 Using a calculator to approximate 11.0125212 1160.754518 The symbol L means “approximately equal to.” $1160.75.

There is $1160.75 in the account after 3 yr.

O Now try Quick Check 6 A calculator with a yx or a ^ key and a ten-digit readout was used to find 11.02212 in Example 7. The number of places on a calculator may affect the accuracy of the answer. Thus, you may occasionally find that your answers do not agree with those at the back of the book, which were found on a calculator with a ten-digit readout. In general, when using a calculator, do all computations and round only at the end, as in Example 7. Usually, your answer will agree to at least four digits. It is usually wise to consult with your instructor on the accuracy required.

Section Summary • Most graphs can be created by plotting points and looking for patterns. A graphing calculator can create graphs rapidly.

• Mathematical equations can serve as models of many kinds of applications.

EXERCISE SET

R.1 Exercises designated by the symbol are Thinking and Writing Exercises. They should be answered using one or two English sentences. Because answers to many such exercises will vary, solutions are not given at the back of the book.

17. y = ƒ x ƒ

Graph. (Unless directed otherwise, assume that “Graph” means “Graph by hand.”)

APPLICATIONS

1. y = x + 4

2. y = x - 1

3. y = - 3x

4. y = - 14 x

5. y = 23 x - 4

6. y = - 53 x + 3

7. x + y = 5

8. x - y = 4

9. 8y - 2x = 4

10. 6x + 3y = - 9

11. 5x - 6y = 12

12. 2x + 5y = 10

13. y = x - 5

14. y = x 2 - 3

15. x = 2 - y 2

16. x = y 2 + 2

2

19. y = 7 - x

18. y = ƒ 4 - x ƒ 2

20. y = 5 - x 2

21. y + 1 = x 3

22. y - 7 = x 3

23. Running records. According to at least one study, the world record in any running race can be modeled by a linear equation. In particular, the world record R, in minutes, for the mile run in year x can be modeled by R = - 0.00582x + 15.3476. Use this model to estimate the world records for the mile run in 1954, 2008, and 2012. Round your answers to the nearest hundredth of a minute. 24. Medicine. Ibuprofen is a medication used to relieve pain. The function A = 0.5t 4 + 3.45t 3 - 96.65t 2 + 347.7t, 0 … t … 6,

can be used to estimate the number of milligrams, A, of ibuprofen in the bloodstream t hours after 400 mg of the medication has been swallowed. (Source: Based on data from Dr. P. Carey, Burlington, VT.) How many milligrams of ibuprofen are in the bloodstream 2 hr after 400 mg has been swallowed? 25. Snowboarding in the half-pipe. Shaun White, “The Flying Tomato,” won a gold medal in the 2010 Winter Olympics for snowboarding in the half-pipe. He soared an unprecedented 25 ft above the edge of the half-pipe. His speed v1t2, in miles per hour, upon reentering the pipe can be approximated by v1t2 = 10.9t, where t is the number of seconds for which he was airborne. White was airborne for 2.5 sec. (Source: “White Rides to Repeat in Halfpipe, Lago Takes Bronze,” Associated Press, 2/18/2010.) How fast was he going when he reentered the half-pipe?

Number of hearing-impaired Americans (in millions)

Exercise Set R.1

11

N 5 4 3 2 1 20

40

60

80

x

Age

Use the graph to answer the following. a) Approximate the number of hearing-impaired Americans of ages 20, 40, 50, and 60. b) For what ages is the number of hearing-impaired Americans approximately 4 million? c) Examine the graph and try to determine the age at which the greatest number of Americans is hearingimpaired. d) What difficulty do you have in making this determination? 28. Life science: incidence of breast cancer. The following graph approximates the incidence of breast cancer y, per 100,000 women, as a function of age x, where x represents ages 25 to 102.

26. Skateboard bomb drop. The distance s1t2, in feet, traveled by a body falling freely from rest in t seconds is approximated by s1t2 = 16t 2. On April 6, 2006, pro skateboarder Danny Way smashed the world record for the “bomb drop” by free-falling 28 ft from the Fender Stratocaster guitar atop the Hard Rock Hotel & Casino in Las Vegas onto a ramp below. (Source: www.skateboardingmagazine.com.) How long did it take until he hit the ramp?

Incidence of breast cancer per 100,000 women

y 400 300 200 100 20

40 60 80 100 120 x Age of women

(Source: National Cancer Institute.)

a) What is the incidence of breast cancer in 40-yr-old women? b) For what ages is the incidence of breast cancer about 400 per 100,000 women? c) Examine the graph and try to determine the age at which the largest incidence of breast cancer occurs. d) What difficulty do you have making this determination? 29. Compound interest. An investor purchases a $100,000 certificate of deposit from Newton Bank, at 2.8%. How much is the investment worth (rounded to the nearest cent) at the end of 1 yr, if interest is compounded: a) annually? b) semiannually? d) daily (use 365 days c) quarterly? e) hourly? for 1 yr)?

27. Hearing-impaired Americans. The number N, in millions, of hearing-impaired Americans of age x can be approximated by the graph that follows. (Source: American Speech-Language Hearing Association.)

30. Compound interest. An investor purchases a $300,000 certificate of deposit from Descartes Bank, at 2.2%. How much is the investment worth (rounded to the nearest cent) at the end of 1 yr, if interest is compounded: a) annually? b) semiannually? c) quarterly? d) daily (use 365 days e) hourly? for 1 yr)?

12

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31. Compound interest. An investor deposits $30,000 in Godel Municipal Bond Funds, at 4%. How much is the investment worth (rounded to the nearest cent) at the end of 1 yr, if interest is compounded: a) annually? b) semiannually? c) quarterly? d) daily (use 365 days e) hourly? for 1 yr)? 32. Compound interest. An investor deposits $1000 in Wiles Municipal Bond Funds, at 5%. How much is the investment worth (rounded to the nearest cent) at the end of 1 yr, if interest is compounded: a) annually? b) semiannually? c) quarterly? d) daily (use 365 days e) hourly? for 1 yr)? Determining monthly payments on a loan. If P dollars are borrowed, the monthly payment M, made at the end of each month for n months, is given by

MP

i i n a1  b 12 12 a1 

n

i b 1 12

,

where i is the annual interest rate and n is the total number of monthly payments. 33. Fermat’s Last Bank makes a car loan of $18,000, at 6.4% interest and with a loan period of 3 yr. What is the monthly payment? 34. At Haken Bank, Ken Appel takes out a $100,000 mortgage at an interest rate of 4.8% for a loan period of 30 yr. What is the monthly payment? Annuities. If P dollars are invested annually in an annuity (investment fund), after n years, the annuity will be worth W  PB

11  i2n  1 i

R,

where i is the interest rate, compounded annually. 35. You invest $3000 annually in an annuity from Mersenne Fund Annuities that earns 6.57% interest. How much is the investment worth after 18 yr? Round to the nearest cent. 36. Suppose that you establish an annuity that earns 7 14% interest, and you want it to be worth $50,000 in 20 yr. How much will you need to invest annually to achieve this goal?

a) In what years was the deer population in Maine at or above 250,000? b) In what years was the deer population at 200,000? c) In what year was the deer population highest? d) In what years was the deer population lowest? 38. Speculate as to why the deer population in Maine was so high in the years 1997–2001.

SYNTHESIS Retirement account. Sally makes deposits into a retirement account every year from the age of 30 until she retires at age 65. 39. a) If Sally deposits $1200 per year and the account earns interest at a rate of 8% per year, compounded annually, how much does she have in the account when she retires? (Hint: Use the annuity formula for Exercises 35 and 36.) b) How much of that total amount is from Sally’s deposits? How much is interest? 40. a) Sally plans to take regular monthly distributions from her retirement account from the time she retires until she is 80 years old, when the account will have a value of $0. How much should she take each month? Assume the interest rate is 8% per year, compounded monthly. (Hint: Use the formula for Exercises 33 and 34 that calculates the monthly payments on a loan.) b) What is the total of the payments she will receive? How much of the total will be her own money (see part b of Exercise 39), and how much will be interest?

TECHNOLOGY CONNECTION The Technology Connection heading indicates exercises designed to provide practice using a graphing calculator. Graph. 41. y = x - 150

42. y = 25 - | x |

43. y = x + 2x - 4x - 13

44. y = 223 - 7x

45. 9.6x + 4.2y = - 100

46. y = - 2.3x 2 + 4.8x - 9

47. x = 4 + y 2

48. x = 8 - y 2

3

Answers to Quick Checks 1. y = 3 - x

Number of deer (in thousands)

2. 3x - 5y = 10 y

y

37. Deer population in Maine. The deer population in Maine from 1986 to 2006 is approximated in the graph below. DEER POPULATION IN MAINE

2

6

4

4

2

2 −4

−2

−4

2

4

−2

250 200

−2

y

y

2

2.0

−1

1 −2

150 100 ’86 ’88 ’90 ’92 ’94 ’96 ’98 ’00 ’02 ’04 ’06 Year

(Source: State of Maine, Department of Inland Fisheries and Wildlife.)

x

3

x

4. x = 1 + y 2

3. y = 2 - x 2

300

4

−4

−2

350

2 −2

x

2

x

1.0

−4

−1 − 1.0

−6

− 2.0

1

2

5. There is $1123.60 in the account after 2 yr. 6. There is $1195.62 in the account after 3 yr.

R.2

R.2 OBJECTIVES • Determine whether or not a correspondence is a function. • Find function values. • Graph functions and determine whether or not a graph is that of a function. • Graph functions that are piecewise-defined.

•

13

Functions and Models

Functions and Models Identifying Functions The idea of a function is one of the most important concepts in mathematics. Put simply, a function is a special kind of correspondence between two sets. Let’s look at the following. To each letter on a telephone keypad there corresponds a number. To each model of cell phone in a store there corresponds its price. To each real number there corresponds the cube of that number. In each of these examples, the first set is called the domain and the second set is called the range. Given a member of the domain, there is exactly one member of the range to which it corresponds. This type of correspondence is called a function.

Correspondence Domain

Range

DEFINITION A function is a correspondence between a first set, called the domain, and a second set, called the range, such that each member of the domain corresponds to exactly one member of the range.

■ EXAMPLE 1 Determine whether or not each correspondence is a function. a) Cumulative number of iPhones sold Domain

Range

2006 2007 2008 2009

0 1,389,000 11,627,000 20,371,000

b) Squaring Domain 3 4 5 -5

Range 9 16 25

(Source: Apple Inc.)

c) Baseball teams

d) Baseball teams

Domain

Range

Arizona Chicago

Diamondbacks Cubs White Sox Orioles

Baltimore

Domain Diamondbacks Cubs White Sox Orioles

Range Arizona Chicago Baltimore

Solution a) The correspondence is a function because each member of the domain corresponds (is matched) to only one member of the range. b) The correspondence is a function because each member of the domain corresponds to only one member of the range, even though two members of the domain correspond to 25. c) The correspondence is not a function because one member of the domain, Chicago, corresponds to two members of the range, the Cubs and the White Sox. d) The correspondence is a function because each member of the domain corresponds to only one member of the range, even though two members of the domain correspond to Chicago.

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Functions, Graphs, and Models

■ EXAMPLE 2 Determine whether or not each correspondence is a function. Domain a) A family

Correspondence

Range

Each person’s weight

A set of positive numbers A set of nonnegative integers: 50, 1, 4, 9, 16, 25, . . .6 The set of all U.S. senators

b) The integers Each number’s square 5. . . , - 3, - 2, - 1, 0, 1, 2, 3, ...6 c) The set of all states

Each state’s members of the U.S. Senate

Solution a) The correspondence is a function because each person has only one weight. b) The correspondence is a function because each integer has only one square. c) The correspondence is not a function because each state has two U.S. Senators. Consistent with the definition on p. 13, we will regard a function as a set of ordered pairs, such that no two pairs have the same first coordinate paired with different second coordinates. The domain is the set of all first coordinates, and the range is the set of all second coordinates. Function names are usually represented by lowercase letters. Thus, if f represents the function in Example 1(b), we have f = 513, 92, 14, 162, 15, 252, 1- 5, 2526

and

Domain of f = 53, 4, 5, - 56;

Range of f = 59, 16, 256.

Finding Function Values Most functions considered in mathematics are described by an equation like y = 2x + 3 or y = 4 - x 2. To graph the function given by y = 2x + 3, we find ordered pairs by performing calculations for selected x values. for x = 4, y = 2x + 3 = 2 # 4 + 3 = 11; The graph includes 14, 112. # for x = - 5, y = 2x + 3 = 2 1- 52 + 3 = - 7; The graph includes for x = 0, y = 2x + 3 = 2 # 0 + 3 = 3; and so on.

1- 5, - 72. The graph includes 10, 32.

For y = 2x + 3, the inputs (members of the domain) are the values of x substituted into the equation. The outputs (members of the range) are the resulting values of y. If we call the function f, we can use x to represent an arbitrary input and f1x2, read “f of x” or “f at x” or “the value of f at x,” to represent the corresponding output. In this notation, the function given by y = 2x + 3 is written as f1x2 = 2x + 3, and the calculations above can be written more concisely as f142 = 2 # 4 + 3 = 11; f1- 52 = 2 # 1- 52 + 3 = - 7; f102 = 2 # 0 + 3 = 3; and so on.

Thus, instead of writing “when x = 4, the value of y is 11,” we can simply write “f142 = 11,” which is most commonly read as “f of 4 is 11.” It helps to think of a function as a machine. Think of f142 = 11 as the result of putting a member of the domain (an input), 4, into the machine. The machine knows

R.2

•

Functions and Models

15

the correspondence f1x2 = 2x + 3, computes 2 # 4 + 3, and produces a member of the range (the output), 11. Function: f1x2 = 2x + 3 Input

Output

4 -5 0 t a + h

11 -7 3 2t + 3 21a + h2 + 3

x

f

f (x)

Remember that f1x2 does not mean “f times x” and should never be read that way.

■ EXAMPLE 3 The squaring function f is given by f1x2 = x 2.

Find f1- 32, f112, f1k2, f12k 2, f11 + t2, and f1x + h2.

Solution We have

f1- 32 = 1- 322 = 9; f112 = 12 = 1; f1k2 = k 2;

f12k 2 = 12k 22 = k; f11 + t2 = 11 + t22 = 1 + 2t + t 2; f1x + h2 = 1x + h22 = x 2 + 2xh + h2. Quick Check 1

O

A function f is given by f1x2 = 3x + 5. Find f142, f1- 52, f102, f1a2, and f1a + h2.

For a review of algebra, see Appendix A on p. 605.

To find f1x + h2, remember what the function does: It squares the input. Thus, f1x + h2 = 1x + h22 = x 2 + 2xh + h2. This amounts to replacing x on both sides of f1x2 = x 2 with x + h.

O Quick Check 1 ■ EXAMPLE 4 A function f is given by f1x2 = 3x 2 - 2x + 8. Find f102, f1- 52, and f17a2.

Solution One way to find function values when a formula is given is to think of the formula with blanks, or placeholders, as follows: f1

2 = 3

2

- 2

+ 8.

To find an output for a given input, we think: “Whatever goes in the blank on the left goes in the blank(s) on the right.” Quick Check 2 A function f is given by f1x2 = 3x 2 + 2x - 7. Find f142, f1- 52, f102, f1a2, and f15a2.

f102 = 3 # 02 - 2 # 0 + 8 = 8 f1- 52 = 31- 522 - 2 # 1- 52 + 8 = 3 # 25 + 10 + 8 = 75 + 10 + 8 = 93 f17a2 = 317a22 - 217a2 + 8 = 3 # 49a 2 - 14a + 8 = 147a 2 - 14a + 8

O Quick Check 2

O

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Functions, Graphs, and Models

TECHNOLOGY CONNECTION The arrow keys, h and e, allow us to scroll up and down the table and extend it to other values not initially shown.

The TABLE Feature The TABLE feature is one way to find ordered pairs of inputs and outputs of functions. To see how, consider the function given by f 1x2 = x 3 - 5x + 1. We enter it as y1 = x 3 - 5x + 1. To use the TABLE feature, we access the TABLE SETUP screen and enter the x-value at which the table will start and an increment for the x-value. For this equation, let’s set TblStart = 0.3 and ¢Tbl = 1. (Other values can be chosen.) This means that the table’s x-values will start at 0.3 and increase by 1.

X 12.3 13.3 14.3 15.3 16.3 17.3 18.3 X = 18.3

Y1 1800 .4 2287.1 2853.7 3506.1 4250.2 5092.2 6038

EXERCISES

Use the function given by f 1x2 = x 3 - 5x + 1 for Exercises 1 and 2.

TABLE SETUP TblStart = .3 ΔTbl = 1 Indpnt: Auto Ask Depend: Auto Ask

1. Use the TABLE feature to construct a table starting with

x = 10 and ¢Tbl = 5. Find the value of y when x is 10. Then find the value of y when x is 35.

2. Adjust the table settings to Indpnt: Ask. How does the

table change? Enter a number of your choice and see what happens. Use this setting to find the value of y when x is 28.

We next set Indpnt and Depend to Auto and then press TABLE. The result is shown below. X

.3 1.3 2.3 3.31 4.3 5.3 6.3 X = .3

Y1 –.473 –3.303 1.667 20.437 59.007 123.38 219.55

■ EXAMPLE 5 A function f subtracts the square of an input from the input: f1x2 = x - x 2. Find f142, f1x + h2, and

f1x + h2 - f1x2 h

.

Solution We have f142 = 4 - 4 2 = 4 - 16 = - 12; f1x + h2 = 1x + h2 - 1x + h22 = x + h - 1x 2 + 2xh + h22 = x + h - x 2 - 2xh - h2 Squaring the binomial f1x + h2 - f1x2 Quick Check 3 A function f is given by f1x2 = 2x - x 2. Find f142, f1x + h2, and f1x + h2 - f1x2 . h

h

= =

x + h - x 2 - 2xh - h2 - 1x - x 22 h

=

h - 2xh - h2 h

h11 - 2x - h2

h = 1 - 2x - h, for h Z 0.

O Quick Check 3 Graphs of Functions Consider again the squaring function. The input 3 is associated with the output 9. The input–output pair 13, 92 is one point on the graph of this function.

O

R.2

DEFINITION

•

17

Functions and Models

y

The graph of a function f is a drawing that represents all the input–output pairs 1x, f1x22. In cases where the function is given by an equation, the graph of the function is the graph of the equation y = f1x2.

Graph of f y = f(x) (x, f(x)) x

It is customary to locate input values (the domain) on the horizontal axis and output values (the range) on the vertical axis.

■ EXAMPLE 6 Graph: f1x2 = x 2 - 1. Solution x -2

f1x2 3

1- 2, 32

-1

0

1- 1, 02

0

-1

10, - 12

1

0

11, 02

2

3

12, 32

(1) Choose any x. (2) Compute y. (3) Form the pair 1x, y2. (4) Plot the points.

y

1x, f1x22

4

(– 2, 3) (–1, 0)

(2, 3)

3 2

f(x) = x 2 – 1

1

(1, 0)

–3 – 2 –1

1

2

3

4

x

–1 –2

(0, –1)

We plot the input–output pairs from the table and, in this case, draw a curve to complete the graph.

Quick Check 4

O Quick Check 4

O

Graph: f1x2 = 2 - x 2.

TECHNOLOGY CONNECTION

Graphs and Function Values We discussed graphing equations in the Technology Connection of Section R.1. Graphing a function makes use of the same procedure. We just change the “f 1x2 = ” notation to “y = .” Thus, to graph f 1x2 = 2x 2 + x, we key in y1 = 2x 2 + x.

EXERCISES

10

1. Graph f 1x2 = x 2 + 3x - 4. Then find f 1- 52, f 1- 4.72,

Y1 = 2X 2 + X

–10

10

X = –2

Another way is to use the TRACE feature. To do so, graph the function, press TRACE, and either move the cursor or enter any x-value that is in the window. The corresponding y-value appears automatically. Function values can also be found using the VALUE or Y-VARS feature. Consult an owner’s manual or the GCM for details.

Y=6 –10

There are several ways in which to find function values. One is to use the TABLE feature, as previously discussed.

f 1112, and f 12>32. (Hint: To find f 1112, be sure that the window dimensions for the x-values include x = 11.)

2. Graph f 1x2 = 3.7 - x 2. Then find f 1- 52, f 1- 4.72,

f 1112, and f 12>32.

3. Graph f 1x2 = 4 - 1.2x - 3.4x 2. Then find f 1- 52,

f 1- 4.72, f 1112, and f 12>32.

18

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Functions, Graphs, and Models y

The Vertical-Line Test Let’s now determine how we can look at a graph and decide whether it is a graph of a function. In the graph at the right, note that the input x 1 has two outputs. Since a function has exactly one output for every input, this fact means that the graph does not represent a function. It also means that a vertical line could intersect the graph in more than one place.

x = y2

(x 1, y1)

y1

x1 y2

x

(x1, y2)

The Vertical-Line Test A graph represents a function if it is impossible to draw a vertical line that intersects the graph more than once.

■ EXAMPLE 7 Determine whether each of the following is the graph of a function. a)

y

y

b)

x

d)

y

c)

x

x

y

e)

y

f)

y

x x

x

Solution a) The graph is that of a function. It is impossible to draw a vertical line that intersects the graph more than once. b) The graph is not that of a function. A vertical line (in fact, many) can intersect the graph more than once. c) The graph is not that of a function. d) The graph is that of a function. e) The graph is that of a function. f) The graph is not that of a function.

Functions Defined Piecewise Sometimes functions are defined piecewise. That is, there are different output formulas for different parts of the domain, as in parts (e) and (f) of Example 7. To graph a piecewisedefined function, we usually work from left to right, paying special attention to the correspondence specified for the x-values on each part of the horizontal axis.

R.2

•

19

Functions and Models

■ EXAMPLE 8 Graph the function defined as follows: ⎫⎢ ⎢ ⎢ ⎢ ⎬ ⎢ ⎢ ⎭

f1x2 =

4,

for x … 0,

3 - x 2,

for 0 6 x … 2,

2x - 6, for x 7 2.

This means that for any input x less than or equal to 0, the output is 4. This means that for any input x greater than 0 and less than or equal to 2, the output is 3 - x 2. This means that for any input x greater than 2, the output is 2x - 6.

Solution Working from left to right along

y

the x-axis, we note that for any x-values less than or equal to 0, the graph is the horizontal line y = 4. Note that for f1x2 = 4,

and

f1- 22 = 4; f1- 12 = 4; f102 = 4.

The solid dot indicates that 10, 42 is part of the graph.

5 4 f(x) = 4, for x ≤ 0 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

and

The open dot at 10, 32 indicates that that point is not part of the graph.

Quick Check 5

f1x2 = 4, for x … 0, c 4 - x 2, for 0 6 x … 2, 2x - 6, for x 7 2.

5 4 3 2 1

and

f12.52 = 2 # 2.5 - 6 = - 1; f142 = 2; f152 = 4.

O Quick Check 5

(0.5, 2.75) (1, 2)

−5 −4 −3 −2 −1 1 2 3 4 5 x −1 (2, −1) −2 −3 4, for x ≤ 0 −4 f(x) = 3 − x2, for 0 < x ≤ 2 −5

y

Finally, note that for x-values greater than 2, the graph is the line y = 2x - 6. Note that for f1x2 = 2x - 6,

Graph the function defined as follows:

x

y

Next, observe that for x-values greater than 0 but not greater than 2, the graph is a portion of the parabola given by y = 3 - x 2. Note that for f1x2 = 3 - x 2, f10.52 = 3 - 0.52 = 2.75; f112 = 2; f122 = - 1.

1 2 3 4 5

5 4 3 2 1

(5, 4) (4, 2)

−5 −4 −3 −2 −1 1 2 −1 −2 −3 −4 f(x) = −5

3 4 5

x

(2.5, −1) 4, for x ≤ 0 3 − x 2, for 0 < x ≤ 2 2x − 6, for x > 2

■ EXAMPLE 9 Graph the function defined as follows: g1x2 = e

3, for x = 1, - x + 2, for x Z 1.

Solution The function is defined such that g112 = 3 and for all other x-values (that is, for x Z 1), we have g1x2 = - x + 2. Thus, to graph this function, we graph the line given by y = - x + 2, but with an open dot at the point above x = 1. To complete the graph, we plot the point 11, 32 since g112 = 3.

O

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Functions, Graphs, and Models

x

g1x2

1x, g1x22

-3

- 1- 32 + 2

1- 3, 52

0

-0 + 2

10, 22

1

3

11, 32

Quick Check 6

2

-2 + 2

12, 02

Graph the function defined as follows:

3

-3 + 2

13, - 12

O

f1x2 = e

1, for x = - 2, 2 - x, for x Z - 2.

y 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

g

1 2 3 4 5

x

O Quick Check 6

TECHNOLOGY CONNECTION

Graphing Functions Defined Piecewise Graphing functions defined piecewise generally involves the use of inequality symbols, which are often accessed using the TEST menu. The function in Example 8 is entered as follows: Scrolled equation Plot 1 Plot 2 Plot 3 Y1 = (4)(X ≤ 0) + (3 − X 2 )(0 < X)

Note that most graphing calculators will not display solid or open dots.

EXERCISES Graph. - x - 2, for x 6 - 2, for - 2 … x 6 2, x + 3, for x Ú 2

1. f 1x2 = c 4 - x 2,

(X ≤ 2) + (2X − 6)(X > 2 ) Y2 Y3 Y4 Y5

2. f 1x2 = e

= = = =

x 2 - 2, for x … 3, 1, for x 7 3 x + 3,

for x … - 2, for - 2 6 x … 3, x 2 - 10, for x 7 3

3. f 1x2 = c 1,

The graph is shown below. 6

–5

5

–3

Some Final Remarks We sometimes use the terminology y is a function of x. This means that x is an input and y is an output. It also means that x is the independent variable because it represents inputs and y is the dependent variable because it represents outputs. We may refer to “a function y = x 2” without naming it using a letter f. We may also simply refer to x 2 (alone) as a function. In calculus we will study how the outputs of a function change when the inputs change.

Section Summary • Functions are a key concept in mathematics.

• The essential trait of a function is that to each number in the domain there corresponds one and only one number in the range.

Exercise Set R.2

21

EXERCISE SET

R.2 Note: A review of algebra can be found in Appendix A on p. 605. Determine whether each correspondence is a function. 1. Domain

Range

2. Domain

Range

5

3

5

4

-3

7

7

8

9

7 -7 3. Domain

Range

4. Domain

-6

5

7

-7

8

3

-3

1

5. Sandwich prices. DOMAIN

RANGE

Hamburger Cheeseburger Filet-O-Fish® Quarter Pounder® with cheese Big N' Tasty® with cheese Big Mac® Crispy Chicken Chicken McGrill® Double Quarter Pounder® with cheese

$ 0.89 $ 0.95 $ 3.00 $ 3.20 $ 3.20 $ 3.40 $ 2.89 $ 3.80

Double Quarter Pounder® with cheese

A set of numbers

11. The set of all real numbers

Square each number and then add 8.

The set of all positive numbers greater than or equal to 8

12. The set of all real numbers

Raise each number to the fourth power.

The set of all nonnegative numbers

13. A set of females

Each person’s biological mother

A set of females

14. A set of males

Each person’s biological father

A set of males

15. A set of avenues

An intersecting road

A set of cross streets

16. A set of textbooks

An even-numbered page in each book

A set of pages

17. A set of shapes

The area of each shape

A set of area measurements

18. A set of shapes

The perimeter of each shape

A set of length measurements

a) Complete this table.

250 300 410 440 380 460 500 540 740

x

5.1

5.01

5.001

5

f 1x2 b) Find f142, f132, f1- 22, f1k2, f11 + t2, and f1x + h2. 20. A function f is given by f 1x2 = 3x + 2.

This function takes a number x, multiplies it by 3, and adds 2.

(Source: www.mcdonalds.com.)

Determine whether each of the following is a function.

Correspondence

Range

7. A set of iPods

Each iPod’s memory in gigabytes

A set of numbers

8. A set of iPods

Each iPod’s owner

A set of people

Domain

The number of Avril Lavigne songs on each iPod

This function takes a number x, multiplies it by 4, and subtracts 3. RANGE

Double Cheeseburger® Filet-O-Fish® Big N' Tasty® McRib® Big Mac®

10. A set of iPods

f 1x2 = 4x - 3.

6. Sandwich calorie content. DOMAIN

A set of numbers

19. A function f is given by

(Source: www.mcdonalds.com.)

Hamburger Cheeseburger Quarter Pounder®

The number of songs on each iPod

Range

6

-5

9. A set of iPods

a) Complete this table. x

4.1

4.01

4.001

4

f 1x2 b) Find f 152, f 1- 12, f 1k2, f 11 + t2, and f 1x + h2.

22

•

CHAPTER R

Functions, Graphs, and Models

32. g1x2 = - x 2 + 1

21. A function g is given by g1x2 = x 2 - 3. This function takes a number x, squares it, and subtracts 3. Find g1- 12, g102, g112, g152, g1u2, g1a + h2, g1a + h2 - g1a2 . and h 22. A function g is given by g1x2 = x 2 + 4.

33. g1x2 = x 3 34. g1x2 = 12 x 3 Use the vertical-line test to determine whether each graph is that of a function. (In Exercises 43–46, the vertical dashed lines are not part of the graph.) 35.

36.

37.

38.

39.

40.

41.

42.

This function takes a number x, squares it, and adds 4. Find g1 - 32, g102, g1- 12, g172, g1v2, g1a + h2, and g1a + h2 - g1a2 . h 23. A function f is given by f 1x2 =

1

1x + 322

.

This function takes a number x, adds 3, squares the result, and takes the reciprocal of that result.

a) Find f 142, f 1- 32, f 102, f 1a2, f 1t + 42, f 1x + h2, and f1x + h2 - f1x2 . If an output is undefined, state h that fact. b) Note that f could also be given by f 1x2 =

1 x + 6x + 9 2

.

Explain what this does to an input number x. 24. A function f is given by f 1x2 =

1

1x - 522

.

This function takes a number x, subtracts 5 from it, squares the result, and takes the reciprocal of the square.

43.

y

44.

a) Find f 132, f 1- 12, f 152, f 1k2, f 1t - 12, f 1t - 42, and f 1x + h2. If an output is undefined, state that fact. b) Note that f could also be given by f 1x2 =

1 x - 10x + 25 2

y

x

x

.

Explain what this does to an input number x.

45.

y

46.

y

Graph each function. 25. f 1x2 = 2x - 5

26. f 1x2 = 3x - 1

x

27. g1x2 = - 4x 28. g1x2 = - 2x

29. f 1x2 = x 2 - 2

47. a) Graph x = y 2 - 2. b) Is this a function?

31. f 1x2 = 6 - x 2

48. a) Graph x = y 2 - 3. b) Is this a function?

30. f 1x2 = x 2 + 4

x

Exercise Set R.2

49. For f 1x2 = x 2 - 3x, find

f1x + h2 - f1x2

50. For f 1x2 = x 2 + 4x, find

f1x + h2 - f1x2

h

h

.

2x  1, for x12x - 52 cannot be calculated when the denominator 2x - 5 is 0, we solve the following equation to find those real numbers that must be excluded from the domain of f:

1. Make a table for

f 1x2 = 1>1x - 42 from x = - 3 to x = 0 and with ¢Tbl set at 0.5. 2

2x - 5 = 0 2x = 5 x = 52 .

2. Create a table for the function

given in Example 5.

Setting the denominator equal to 0 Adding 5 to both sides Dividing both sides by 2

Thus, 52 is not in the domain, whereas all other real numbers are. We say that f is not defined at 52 , or f A 52 B does not exist. The domain of f is E x ƒ x is a real number and x Z 52 F , or, in interval notation, A - q , 52 B ´ A 52 , q B . The symbol ´ indicates the union of two sets and means that all elements in both sets are included in the domain.

■ EXAMPLE 6 Find the domain: f1x2 = 24 + 3x . Solution We ask, “What can we substitute?” Is there any number x for which we cannot calculate 24 + 3x? We recall that radicands in even roots cannot be negative. Since 24 + 3x is not a real number when the radicand 4 + 3x is negative, the domain is all real numbers for which 4 + 3x Ú 0. We find the domain by solving the inequality. (See Appendix A for a review of inequality solving.)

Quick Check 3

O

Find the domain of each function. Express your answers in interval notation. 5 a) f1x2 = x - 8 b) f1x2 = x 3 + ƒ 2x ƒ c) f1x2 = 22x - 8

4 + 3x Ú 0 3x Ú - 4 x Ú - 34

The domain is C - 43 , q B .

Simplifying Dividing both sides by 3

O Quick Check 3

TECHNOLOGY CONNECTION

Determining Domain and Range Graph each function in the given viewing window. Then determine the domain and the range. a) f 1x2 = 3 - ƒ x ƒ, 3- 10, 10, - 10, 104 b) f 1x2 = x 3 - x, 3- 3, 3, - 4, 44

c) f 1x2 =

12 , or 12x -1, 3- 14, 14, - 14, 144 x

d) f 1x2 = x 4 - 2x 2 - 3, 3- 4, 4, - 6, 64

(continued)

•

R.3

We have the following. a)

d)

6 10

Range

Domain –10

4

–4

10

Domain

Range

–6

–10

Domain =  Range = 3- 4, q 2

Domain =  (the real numbers), or 1- q , q 2 Range = 1 - q , 34 y = x3 – x

29

y = x 4 – 2x 2 – 3

y = 3 – |x|

b)

Finding Domain and Range

We can confirm our results using the TRACE feature, moving the cursor along the curve or entering any x-value in which we have interest. We can also use the TABLE feature. In Example (d), it might not appear as though the domain is all real numbers because the graph seems “thin,” but careful examination of the formula shows that we can indeed substitute any real number.

4

Range

–3

3 Domain

EXERCISES –4

Graph each function in the given viewing window. Then determine the domain and the range.

Domain =  Range = 

1. f 1x2 = ƒ x ƒ - 4,

14

3. f 1x2 =

Range

–14

3- 10, 10, - 10, 104

2. f 1x2 = 2 + 3x - x 3,

12 y = x , or 12x –1

c)

Domain Domain

3- 5, 5, - 5, 54

-3 , or - 3x -1, 3- 20, 20, - 20, 204 x

4. f 1x2 = x 4 - 2x 2 - 7,

14

5. f 1x2 = 2x + 4,

Range

6. f 1x2 = 29 - x 2,

3- 8, 8, - 8, 84

7. f 1x2 = - 29 - x ,

–14

2

Xscl = 2, Yscl = 2

3- 4, 4, - 9, 94

3- 5, 5, - 5, 54

3- 5, 5, - 5, 54

8. f 1x2 = x 3 - 5x 2 + x - 4,

The number 0 is excluded as an input. Domain = 5x ƒ x is a real number and x Z 06, or 1 - q , 02 ´ 10, q 2; Range = 5 y ƒ y is a real number and y Z 06, or 1- q , 02 ´ 10, q 2

3- 10, 10, - 20, 104

Domains and Ranges in Applications The domain and the range of a function given by a formula are sometimes affected by the context of an application. Let’s look again at Exercise 65 in Section R.2.

■ EXAMPLE 7 Business: Compound Interest. Suppose that $500 is invested at 6%, compounded quarterly for t years. From Theorem 2 in Section R.1, we know that the amount in the account is given by A1t2 = 500a1 +

0.06 4t b 4

= 50011.01524t. The amount A is a function of the number of years for which the money is invested. Determine the domain.

Solution We can substitute any real number for t into the formula, but a negative number of years is not meaningful. The context of the application excludes negative numbers. Thus, the domain is the set of all nonnegative numbers, 30, q 2.

30

CHAPTER R

•

Functions, Graphs, and Models

■ EXAMPLE 8 Cellphone Calling Plans. Recently, Sprint® offered a cellphone

Monthly bill (in dollars)

calling plan in which a customer’s monthly bill could be modeled by the graph below. Find the range of the function shown.

70

f

60 50 40 30 20 10 100

200

300

400

500

600

700

800

Phone use (in minutes) (Source: The New York Times. Bill does not include taxes and fees.)

Solution The range is the set of all outputs—in this case, the different monthly bill

Monthly bill (in dollars)

amounts—shown in the graph. We see that only six different outputs are used, highlighted in blue. Thus, the range of the function shown is 535, 40, 45, 50, 55, 606.

70

f

60 55 50 45 40 35 30 20 10 100

200

300

400

500

600

700

800

Phone use (in minutes)

We will continue to determine the domain and the range of a function as we progress through this book.

Section Summary The following is a review of the function concepts considered in Sections R.1–R.3. Function Concepts • • • • • •

Graph

Formula for f: f1x2 = x - 7 For every input of f, there is exactly one output. For the input 1, - 6 is the output. 2

f112 = - 6 11, - 62 is on the graph. Domain = The set of all inputs = The set of all real numbers, 

• Range = The set of all outputs = 3- 7, q 2

y 10 8 6 4 2 −5 −4 −3 −2 −1 −2 −4 f(x) = x2 − 7 −6 −8 −10

1 2 3 4 5

(1, −6)

x

Exercise Set R.3

31

EXERCISE SET

R.3 23.

In Exercises 1–10, write interval notation for each graph. 1. 2. 3. 4. 5. 6. 7.

−5 −4

−3

−2

−1

0

1

2

3

4

5

−5 −4

−3

−2

−1

0

1

2

3

4

5

24.

y 5 4 3 2 1

5 4 3 2 1

54321 1 1 2 3 4 5

–1

0

1

2

3

4

5

6

–1

0

1

2

3

4

5

6

–10

–9

–8

–7

–6

–5

–4

–3

–10

–9

–8

–7

–6

–5

–4

–3

x x p

10.

1 2 3 4 5

x

1 2 3 4 5

x

54321 1 1 2 3 4 5

x

2 3 4 5

26.

y

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x+h

9.

54321 1

27.

q

Write interval notation for each of the following. Then graph the interval on a number line.

2 1 1 2 3 4 5

x

11. The set of all numbers x such that - 2 … x … 2 12. The set of all numbers x such that - 5 6 x 6 5

54321 1

28.

14. 5x ƒ 6 6 x … 206

y 5 4 3 1

1 2 3 4 5

x

2 3 4 5

13. 5x ƒ - 4 … x 6 - 16

54321 1 2 3 4 5

y 5 4 3 2 1

y 5 4

5 4 3 2 1

x+h

8.

x

2 3 4 5

25.

y

2 3 4 5

15. 5x ƒ x … - 26

16. 5x ƒ x 7 - 36

29.

17. 5x ƒ - 2 6 x … 36

5 4 3 2 1

18. 5x ƒ - 10 … x 6 46 19. 5x ƒ x 6 12.56

54321 1

20. 5x ƒ x Ú 12.56

y

22.

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

31.

y 5 4 3 2 1

1 2 3 4 5

x

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

x

x

32.

y

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

54321 1 2 3 4 5

x

2 3 4 5

y 5 4

5 4 3 2 1 1 2 3 4 5

y 5 4 3 2 1

2 3 4 5

In Exercises 21–32, the graph is that of a function. Determine for each one (a) f 112; (b) the domain; (c) all x-values such that f 1x2 = 2; and (d) the range. 21.

30.

y

2 1 1 2 3 4 5

x

54321 1 2 3 4 5

1 2 3 4 5

x

32

•

CHAPTER R

Functions, Graphs, and Models

Find the domain of each function given below. 33. f 1x2 =

34. f 1x2 =

6 2 - x

35. f 1x2 = 22x

a) The amount A that is owed is a function of time. Find an equation for this function. b) Determine the domain of the function in part (a).

38. f 1x2 = x 2 + 3

x - 2 6x - 12

40. f 1x2 =

3x - 1 7 - 2x

44. f 1x2 =

41. f 1x2 = | x - 4 | 43. f 1x2 =

2 x + 3

36. f 1x2 = 2x - 2

37. f 1x2 = x 2 - 2x + 3 39. f 1x2 =

58. Compound interest. Suppose that $3000 is borrowed as a college loan, at 5% interest, compounded daily, for t years.

Life and Physical Sciences

8 3x - 6

59. Hearing-impaired Americans. The following graph (considered in Exercise Set R.1) approximates the number N, in millions, of hearing-impaired Americans as a function of age x. (Source: American Speech-Language Hearing Association.) The equation for this graph is the function given by

42. f 1x2 = | x | - 4

45. g1x2 = 24 + 5x

2x - 1 9 - 2x

46. g1x2 = 22 - 3x

N1x2 = - 0.00006x 3 + 0.006x 2 - 0.1x + 1.9.

47. g1x2 = x 2 - 2x + 1

N(x)

48. g1x2 = 4x + 5x - 2x 49. g1x2 = 50. g1x2 =

2

2x x - 25 2

x - 1 x 2 - 36

(Hint: Factor the denominator.) (Hint: Factor the denominator.)

51. g1x2 = | x | + 1 53. g1x2 = 54. g1x2 =

5

Number of hearing-impaired Americans (in millions)

3

2x - 6 x 2 - 6x + 5 3x - 10 x 2 - 4x - 5

52. g1x2 = | x + 7 | (Hint: Factor the denominator.)

4

N

3 2 1

0

20

40

60

Age

(Hint: Factor the denominator.)

55. For the function f whose graph is shown to the right, find all x-values for which f 1x2 … 0.

y (0, 0) −2

−1

(−1, −8)

f

2 1 −1 −2 −3 −4 −5 −6 −7 −8

1

2

x

(1, −2)

84.7

a) Use the graph to determine the domain of N. b) Use the graph to determine the range of N. c) If you were marketing a new type of hearing aid, at what age group (expressed as a 10-yr interval) would you target advertisements? Why? 60. Incidence of breast cancer. The following graph (considered in Exercise 28 of Exercise Set R.1 without an equation) approximates the incidence of breast cancer I, per 100,000 women, as a function of age x. The equation for this graph is the function given by I1x2 = - 0.0000554x 4 + 0.0067x 3 - 0.0997x 2

y

56. For the function g whose graph is shown to the right, find all x-values for which g1x2 = 1.

100 x

80

- 0.84x - 0.25.

g

2

I(x) 500

−4 −3 −2 −1

1

2

3

4

APPLICATIONS Business and Economics

x

Incidence of breast cancer per 100,000 women

1

400 300

I

200 100

57. Compound interest. Suppose that $5000 is invested at 8% interest, compounded semiannually, for t years. a) The amount A in the account is a function of time. Find an equation for this function. b) Determine the domain of the function in part (a).

0

10 20 30 40 50 60 70 80 90 100 110 120

Age (Source: Based on data from the National Cancer Institute.)

x

R.4

a) Use the graph to determine the domain of I. b) Use the graph to determine the range of I. c) What 10-yr age interval sees the greatest increase in the incidence of breast cancer? Explain how you determined this. 61. Lung cancer. The following graph approximates the incidence of lung and bronchus cancer L, per 100,000 males, as a function of t, the number of years since 1940. The equation for this graph is the function given by L1t2 = - 0.00054t 3 + 0.02917t 2 + 1.2329t + 8. Number of lung and bronchus cancer patients (per 100,000 males)

L(t)

Slope and Linear Functions

33

62. See Exercise 61. a) Use the graph to approximate all the x-values (ages) where the cancer rate is 50 per 100,000. b) Use the graph to approximate all the x-values (ages) where the cancer rate is 70 per 100,000. c) Use the formula to approximate the lung and bronchus cancer rate in 2010.

SYNTHESIS

63. For a given function, f 122 = - 5. Give as many interpretations of this fact as you can. 64. Explain how it is possible for the domain and the range of a function to be the same set.

80 70

65. Give an example of a function for which the number 3 is not in the domain, and explain why it is not.

60

L 50

TECHNOLOGY CONNECTION

40 30

66. Determine the range of each of the functions in Exercises 33, 35, 39, 40, and 47.

20

67. Determine the range of each of the functions in Exercises 34, 36, 48, 51, and 54.

10 0

•

10 20 (1950)

30

40

50

60

70 t (2010)

Answers to Quick Checks

Number of years since 1940 (Source: Based on data from the American Cancer Society Surveillance Research, 2005.)

a) Use the graph to estimate the domain of L. b) Use the graph to estimate the range of L.

R.4 OBJECTIVES • Graph equations of the types y  f 1x2  c and x = a. • Graph linear functions. • Find an equation of a line when given its slope and one point on the line and when given two points on the line. • Solve applied problems involving slope and linear functions.

1. (a) 3- 2, 54, (b) 3- 2, 52, (c) 1- 2, 54, (d) 1- 2, 52 2. f 1- 12 = 4, f 112 = 2; domain is 3- 2, 14, and range is 3- 1, 44 3. (a) 5x | x is a real number and x Z 86 (b)  (c) 34, q 2

Slope and Linear Functions Horizontal and Vertical Lines Let’s consider graphs of equations y = c and x = a, where c and a are real numbers.

■ EXAMPLE 1 a) Graph y = 4. b) Decide whether the graph represents a function.

Solution a) The graph consists of all ordered pairs whose second coordinate is 4. To see how a pair such as 1- 2, 42 could be a solution of y = 4, we can consider the equation above in the form y = 0x + 4.

Then 1- 2, 42 is a solution because

y (–2, 4)

5

(1, 4) y=4

3

01- 22 + 4 = 4

2

is true.

1 –4 – 3 –2 –1

1

2

b) The vertical-line test holds. Thus, the graph represents a function.

3

4 x

34

CHAPTER R

•

Functions, Graphs, and Models

■ EXAMPLE 2 a) Graph x = - 3. b) Decide whether the graph represents a function.

Solution a) The graph consists of all ordered pairs whose first coordinate is - 3. To see how a pair such as 1- 3, 42 could be a solution of x = - 3, we can consider the equation in the form x + 0y = - 3.

y x = –3 5

Then 1- 3, 42 is a solution because

(–3, 4) 4

1- 32 + 0142 = - 3

3 2

(–3, 1) 1

is true.

–4 – 3 –2 –1 –1

1

2

3 x

–2

Quick Check 1 b) This graph does not represent a function because it fails the vertical-line test. The line itself meets the graph more than once—in fact, infinitely many times.

O

Graph each equation: a) x = 4; b) y = - 3.

O Quick Check 1 In general, we have the following.

THEOREM 3 The graph of y = c, or f1x2 = c, a horizontal line, is the graph of a function. Such a function is referred to as a constant function. The graph of x = a is a vertical line, and x = a is not a function.

TECHNOLOGY CONNECTION

Visualizing Slope Exploratory: Squaring a Viewing Window

The standard 3- 10, 10, - 10, 104 viewing window shown below is not scaled identically on both axes. Note that the intervals on the y-axis are about two-thirds the length of those on the x-axis.

If we change the dimensions of the window to 3- 6, 6, - 4, 44, we get a graph for which the units are visually about the same on both axes. 4

–6

6

10

–4 –10

10

–10

Creating such a window is called squaring the window. On many calculators, this is accomplished automatically by selecting the ZSquare option of the ZOOM menu. (continued)

R.4

Each of the following is a graph of y = 2x - 3, but with different viewing windows. When the window is square, as shown in the last graph, we get the most accurate representation of the slope of the line. y = 2x – 3

y = 2x – 3

12

•

Slope and Linear Functions

EXERCISES Use a squared viewing window for each of these exercises. 1. Graph y = x + 1, y = 2x + 1, y = 3x + 1, and

y = 10x + 1. What do you think the graph of y = 247x + 1 will look like?

10

7

2. Graph y = x, y = 8 x, y = 0.47x, and y = –4

–10

4

35

2 31 x.

What do you think the graph of y = 0.000018x will look like?

10

3. Graph y = - x, y = - 2x, y = - 5x, and y = - 10x.

What do you think the graph of y = - 247x will look like?

–10

–12

y = 2x – 3

6

3

4. Graph y = - x - 1, y = - 4 x - 1, y = - 0.38x - 1, 5 x - 32

–9

- 1. What do you think the graph of and y = y = - 0.000043x - 1 will look like?

9

Squared window

–6

The Equation y  mx Consider the following table of numbers and look for a pattern. y (2, 6)

6

x

1

-1

- 12

2

-2

3

-7

5

y

3

-3

- 32

6

-6

9

- 21

15

5 4

(1, 3)

3 2

Note that the ratio of the y-value to the x-value is 3. That is,

1

⎧ ⎨ ⎩

x

⎧ y ⎨ ⎩

1

y = 3, or y = 3x. x

x

2

(0, 0)

Ordered pairs from the table can be used to graph the equation y = 3x (see the figure at the left). Note that this is a function.

(x, y)

THEOREM 4 y = 3x y = 2x y=x

y (1, 3)

4 3

(1, 2)

2 1 – 2 –1

(1, 1) 1

2

3

The graph of the function given by y = mx

or

f1x2 = mx

is the straight line through the origin 10, 02 and the point 11, m2. The constant m is called the slope of the line.

4 x

–1 –2 –3

Various graphs of y = mx for positive values of m are shown to the left. Note that such graphs slant up from left to right. A line with large positive slope rises faster than a line with smaller positive slope.

36

CHAPTER R

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Functions, Graphs, and Models

When m = 0, y = 0x, or y = 0. On the left below is a graph of y = 0. Note that this is both the x-axis and a horizontal line. y

y = – 3x y = – 2x y = –x

4 3 2 1

y=0 (1, 0)

y

x

–3 – 2 –1

1 –1 –2 –3

2

3

4 x

(1, –1) (1, – 2)

(1, –3)

Quick Check 2 The rows of flowers form lines of equal slope.

Graph each equation: a) y =

1 x; 2

1 b) y = - x. 2

Graphs of y = mx for negative values of m are shown on the right above. Note that such graphs slant down from left to right.

O Quick Check 2 Direct Variation There are many applications involving equations like y = mx, where m is some positive number. In such situations, we say that we have direct variation, and m (the slope) is called the variation constant, or constant of proportionality. Generally, only positive values of x and y are considered.

DEFINITION The variable y varies directly as x if there is some positive constant m such that y = mx. We also say that y is directly proportional to x.

■ EXAMPLE 3 Life Science: Weight on Earth and the Moon. The weight M, in pounds, of an object on the moon is directly proportional to the weight E of that object on Earth. An astronaut who weighs 180 lb on Earth will weigh 28.8 lb on the moon. a) Find an equation of variation. b) An astronaut weighs 120 lb on Earth. How much will the astronaut weigh on the moon?

O

R.4

•

37

Slope and Linear Functions

Solution a) The equation has the form M = mE. To find m, we substitute:

TECHNOLOGY CONNECTION

Exploring b Let’s explore the effect of b when graphing y = mx + b.

EXERCISES 1. Graph y1 = x. Then, using

the same viewing window, compare the graphs of y2 = x + 3 and y3 = x - 4. Next, without drawing them, describe how the graphs of y = x and y = x - 5 compare.

= mE = m # 180

M 40 Weight on moon (in pounds)

M 28.8 28.8 180 0.16

= m = m.

Thus, M = 0.16E is the equation of variation. b) To find the weight on the moon of an astronaut who weighs 120 lb on Earth, we substitute 120 for E in the equation of variation, M = 0.16 # 120,

30 M = 0.16E

20 10

40

80

120

160

200

E

Weight on Earth (in pounds)

Substituting 120 for E

and get M = 19.2. Thus, an astronaut who weighs 120 lb on Earth weighs 19.2 lb on the moon.

2. Use the TABLE feature to

compare the values of y1, y2, and y3 when x = 0. Then scroll through other values and describe a pattern.

The Equation y  mx  b Compare the graphs of the equations y = 3x and y = 3x - 2 (see the following figure). Note that the graph of y = 3x - 2 is a shift 2 units down of the graph of y = 3x, and that y = 3x - 2 has y-intercept 10, - 22. Both graphs represent functions. y 5 4 3 2 1

y = 3x − 2

(0, 0)

−5 −4 −3 −2 −1 1 2 3 4 5 −1 −2 (0, −2) −3 y = 3x −4 −5

x

y

DEFINITION

y = mx + b

A linear function is given by

(0, b) y = mx (0, 0)

x

y = mx + b or f1x2 = mx + b and has a graph that is the straight line parallel to the graph of y = mx and crossing the y-axis at 10, b2. The point 10, b2 is called the y-intercept. (See the figure at the left.)

As before, the constant m is the slope of the line. When m = 0, y = 0x + b = b, and we have a constant function (see Theorem 3 at the beginning of this section). The graph of such a function is a horizontal line.

38

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The Slope–Intercept Equation

Every nonvertical line l is uniquely determined by its slope m and its y-intercept 10, b2. In other words, the slope describes the “slant” of the line, and the y-intercept locates the point at which the line crosses the y-axis. Thus, we have the following definition.

DEFINITION y = mx + b is called the slope–intercept equation of a line.

■ EXAMPLE 4 Find the slope and the y-intercept of the graph of 2x - 4y - 7 = 0. Solution We solve for y: 2x - 4y - 7 = 0 4y = 2x - 7

Adding 4y to both sides

2 7 x 4 4

Dividing both sides by 4

y = Quick Check 3 Find the slope and the y-intercept of the graph of 3x - 6y - 7 = 0.

Slope: 12

y-intercept: A 0, - 74 B

O Quick Check 3 The Point–Slope Equation Suppose that we know the slope of a line and some point on the line other than the y-intercept. We can still find an equation of the line.

■ EXAMPLE 5 Find an equation of the line with slope 3 containing the point 1- 1, - 52.

Solution The slope is given as m = 3. From the slope–intercept equation, we have y = 3x + b,

(1)

so we must determine b. Since 1- 1, - 52 is on the line, we substitute - 5 for y and - 1 for x:

so

- 5 = 31- 12 + b - 5 = - 3 + b, -2 = b

Then, replacing b in equation (1) with - 2, we get y = 3x - 2. More generally, if a point 1x 1, y12 is on the line given by y = mx + b,

(2)

it must follow that y1 = mx 1 + b.

(3)

O

•

R.4

Slope and Linear Functions

39

y (x, y)

y

( y – y1) y1

(x1, y1) (x – x1)

x1

x

x

Subtracting the left and right sides of equation (3) from the left and right sides, respectively, of equation (2), we have y - y1 = = = =

1mx + b2 - 1mx 1 + b2 mx + b - mx 1 - b mx - mx 1 m1x - x 12.

Multiplying by - 1 Combining like terms Factoring

DEFINITION

y - y1 = m1x - x 12 is called the point–slope equation of a line. The point is 1x 1, y12, and the slope is m. This definition allows us to write an equation of a line given its slope and the coordinates of any point on the line.

■ EXAMPLE 6 Find an equation of the line with slope 1- 1, - 52.

2 3

containing the point

Solution Substituting in

y - y1 = m1x - x 12,

we get

y - 1- 52 = 233x - 1- 124 y + 5 = 231x + 12 y + 5 = 23 x + y = 23 x +

Quick Check 4

O

Find the equation of the line with slope - 23 containing the point 1- 3, 62.

y = y =

2 3x 2 3x

+ -

2 3 2 3 - 5 2 15 3 - 3 13 3.

Multiplying by 23 Subtracting 5

Combining like terms

O Quick Check 4 Computing Slope We now determine a method of computing the slope of a line when we know the coordinates of two of its points. Suppose that 1x 1, y12 and 1x 2, y22 are the coordinates of two different points, P1 and P2, respectively, on a line that is not vertical. Consider a right triangle with legs parallel to the axes, as shown in the following figure.

Which lines have the same slope?

40

CHAPTER R

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Functions, Graphs, and Models y P2(x 2, y2)

P1(x1, y1) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ (x 2 – x1) the change in x

⎫ ⎪ (y – y ) 2 1 ⎪ ⎬ the change in y ⎪ ⎭ (x 2, y1) x

Note that the change in y is y2 - y1 and the change in x is x 2 - x 1. The ratio of these changes is the slope. To see this, consider the point–slope equation, y - y1 = m1x - x 12.

Since 1x 2, y22 is on the line, it must follow that y2 - y1 = m1x 2 - x 12.

Substituting

Since the line is not vertical, the two x-coordinates must be different; thus, x 2 - x 1 is nonzero, and we can divide by it to get the following theorem.

THEOREM 5

The slope of a line containing points 1x 1, y12 and 1x 2, y22 is m =

y2 - y1 change in y = . x2 - x1 change in x

■ EXAMPLE 7 Find the slope of the line containing the points 1- 2, 62 and 1- 4, 92. Solution We have y2 - y1 6 - 9 = x2 - x1 - 2 - 1- 42 3 -3 = - . = 2 2

m =

We treated 1- 2, 62 as P2 and 1- 4, 92 as P1.

Note that it does not matter which point is taken first, so long as we subtract the coordinates in the same order. In this example, we can also find m as follows: Quick Check 5 Find the slope of the line containing the points 12, 32 and 11, - 42.

m =

9 - 6 3 3 = = - . - 4 - 1- 22 -2 2

Here, 1- 4, 92 serves as P2, and 1- 2, 62 serves as P1.

O Quick Check 5 If a line is horizontal, the change in y for any two points is 0. Thus, a horizontal line has slope 0. If a line is vertical, the change in x for any two points is 0. Thus, the slope is not defined because we cannot divide by 0. A vertical line has undefined slope. Thus, “0 slope” and “undefined slope” are two very different concepts.

O

•

R.4 y

y

(x1, y1)

y2

(x 2, y2)

y1

(x1, y1)

(x 2, y2)

y2 – y1 0 ——— m = ——— x 2 – x1 = x 2 – x1 = 0 x1

x2

41

Slope and Linear Functions y2 – y1 m = ——— x –x 2

1

y2 – y1 = ——— 0 not defined

x

x

Applications of Slope Slope has many real-world applications. For example, numbers like 2%, 3%, and 6% are often used to represent the grade of a road, a measure of how steep a road on a hill is. 3 A 3% grade A 3% = 100 B means that for every horizontal distance of 100 ft, the road rises 3 ft. In architecture, the pitch of a roof is a measure of how steeply it is angled—a steep pitch sheds more snow than a shallow pitch. Wheelchair-ramp design also involves 1 slope: Building codes rarely allow the steepness of a wheelchair ramp to exceed 12 . a Road grade  b (expressed as a percent)

a b 2 ft 24 ft

a b

Ski trail difficulty ratings, or gradients, are yet another application of slope. The following table presents examples. Ski Trail Difficulty Ratings in North America Trail Rating

Symbol

Level of Difficulty

Description

Green Circle

Easiest

A Green Circle trail is the easiest. These trails are generally wide and groomed, typically with slope gradients ranging from 6% to 25% (a 100% slope is a 45° angle).

Blue Square

Intermediate

A Blue Square trail is of intermediate difficulty. These trails have gradients ranging from 25% to 40%. They are usually groomed and are usually among the most heavily used.

Black Diamond

Difficult

Black Diamond trails tend to be steep (typically 40% and up), may or may not be groomed, and are among the most difficult.

CHAPTER R

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Functions, Graphs, and Models

There are even more difficult ski trails. The rating of a trail is done at the discretion of the ski resort operators. There are a number of iPod and iPhone apps that skiers can use to estimate difficulty ratings. To estimate a gradient, hold your arm parallel to the ground out from your side—that is a 0% gradient. Hold it at a 45° angle—that is a 100% gradient. A 22.5° angle is a 41% gradient. And, surprisingly, an angle of only 3.5° constitutes a 6% gradient. What do you think the slope is of the steep road at the top of the mountain in the photo?

Frenchman Mountain near Las Vegas, Nevada.

Slope can also be considered as an average rate of change.

■ EXAMPLE 8 Life Science: Amount Spent on Cancer Research. The amount spent on cancer research has increased steadily over the years and is approximated in the following graph. Find the average rate of change of the amount spent on research. CANCER RESEARCH National Cancer Institute’s annual research and development budget (in billions of dollars)

42

y $5.0 4.8 4.6 4.4 4.2 4.0 3.8 3.6 3.4 3.2

(2008, $4.828)

(2000, $3.311)

’00 ’01 ’02 ’03 ’04 ’05 ’06 ’07 ’08 ’09 ’10 Year

x

(Source: National Cancer Institute.)

Solution First, we determine the coordinates of two points on the graph. In this case, they are given as 12000, $3.3112 and 12008, $4.8282. Then we compute the slope, or rate of change, as follows: Slope = average rate of change = =

change in y change in x

$4.828 - $3.311 1.517 = L $0.1896 billion>yr. 2008 - 2000 8

•

R.4

43

Slope and Linear Functions

Applications of Linear Functions Many applications are modeled by linear functions.

■ EXAMPLE 9 Business: Total Cost. Raggs, Ltd., a clothing firm, has fixed costs of $10,000 per year. These costs, such as rent, maintenance, and so on, must be paid no matter how much the company produces. To produce x units of a certain kind of suit, it costs $20 per suit (unit) in addition to the fixed costs. That is, the variable costs for producing x of these suits are 20x dollars. These costs are due to the amount produced and stem from items such as material, wages, fuel, and so on. The total cost C1x2 of producing x suits in a year is given by a function C: C1x2 = 1Variable costs2 + 1Fixed costs2 = 20x + 10,000.

a) Graph the variable-cost, the fixed-cost, and the total-cost functions. b) What is the total cost of producing 100 suits? 400 suits?

Solution a) The variable-cost and fixed-cost functions appear in the graph on the left below. The total-cost function is shown in the graph on the right. From a practical standpoint, the domains of these functions are nonnegative integers 0, 1, 2, 3, and so on, since it does not make sense to make either a negative number or a fractional number of suits. It is common practice to draw the graphs as though the domains were the entire set of nonnegative real numbers. y

y

$20,000

$20,000

$10,000

Fixed costs = 10,000

Total cost = C(x) = 20x + 10,000

⎫ ⎬ C(400) – C(100) ⎭

$10,000 Variable costs = 20x

100

200

300 400 Suits produced

500

x

100

200

300 400 Suits produced

500

x

b) The total cost of producing 100 suits is

C11002 = 20 # 100 + 10,000 = $12,000.

The total cost of producing 400 suits is C14002 = 20 # 400 + 10,000 = $18,000.

■ EXAMPLE 10 Business: Profit-and-Loss Analysis. When a business sells an item, it receives the price paid by the consumer (this is normally greater than the cost to the business of producing the item). a) The total revenue that a business receives is the product of the number of items sold and the price paid per item. Thus, if Raggs, Ltd., sells x suits at $80 per suit, the total revenue R1x2, in dollars, is given by R1x2 = Unit price # Quantity sold = 80x.

If C1x2 = 20x + 10,000 (see Example 9), graph R and C using the same set of axes. b) The total profit that a business receives is the amount left after all costs have been subtracted from the total revenue. Thus, if P1x2 represents the total profit when x items are produced and sold, we have P1x2 = 1Total revenue2 - 1Total costs2 = R1x2 - C1x2.

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Determine P1x2 and draw its graph using the same set of axes as was used for the graph in part (a). c) The company will break even at that value of x for which P1x2 = 0 (that is, no profit and no loss). This is the point at which R1x2 = C1x2. Find the break-even value of x.

Solution a) The graphs of R1x2 = 80x and C1x2 = 20x + 10,000 are shown below. When C1x2 is above R1x2, a loss will occur. This is shown by the region shaded red. When R1x2 is above C1x2, a gain will occur. This is shown by the region shaded gray. y

R(x) = 80x Gain

$20,000

C(x) = 20x + 10,000 $10,000

Loss

100

200

300 400 Suits produced

500

x

–$10,000

b) To find P, the profit function, we have

P1x2 = R1x2 - C1x2 = 80x - 120x + 10,0002 = 60x - 10,000.

The graph of P1x2 is shown by the heavy line.The red portion of the line shows a “negative” profit, or loss. The black portion of the heavy line shows a “positive” profit, or gain. y

R(x) = 80x Gain

$20,000

C(x) = 20x + 10,000

Break even $10,000 Loss

P(x) = R(x) – C(x) = 60x – 10,000 100

200 166 23-

300

400

500

x

Suits produced

–$10,000

c) To find the break-even value, we solve R1x2 = C1x2:

Quick Check 6 Business. Suppose that in Examples 9 and 10 fixed costs are increased to $20,000. Find: a) the total-cost, total-revenue, and total-profit functions; b) the break-even value.

R1x2 80x 60x x

= = = =

C1x2 20x + 10,000 10,000 166 23.

How do we interpret the fractional answer, since it is not possible to produce 23 of a suit? We simply round up to 167. Estimates of break-even values are usually sufficient since companies want to operate well away from break-even values in order to maximize profit.

O Quick Check 6

O

45

Exercise Set R.4

Section Summary • Graphs of functions that are straight lines (linear functions) are characterized by an equation of the type

f1x2 = mx + b, where m is the slope and 10, b2 is the yintercept, the point at which the graph crosses the y-axis.

EXERCISE SET

R.4 39. 12, - 32 and 1- 1, - 42

Graph.

41. 13, - 72 and 13, - 92

1. x = 3

2. x = 5

3. y = - 2

4. y = - 4

43.

5. x = - 4.5

6. x = - 1.5

45. 12, 32 and 1- 1, 32

7. y = 3.75

8. y = 2.25

Graph. List the slope and y-intercept. 9. y = - 2x

10. y = - 3x

11. f 1x2 = 0.5x

12. f 1x2 = - 0.5x

13. y = 3x - 4

14. y = 2x - 5

15. g1x2 = - x + 3

16. g1x2 = x - 2.5

17. y = 7

18. y = - 5

40. 1- 3, - 52 and 11, - 62

A - 3 B and A 4 5,

1 2 2, 5

42. 1- 4, 22 and 1- 4, 102

B

44. 46.

47. 1x, 3x2 and 1x + h, 31x + h22

A - 163 , - 12 B and A 58, - 34 B A - 6, 12 B and A - 7, 12 B

48. 1x, 4x2 and 1x + h, 41x + h22

49. 1x, 2x + 32 and 1x + h, 21x + h2 + 32

50. 1x, 3x - 12 and 1x + h, 31x + h2 - 12 51–60. Find an equation of the line containing the pair of points in each of Exercises 37–46. 61. Find the slope of the skateboard ramp.

Find the slope and y-intercept. 19. y - 3x = 6

20. y - 4x = 1

21. 2x + y - 3 = 0

22. 2x - y + 3 = 0

23. 2x + 2y + 8 = 0

24. 3x - 3y + 6 = 0

25. x = 3y + 7

26. x = - 4y + 3

Find an equation of the line:

27. with m = - 5, containing 1- 2, - 32.

2 ft

5 ft

62. Find the slope (or grade) of the treadmill.

28. with m = 7, containing 11, 72.

29. with m = - 2, containing 12, 32.

30. with m = - 3, containing 15, - 22. 31. with slope 2, containing 13, 02.

32. with slope - 5, containing 15, 02.

33. with y-intercept 10, - 62 and slope 12.

34. with y-intercept 10, 72 and slope 43.

0.4 ft 5 ft

63. Find the slope (or head) of the river. Express the answer as a percentage.

35. with slope 0, containing 12, 32. 36. with slope 0, containing 14, 82.

Find the slope of the line containing the given pair of points. If a slope is undefined, state that fact. 37. 15, - 32 and 1- 2, 12

38. 1 - 2, 12 and 16, 32

43.33 ft 1238 ft

CHAPTER R

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Functions, Graphs, and Models

APPLICATIONS Business and Economics 64. Highway tolls. It has been suggested that since heavier vehicles are responsible for more of the wear and tear on highways, drivers should pay tolls in direct proportion to the weight of their vehicles. Suppose that a Toyota Camry weighing 3350 lb was charged $2.70 for traveling an 80-mile stretch of highway. a) Find an equation of variation that expresses the amount of the toll T as a function of the vehicle’s weight w. b) What would the toll be if a 3700-lb Jeep Cherokee drove the same stretch of highway? 65. Inkjet cartridges. A registrar’s office finds that the number of inkjet cartridges, I, required each year for its copiers and printers varies directly with the number of students enrolled, s. a) Find an equation of variation that expresses I as a function of s, if the office requires 16 cartridges when 2800 students enroll. b) How many cartridges would be required if 3100 students enrolled? 66. Profit-and-loss analysis. Boxowitz, Inc., a computer firm, is planning to sell a new graphing calculator. For the first year, the fixed costs for setting up the new production line are $100,000. The variable costs for producing each calculator are estimated at $20. The sales department projects that 150,000 calculators can be sold during the first year at a price of $45 each. a) Find and graph C1x2, the total cost of producing x calculators. b) Using the same axes as in part (a), find and graph R1x2, the total revenue from the sale of x calculators. c) Using the same axes as in part (a), find and graph P1x2, the total profit from the production and sale of x calculators. d) What profit or loss will the firm realize if the expected sale of 150,000 calculators occurs? e) How many calculators must the firm sell in order to break even? 67. Profit-and-loss analysis. Red Tide is planning a new line of skis. For the first year, the fixed costs for setting up production are $45,000. The variable costs for producing each pair of skis are estimated at $80, and the selling price will be $255 per pair. It is projected that 3000 pairs will sell the first year. a) Find and graph C1x2, the total cost of producing x pairs of skis. b) Find and graph R1x2, the total revenue from the sale of x pairs of skis. Use the same axes as in part (a). c) Using the same axes as in part (a), find and graph P1x2, the total profit from the production and sale of x pairs of skis. d) What profit or loss will the company realize if the expected sale of 3000 pairs occurs? e) How many pairs must the company sell in order to break even?

68. Straight-line depreciation. Quick Copy buys an office machine for $5200 on January 1 of a given year. The machine is expected to last for 8 yr, at the end of which time its salvage value will be $1100. If the company figures the decline in value to be the same each year, then the book value, V1t2, after t years, 0 … t … 8, is given by V1t2 = C - ta

C - S b, N

where C is the original cost of the item, N is the number of years of expected life, and S is the salvage value. a) Find the linear function for the straight-line depreciation of the office machine. b) Find the book value after 0 yr, 1 yr, 2 yr, 3 yr, 4 yr, 7 yr, and 8 yr. 69. Profit-and-loss analysis. Jimmy decides to mow lawns to earn money. The initial cost of his lawnmower is $250. Gasoline and maintenance costs are $4 per lawn. a) Formulate a function C1x2 for the total cost of mowing x lawns. b) Jimmy determines that the total-profit function for the lawnmowing business is given by P1x2 = 9x - 250. Find a function for the total revenue from mowing x lawns. How much does Jimmy charge per lawn? c) How many lawns must Jimmy mow before he begins making a profit? 70. Straight-line depreciation. (See Exercise 68.) A business tenant spends $40 per square foot on improvements to a 25,000-ft2 office space. Under IRS guidelines for straightline depreciation, these improvements will depreciate completely—that is, have zero salvage value—after 39 yr. Find the depreciated value of the improvements after 10 yr. 71. Book value. (See Exercise 68.) The Video Wizard buys a new computer system for $60,000 and projects that its book value will be $2000 after 5 yr. Using straight-line depreciation, find the book value after 3 yr. 72. Book value. Tyline Electric uses the function B1t2 = - 700t + 3500 to find the book value, B1t2, in dollars, of a photocopier t years after its purchase. B(t) Book value of machine (in dollars)

46

4000 3500 3000 2500

B(t) = – 700t + 3500

2000 1500 1000 500 0

1

2

3

4

5

6

t

Number of years of use

a) What do the numbers - 700 and 3500 signify? b) How long will it take the copier to depreciate completely? c) What is the domain of B? Explain.

Exercise Set R.4

73. Stair requirements. A North Carolina state law requires that stairs have minimum treads of 9 in. and maximum risers of 8.25 in. (Source: North Carolina Office of the State Fire Marshal.) See the illustration below. According to this law, what is the maximum grade of stairs in North Carolina? 9 in.

8.25 in

.

76. Two-year college tuitions. Find the average rate of change of the tuition and fees at public two-year colleges.

Public two-year college tuition and fees (in dollars)

General Interest

2200

(2008, $2063) 2000 1800

(2004, $1702)

1600 1400

2004

2008

Year (Source: U.S. National Center for Education Statistics, Digest of Education Statistics, annual.)

77. Wedding cost. Find the average rate of change of the cost of a formal wedding. 15,000

(2009, $13,375) 12,000

(2005, $10,880) 9,000

2003 2004 2005 2006 2007 2008 2009 2010

Year

30,000

(2005, $26,450) 25,000

20,000

(2009, $19,580)

15,000

(Source: The Kaiser Family Foundation; Health Research and Education Trust.)

75. Health insurance premiums. Find the average rate of change in the annual premium for a single person. Annual health insurance premium for single person (in dollars)

Cost of a formal wedding (in dollars)

Annual health insurance premium for family (in dollars)

74. Health insurance premiums. Find the average rate of change in the annual premium for a family’s health insurance.

47

2005

2009

Year (Source: The Fairchild Bridal Group.)

78. Energy conservation. The R-factor of home insulation is directly proportional to its thickness T. 6,000

(2009, $4824) 3,000

(2005, $4024)

a) Find an equation of variation if R = 12.51 when T = 3 in. b) What is the R-factor for insulation that is 6 in. thick? 79. Nerve impulse speed. Impulses in nerve fibers travel at a speed of 293 ft> sec. The distance D, in feet, traveled in t sec is given by D = 293t. How long would it take an impulse to travel from the brain to the toes of a person who is 6 ft tall?

2003 2004 2005 2006 2007 2008 2009 2010

Year (Source: The Kaiser Family Foundation; Health Research and Education Trust.)

80. Muscle weight. The weight M of the muscles in a human is directly proportional to the person’s body weight W.

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a) It is known that a person who weighs 200 lb has 80 lb of muscles. Find an equation of variation expressing M as a function of W. b) Express the variation constant as a percent, and interpret the resulting equation. c) What is the muscle weight of a person who weighs 120 lb?

a) Find D152, D1102, D1202, D1502, and D1652. b) Graph D1r2. c) What is the domain of the function? Explain.

r D

84. Estimating heights. An anthropologist can use certain linear functions to estimate the height of a male or female, given the length of certain bones. The humerus is the bone from the elbow to the shoulder. Let x = the length of the humerus, in centimeters. Then the height, in centimeters, of a male with a humerus of length x is given by M1x2 = 2.89x + 70.64. Muscle weight is directly proportional to body weight.

81. Brain weight. The weight B of a human’s brain is directly proportional to a person’s body weight W. a) It is known that a person who weighs 120 lb has a brain that weighs 3 lb. Find an equation of variation expressing B as a function of W. b) Express the variation constant as a percent and interpret the resulting equation. c) What is the weight of the brain of a person who weighs 160 lb?

The height, in centimeters, of a female with a humerus of length x is given by F1x2 = 2.75x + 71.48.

x

Humerus

82. Stopping distance on glare ice. The stopping distance (at some fixed speed) of regular tires on glare ice is given by a linear function of the air temperature F, D1F2 = 2F + 115, where D1F2 is the stopping distance, in feet, when the air temperature is F, in degrees Fahrenheit.

A 26-cm humerus was uncovered in some ruins. a) If we assume it was from a male, how tall was he? b) If we assume it was from a female, how tall was she?

D 200 150 100

D(F) = 2F + 115

50 –60 –40 –20

20

40

F

a) Find D10°2, D1- 20°2, D110°2, and D132°2. b) Explain why the domain should be restricted to the interval 3- 57.5°, 32°4. 83. Reaction time. While driving a car, you see a child suddenly crossing the street. Your brain registers the emergency and sends a signal to your foot to hit the brake. The car travels a distance D, in feet, during this time, where D is a function of the speed r, in miles per hour, that the car is traveling when you see the child. That reaction distance is a linear function given by 11r + 5 . D1r2 = 10

85. Percentage of young adults using the Internet. In 2000, the percentage of 18- to 29-year-olds who used the Internet was 72%. In 2009, that percentage had risen to 92%. a) Use the year as the x-coordinate and the percentage as the y-coordinate. Find the equation of the line that contains the data points. b) Use the equation in part (a) to estimate the percentage of Internet users in 2010. c) Use the equation in part (a) to estimate the year in which the percentage of Internet users will reach 100%. d) Explain why a linear equation cannot be used for years after the year found in part(c). 86. Manatee population. In January 2001, 3300 manatees were counted in an aerial survey of Florida. In January 2005, 3143 manatees were counted. (Source: Florida Fish and Wildlife Conservation Commission.) a) Using the year as the x-coordinate and the number of manatees as the y-coordinate, find an equation of the line that contains the two data points.

49

Exercise Set R.4

a) Assuming that N is directly proportional to P, find an equation of variation. b) Find N when P = 200,000. c) Find P when N = 367,200. 88. Median age of women at first marriage. In general, people in our society are marrying at a later age. The median age, A1t2, of women at first marriage can be approximated by the linear function A1t2 = 0.08t + 19.7,

IV $9 8 7 6 5 4 3 2 1 Dec. 1

a) Find A102, A112, A1102, A1302, and A1502. b) What was the median age of women at first marriage in 2008? c) Graph A1t2.

Answers to Quick Checks y

1.

SYNTHESIS −4

89. Explain and compare the situations in which you would use the slope–intercept equation rather than the point–slope equation.

−2

y

x4

4

91. Business: daily sales. Match each sentence below with the most appropriate of the following graphs (I, II, III, or IV). a) After January 1, daily sales continued to rise, but at a slower rate. b) After January 1, sales decreased faster than they ever grew. c) The rate of growth in daily sales doubled after January 1. d) After January 1, daily sales decreased at half the rate that they grew in December.

Jan. 1

Feb. 1

Daily sales (in thousands)

II $9 8 7 6 5 4 3 2 1 Dec. 1

Jan. 1

Feb. 1

2 2

4

x

−2

2 −2

−4

−4

4 2 −4

−4

−2

−2

1 y  x 2 2

4

4

x

y  3

y 1 y  x 4 2 2

y

2.

90. Discuss and relate the concepts of fixed cost, total cost, total revenue, and total profit.

$9 8 7 6 5 4 3 2 1

Jan. 1

Feb. 1

93. Graph some of the total-revenue, total-cost, and totalprofit functions in this exercise set using the same set of axes. Identify regions of profit and loss.

2

I

Dec. 1

92. Business: depreciation. A large crane is being depreciated according to the model V1t2 = 900 - 60t, where V1t2 is measured in thousands of dollars and t is the number of years since 2005. If the crane is to be depreciated until its value is $0, what is the domain of the depreciation model?

4

Daily sales (in thousands)

Feb. 1

$9 8 7 6 5 4 3 2 1

TECHNOLOGY CONNECTION

where t is the number of years after 1950. Thus, A102 is the median age of women at first marriage in 1950, A1502 is the median age in 2000, and so on.

Dec. 1

Jan. 1

Daily sales (in thousands)

87. Urban population. The population of Woodland is P. After a growth of 2%, its new population is N.

III Daily sales (in thousands)

b) Use the equation in part (a) to estimate the number of manatees counted in January 2010. c) The actual number counted in January 2010 was 5067. Does the equation found in part (a) give an accurate representation of the number of manatees counted each year?

x

−4

−2

2

−2

−2

−4

−4

4

x

7 1 , y-intercept: a0, - b 2 6 2 4. y = - x + 4 5. 7 3 6. (a) C1x2 = 20x + 20,000; R1x2 = 80x; P1x2 = R1x2 - C1x2 = 60x - 20,000 (b) 333 suits 3. m =

50

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Functions, Graphs, and Models

R.5 OBJECTIVES • Graph functions and solve applied problems.

Nonlinear Functions and Models There are many functions that have graphs that are not lines. In this section, we study some of these nonlinear functions that we will frequently encounter throughout this course.

Quadratic Functions

• Manipulate radical expressions and rational exponents.

DEFINITION A quadratic function f is given by

• Determine the domain of a rational function and graph certain rational functions. • Find the equilibrium point given a supply function and a demand function.

f1x2 = ax 2 + bx + c, where a Z 0. We have already used quadratic functions—for example, f1x2 = x 2 and g1x2 = x - 1. We can create hand-drawn graphs of quadratic functions using the following information. 2

The graph of a quadratic function f1x2 = ax 2 + bx + c is called a parabola. a) It is always a cup-shaped curve, like those in Examples 1 and 2 that follow. b) It opens upward if a 7 0 or opens downward if a 6 0. c) It has a turning point, or vertex, whose first coordinate is x = -

b . 2a

d) The vertical line x = - b>12a2 (not part of the graph) is the line of symmetry.

■ EXAMPLE 1 Graph: f1x2 = x 2 - 2x - 3. Solution Note that for f1x2 = 1x 2 - 2x - 3, we have a = 1, b = - 2, and

c = - 3. Since a 7 0, the graph opens upward. Let’s next find the vertex, or turning point. The x-coordinate of the vertex is x = = -

b 2a -2 = 1. 2112

Substituting 1 for x, we find the second coordinate of the vertex, f112: f112 = 12 - 2112 - 3 = 1 - 2 - 3 = - 4.

The vertex is 11,- 42. The vertical line x = 1 is the line of symmetry of the graph. We choose some x-values on each side of the vertex, compute y-values, plot the points, and graph the parabola.

R.5

f 1x2

x

y

Vertex

0

-3

y-intercept

3

4

5

-1

0

-2

5

2

(–1, 0)

-3 0

(4, 5)

5 4

-4

3

51

Nonlinear Functions and Models

(–2, 5)

1

2

•

(3, 0)

1

–3 – 2 –1

1

2

3

4

5

x

–1

f(x) = x 2 – 2x – 3

–2

(2, – 3)

–3

(0, – 3)

–4

(1, –4)

–5

x = 1 (the line of symmetry)

■ EXAMPLE 2 Graph: f1x2 = - 2x 2 + 10x - 7. TECHNOLOGY CONNECTION

EXERCISES Using the procedure of Examples 1 and 2, graph each of the following by hand, using the TABLE feature to create an input–output table for each function. Then press D to check your sketch. 1. f 1x2 = x 2 - 6x + 4

Solution We first note that a = - 2, and since a 6 0, the graph will open downward. Let’s next find the vertex, or turning point. The x-coordinate of the vertex is b 2a 10 5 = = . 21- 22 2

x = -

Substituting 52 for x in the equation, we find the second coordinate of the vertex: y = f A 52 B = - 2 A 52 B + 10 A 52 B - 7 2

2. f 1x2 = - 2x + 4x + 1

= - 2 A 25 4 B + 25 - 7 = 11 . 2

2

5 The vertex is A 52 ,11 2 B , and the line of symmetry is x = 2 . We choose some x-values on each side of the vertex, compute y-values, plot the points, and graph the parabola:

x

f 1x2

5 2

11 2

0

-7

1

1

2

5

3

Quick Check 1 Graph each function: a) f1x2 = x 2 + 2x - 3; b) f1x2 = - 2x2 - 10x - 5.

5

`

4

1

5

-7

y

⎛–5– , ⎝2

6

11 –––⎛ 2⎝

5

Vertex

4 3

f(x) = –2x 2 + 10x – 7

2 1 –6 –5 –4 – 3 –2 –1 –1

1

2

3

4

5

6

7

8

9 x

–2 –3

x = –5– 2

–4 –5 –6

O Quick Check 1 First coordinates of points at which a quadratic function intersects the x-axis (x-intercepts), if they exist, can be found by solving the quadratic equation ax2 + bx + c = 0. If real-number solutions exist, they can be found using the quadratic formula. See Appendix A at the end of the book for additional review of this important result.

O

52

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THEOREM 6

The Quadratic Formula

The solutions of any quadratic equation ax2 + bx + c = 0, a Z 0, are given by x =

- b ; 2b2 - 4ac . 2a

When solving a quadratic equation, ax2 + bx + c = 0, a Z 0, first try to factor and use the Principle of Zero Products (see Appendix A). When factoring is not possible or seems difficult, use the quadratic formula. It will always give the solutions. When b2 - 4ac 6 0, there are no real-number solutions and thus no x-intercepts. There are solutions in an expanded number system called the complex numbers. In this text, we will work only with real numbers.

■ EXAMPLE 3 Solve: 3x2 - 4x = 2. Solution We first find the standard form ax2 + bx + c = 0, and then determine a, b, and c: 3x2 - 4x - 2 = 0, a = 3, b = - 4, c = - 2. We then use the quadratic formula: x =

- b ; 2b2 - 4ac 2a - 1- 42 ; 21- 422 - 41321- 22

2#3 4 ; 216 + 24 4 ; 240 = = 6 6 # 4 ; 24 10 4 ; 2210 = = 6 6 2 A 2 ; 210 B = 2#3 =

=

Quick Check 2 Solve: 3x2 + 2x = 7.

Substituting Simplifying

2 ; 210 . 3

The solutions are A 2 + 210 B >3 and A 2 - 210 B >3, or approximately 1.721 and - 0.387.

O Quick Check 2 Algebraic–Graphical Connection Let’s make an algebraic–graphical connection between the solutions of a quadratic equation and the x-intercepts of a quadratic function. We just graphed equations of the form f1x2 = ax2 + bx + c, a Z 0. Let’s look at the graph of f1x2 = x2 + 6x + 8 and its x-intercepts, which follows.

O

R.5

TECHNOLOGY CONNECTION

f(x) = x2 + 6x + 8

EXERCISE f 1x2 = x2 - 6x + 8.

Using only the graph, find the solutions of x2 - 6x + 8 = 0.

7 6 5

1 1 2 x-intercepts: 3 (4, 0), (2, 0) 4 5

y = x 2 – 6x + 8 10

8

–8

b) Using only a graph, find the solutions of x2 + 3x - 10 = 0. c) Use the TABLE feature to check your answers to parts (a) and (b). d) Use a graph and a table to check the solutions of x2 + 6x + 8 = 0, solved algebraically on the right.

3

Nonlinear Functions and Models

53

y 5 4 3 2 1

1. a) Below is the graph of

–4

•

g(x) = 0 1 2 3

x

The x-intercepts, 1- 4, 02 and 1- 2, 02, are the points at which the graph crosses the x-axis. These pairs are also the points of intersection of the graphs of f1x2 = x 2 + 6x + 8 and g1x2 = 0 (the x-axis). The x-values, - 4 and - 2, can be found by solving f1x2 = g1x2: x2 + 6x + 8 1x + 421x + 22 x + 4 x

= = = =

0 0 Factoring; there is no need for the quadratic formula here. Principle of Zero Products 0 or x + 2 = 0 - 4 or x = - 2.

The solutions of x2 + 6x + 8 = 0 are - 4 and - 2, which are the first coordinates of the x-intercepts, 1- 4, 02 and 1- 2, 02, of the graph of f1x2 = x2 + 6x + 8. A brief review of factoring can be found in Appendix A at the end of the book.

Polynomial Functions Linear and quadratic functions are part of a general class of polynomial functions.

DEFINITION A polynomial function f is given by f1x2 = anxn + an - 1xn - 1 + Á + a2x2 + a1x1 + a0, where n is a nonnegative integer and an, an - 1, Á , a1, a0 are real numbers, called the coefficients.

The following are examples of polynomial functions: f1x2 f1x2 f1x2 f1x2

= = = =

- 5, 4x + 3, - x2 + 2x + 3, 2x 3 - 4x 2 + x + 1.

(A constant function) (A linear function) (A quadratic function) (A cubic, or third-degree, function)

In general, creating graphs of polynomial functions other than linear and quadratic functions is difficult without a calculator. We use calculus to sketch such graphs in Chapter 2. Some power functions, of the form f1x2 = ax n, are relatively easy to graph.

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■ EXAMPLE 4 Using the same set of axes, graph f1x2 = x 2 and g1x2 = x 3. Solution We set up a table of values, plot the points, and then draw the graphs. y

x

x2

x3

-2

4

-8

-1

1

-1

- 12

1 4

- 18

0

0

0

1 2

1 4

1 8

⎛– –1– , – –1–⎛ ⎝ 2 8⎝

1

1

1

(–1, –1)

(– 2, 4)

3

f(x) = x2

2

(–1, 1) ⎛– –1– , –1–⎛ ⎝ 2 4⎝

1

–2

Quick Check 3

2

O

Graph each function using the same set of axes: f1x2 = 4 - x2 and g1x2 = x3 - 1.

4

(2, 4)

4

–1

– –1– 2

–1

⎛–1– , –1–⎛ ⎝ 2 4⎝ (1, 1)

1 –– 2

1

⎛–1– , –1–⎛ ⎝ 2 8⎝

2

x

(0, 0)

–2

g(x) = x3

8

–3

O Quick Check 3

TECHNOLOGY CONNECTION 8

Solving Polynomial Equations The INTERSECT Feature Consider solving the equation

–3

x = 3x + 1. 3

3 Intersection X = –1.532089

Doing so amounts to finding the x-coordinates of the point(s) of intersection of the graphs of f 1x2 = x 3 and g1x2 = 3x + 1. We enter the functions as y1 = x 3 and y2 = 3x + 1

and then graph. We use a 3- 3, 3, - 5, 84 window to see the curvature and possible points of intersection. y1 = x3, y2 = 3x + 1 8

–3

y1 y2

3

–5

There appear to be at least three points of intersection. Using the INTERSECT feature in the CALC menu, we see that the point of intersection on the left is about 1- 1.53, - 3.602.

Y = –3.596267 –5

In a similar manner, we find the other points of intersection to be about 1- 0.35, - 0.042 and 11.88, 6.642. The solutions of x 3 = 3x + 1 are the x-coordinates of these points, approximately - 1.53, - 0.35, and 1.88.

The ZERO Feature A ZERO, or ROOT, feature can be used to solve an equation. The word “zero” in this context refers to an input, or x-value, for which the output of a function is 0. That is, c is a zero of the function f if f 1c2 = 0. To use such a feature requires a 0 on one side of the equation. Thus, to solve x 3 = 3x + 1, we obtain x 3 - 3x - 1 = 0 by subtracting 3x + 1 from both sides. Graphing y = x 3 - 3x - 1 and using the ZERO feature to find the zero on the left, we view a screen like the following.

(continued)

R.5

•

Nonlinear Functions and Models

55

Using the ZERO feature, solve each equation.

y = x 3 – 3x – 1

8

5. 0.4x 2 = 280x

(Hint: Use 3- 200, 800,- 100,000, 200,0004.)

–3

3

6.

1 3 3x

- 12 x 2 = 2x - 1

7. x 2 = 0.1x 4 + 0.4

Zero X = –1.532089

8. 0 = 2x 4 - 4x 2 + 2

–5

We see that x 3 - 3x - 1 = 0 when x L - 1.53, so - 1.53 is an approximate solution of the equation x 3 = 3x + 1. Proceeding in a similar manner, we can approximate the other solutions as - 0.35 and 1.88. Note that the points of intersection of the graphs of f and g have the same x-values as the zeros of x 3 - 3x - 1.

Find the zeros of each function. 9. f 1x2 = 3x 2 - 4x - 2

10. f 1x2 = - x 3 + 6x 2 + 5 11. g1x2 = x 4 + x 3 - 4x 2 - 2x + 4 12. g1x2 = - x 4 + x 3 + 11x 2 - 9x - 18

EXERCISES Using the INTERSECT feature, solve each equation. 1. x 2 = 10 - 3x

2. 2x + 24 = x 2

3. x 3 = 3x - 2

4. x 4 - 2x 2 = 0

TECHNOLOGY CONNECTION

Apps for the iPhone and iPod Touch The advent of the iPhone and other sophisticated mobile phones has made available many inexpensive mathematics applications. Two useful apps for the iPhone and iPod Touch are iPlot and Graphicus (which can be purchased at the iTunes Store). Each has more visually appealing displays than standard graphing calculators, but each also has limited capability. For example, neither iPlot nor Graphicus can do regression, as described in Section R.6.

iPlot Among the features of this app is the ability to graph most of the functions we encounter in this book. There is a Zoom feature, a Trace feature that can be used to find func-

FIGURE 1

FIGURE 2

tion values, roots (zeros), and points of intersection of graphs, and a separate Root feature for finding zeros. Let’s consider the function f 1x2 = 2x 3 - x 4 as an example. To graph and find roots, first open the iPlot app. You will see a screen like that in Fig. 1. Notice the four icons at the bottom. The Functions icon is highlighted. Press + in the upper right; then enter the function as 2*x^3-x^4 (Fig. 2). Slide to the bottom and change the graph color if desired. Next, press Done in the upper right, followed by the Plot icon to obtain the graph (Fig. 3). Below the graph are buttons for various options. To find a root of the function, press Root (firmly) so that it becomes highlighted. Then move the cursor close to a potential root (Fig. 4). Sometimes it is difficult to position the cursor directly on the root. But if you press the Apply icon in the

FIGURE 3 (continued)

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CHAPTER R

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Functions, Graphs, and Models

Apps for the iPhone and iPod Touch (continued)

set of axes. Try graphing g1x2 = x 3 - 1 along with f 1x2 = 2x 3 - x 4, and then pressing Trace to find points of intersection. Sometimes iPlot will crash or lock up, giving an error message like “Unexpected End of Formula.” If this occurs, go to the line where the function formula occurs, then press Edit function at the top to delete the function and enter it again. For more information, consult the iPlot page at the iTunes Store or visit www.posimotion.com.

EXERCISES Using iPlot, repeat Exercises 1–12 in the Technology Connection on page 55.

Graphicus FIGURE 4

lower right once you are close, the cursor will jump to the answer, in this case, 2 (Fig. 5).

FIGURE 5

You will likely find it necessary to experiment with iPlot. Press some of the options buttons to explore other features. Use the Settings icon to change the window settings and make other modifications to the appearance. Missing symbols can be found by clicking on #+= at the left of the keypad. Piecewise-defined functions can be entered, but an inappropriate vertical line may show up on the graph. The graph x2 - 9 of a function like f 1x2 = will not show the hole at x + 3 x = - 3. iPlot can graph more than one function on the same

FIGURE 6

FIGURE 7

Graphicus is also very appealing visually and has the ability to graph most of the functions we encounter in this book. It can find roots and intersections, and it excels at many aspects of calculus, as we’ll see later in this book. Let’s look again at the function f 1x2 = 2x 3 - x 4. To graph and find roots, first open Graphicus. Touch the blank rectangle at the top of the screen and enter the function as y(x)=2x^3-x^4. Press + in the upper right, and you will see the graph in Fig. 6. Notice the seven icons at the bottom. The one at the far left is for zooming. The fourth icon from the left is for finding roots. Touch it and note how quickly the roots of the function are highlighted (Fig. 7). Touch the symbol marking each root, and a box appears with its value; see Figs. 8 and 9. Piecewise-defined functions cannot be entered. The x2 - 9 , will graph of a function with a hole, such as f 1x2 = x + 3 not show the hole at x = - 3. Graphicus can graph more than one function on the same set of axes. Graph g1x2 = x 3 - 1 along with f 1x2 = 2x 3 - x 4, and then press the fourth icon to find the points of intersection. Press some of the other icons to explore the uses of Graphicus. For more information, consult the Graphicus page at the iTunes Store or visit www.facebook.com/pages/Graphicus/ 189699869029.

EXERCISES Using Graphicus, repeat Exercises 1–12 in the Technology Connection on p. 55.

FIGURE 8

FIGURE 9

R.5

•

Nonlinear Functions and Models

57

Rational Functions DEFINITION Functions given by the quotient, or ratio, of two polynomials are called rational functions.

The following are examples of rational functions: x2 - 9 , x - 3 3x 2 - 4x g1x2 = , 2x + 10 f1x2 =

x - 3 , x - x - 2 x 3 - 2x + 7 k1x2 = = x 3 - 2x + 7. 1 h1x2 =

2

Note that as the function k illustrates, every polynomial function is also a rational function. The domain of a rational function is restricted to those input values that do not result in division by zero. Thus, for f above, the domain consists of all real numbers except 3. To determine the domain of h, we set the denominator equal to 0 and solve: x2 - x - 2 = 0 1x + 121x - 22 = 0 x = - 1 or x = 2. Therefore, - 1 and 2 are not in the domain. The domain of h consists of all real numbers except - 1 and 2. The numbers - 1 and 2 “split,” or “separate,” the intervals in the domain. The graphing of most rational functions is rather complicated and is best dealt with using the tools of calculus that we will develop in Chapters 1 and 2. For now we will focus on graphs that are fairly basic and leave the more complicated graphs for Chapter 2.

■ EXAMPLE 5 Graph: f1x2 =

x2 - 9 . x - 3

Solution This particular function can be simplified before we graph it. We do so by factoring the numerator and removing a factor of 1 as follows: f1x2 = =

x2 - 9 x - 3 1x - 321x + 32

x - 3 x - 3 #x + 3 = x - 3 1 = x + 3, x Z 3.

Note that 3 is not in the domain of f.

We must specify x Z 3.

This simplification assumes that x is not 3. By writing x Z 3, we indicate that for any x-value other than 3, the equation f1x2 = x + 3 is used: f1x2 = x + 3, x Z 3.

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TECHNOLOGY CONNECTION If y1 = 1x 2 - 92>1x - 32 and y2 = x + 3 are both graphed, the two graphs appear indistinguishable. To see this, graph both lines and use the arrow keys, h and e, and the TRACE feature to move the cursor from line to line.

To find function values, we substitute any value for x other than 3. We make calculations as in the following table and draw the graph. The open circle at 13, 62 indicates that this point is not part of the graph. y

EXERCISES 1. Compare the results of using

x

f1x2

-3

0

-2

1

-1

2

0

3

1

4

2

5

4

7

7 6 5 4 3 2 1

TRACE and entering the value 3.

2. Use the TABLE feature with

TblStart set at - 1 and ¢ Tbl set at 1. How do y1 and y2 differ in the resulting table of values?

–7 –6 –5 –4 – 3 –2 –1 –1 –2 2 – 9 –3 f(x) = –x–—— x – 3 –4 –5 –6 –7

(3, 6)

1 2 3 4 5 6 7

x

One important class of rational functions is given by f1x2 = k>x, where k is a constant.

■ EXAMPLE 6 Graph: f1x2 = 1>x. Solution We make a table of values, plot the points, and then draw the graph. y

x

f1x2

4

-3

- 13

3

-2

- 12

-1

-1

⎛–1– , 4⎛ ⎝4 ⎝ f(x) = –1x– ⎛–1– , 2⎛ ⎝2 ⎝

2

(1, 1) 1

- 12

-2

- 14

-4

1 4

4

1 2

2

Quick Check 4

1

1

Graph each function:

2

1 2

3

1 3

x - 9 a) f1x2 = ; x + 3 1 b) f1x2 = - . x 2

O Quick Check 4

–3

⎛– 3, – –1–⎛ 3⎝ ⎝ ⎛– 2, – –1–⎛ 2⎝ ⎝

–2

–1 – –1– 2

(–1, –1) ⎛– –1– , – 2⎛ ⎝ 2 ⎝

1 –– 2

–1

–2

–3

⎛– –1– , – 4⎛ ⎝ 4 ⎝

–4

1

⎛2, –1–⎛ ⎝ 2⎝ ⎛3, –1–⎛ ⎝ 3⎝ 2

3 x

O

R.5

•

Nonlinear Functions and Models

59

TECHNOLOGY CONNECTION

Graphs of Rational Functions

Here use of CONNECTED mode can lead to an incorrect graph. Because, in CONNECTED mode, points are joined with line segments, both branches of the graph are connected, making it appear as though the vertical line x = 3 is part of the graph. On the other hand, in DOT mode, the calculator simply plots dots representing coordinates of points. When graphing rational functions, it is usually best to use DOT mode.

Consider two graphs of the function given by f 1x2 =

2x + 1 . x - 3

CONNECTED mode: 8

EXERCISES Graph each of the following using DOT mode. –10

15

1. f 1x2 = 3. f 1x2 =

–8

DOT mode:

5. f 1x2 =

8

–10

7. f 1x2 =

15

4 x - 2 x2 - 1 x + x - 6 2

10 x2 + 4 2x + 3 3x 2 + 7x - 6

2. f 1x2 = 4. f 1x2 = 6. f 1x2 = 8. f 1x2 =

x x + 2 x2 - 4 x - 1 8 x2 - 4 2x 3 x2 + 1

–8

In Example 6, note that 0 is not in the domain of f because it would yield a denominator of zero.The function is decreasing over the intervals 1- q , 02 and 10, q 2. The function f1x2 = 1>x is an example of inverse variation.

DEFINITION y varies inversely as x if there is some positive number k such that y = k>x. We also say that y is inversely proportional to x.

■ EXAMPLE 7 Business: Stocks and Gold. Certain economists theorize that stock prices are inversely proportional to the price of gold. That is, when the price of gold goes up, the prices of stocks go down; and when the price of gold goes down, the prices of stocks go up. Let’s assume that the Dow Jones Industrial Average D, an index of the overall prices of stocks, is inversely proportional to the price of gold G, in dollars per ounce. One day the Dow Jones was 10,619.70 and the price of gold was $1129.60 per ounce. What will the Dow Jones Industrial Average be if the price of gold rises to $1400?

Solution We assume that D = k>G, so 10,619.7 = k>1129.6 and k = 11,996,013.12. Thus, D =

11,996,013.12 . G

We substitute 1400 for G and compute D: D =

11,996,013.12 L 8568.6. 1400

CHAPTER R

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Functions, Graphs, and Models

Dow Jones Industrial Average

60

Quick Check 5

O

Business: Stocks and Gold. Suppose in Example 7 that the price of gold drops to $1000 per ounce. Predict what happens to the Dow Jones Industrial Average. Then compute the value of D.

D 100,000 90,000 80,000 70,000 60,000 50,000 40,000 30,000 20,000 10,000 0

200 400 600 800 1000 1200 1400 G Price of gold (in dollars per ounce)

Warning! Do not put too much “stock” in the equation of this example. It is meant only to give an idea of economic relationships. An equation for predicting the stock market accurately has not been found!

O Quick Check 5 Absolute-Value Functions The absolute value of a number is its distance from 0 on the number line. We denote the absolute value of a number x as | x | . The absolute-value function, given by f1x2 = | x | , is very important in calculus, and its graph has a distinctive V shape.

■ EXAMPLE 8 Graph: f1x2 = | x | . TECHNOLOGY CONNECTION Absolute value is entered as abs and is often accessed from the NUM option of the MATH menu.

f1x2 3

1 2x

ƒ

2. g1x2 = | x - 4 |

-2

2

-1

1

0

0

1

1

2

2

3

3

f(x) = |x|

3

(– 3, 3) (– 2, 2) (–1, 1)

Graph.

ƒ

y

x -3

EXERCISES 1. f 1x2 =

Solution We make a table of values, plot the points, and then draw the graph.

–4 –3 – 2 –1

(3, 3)

2

(2, 2)

1

(1, 1) 1

2

3

4

x

2

We can think of this function as being defined piecewise by considering the definition of absolute value: f1x2 = | x | = e

x, if x Ú 0, - x, if x 6 0.

R.5

•

61

Nonlinear Functions and Models

Square-Root Functions The following is an example of a square-root function and its graph.

■ EXAMPLE 9 Graph: f1x2 = - 2x. Solution The domain of this function is the set of all nonnegative numbers—the interval 30, q 2. You can find approximate values of square roots on your calculator. We set up a table of values, plot the points, and then draw the graph. y

x f1x2 = - 1x

0 0

1 -1

2 - 1.4

3

4

- 1.7

5

-2

1

- 2.2

3

4

5 x

(1, –1) (2, –1.4) (3, –1.7) (4, –2) (5, – 2.2) f(x) = – √x

–1 –2

Quick Check 6 –3

Graph each function: a) f1x2 = ƒ x + 2 ƒ ; b) f1x2 = 2x + 4.

2

(0, 0)

O Quick Check 6 Power Functions with Rational Exponents We are motivated to define rational exponents so that the following laws of exponents still hold (also see Appendix A): For any nonzero real number a and any integers n and m, a n # a m = a n + m;

an 1 # = a n - m; 1a n2m = a n m; a -m = m . am a

#

This suggests that a 1/2 be defined so that 1a 1/222 = a 11/22 2 = a 1. Thus, we define a 1/2 as # 3 2a. Similarly, in order to have 1a 1/323 = a 11/32 3 = a 1, we define a 1/3 as 2a. In general, n

n

a 1>n = 2a, provided 2a is defined. n

Again, for the laws of exponents to hold, we have, assuming that 2a exists, a m/n = 1a m21/n = 2a m = 12a2m, n

n

and a -m/n is defined by a -m/n =

1 a

m/n

=

1 n

2a m

.

■ EXAMPLE 10 Rewrite each of the following as an equivalent expression with rational exponents: a) 2 4x

b) 2 3 r2

c) 2x 10, for x Ú 0

d)

1 2 3 b5

Solution a) 2 4 x = x 1/4

b) 2 3 r 2 = r 2>3

c) 2x 10 = x 10/2 = x 5, x Ú 0

d)

1 2 3b

5

=

1 = b -5>3 b 5>3

O

62

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■ EXAMPLE 11 Rewrite each of the following as an equivalent expression using radical notation: a) x 1>3; b) t 6>7; c) x -2>3; d) r -1>4.

Solution a) x 1>3 = 2 3 x c) x -2>3 =

1 x

2>3

b) t 6>7 = 2 7 t6 =

1

d) r -1>4 =

2 3 x2

1 r

1>4

=

1 2 4r

■ EXAMPLE 12 Simplify: a) 85>3; b) 813>4. Solution

a) 85/3 = 181/325 = A 2 3 8 B = 2 5 = 32 3 b) 813/4 = 1811/423 = A 2 4 81 B = 33 = 27 5

Because even roots (square roots, fourth roots, sixth roots, and so on) of negative numbers are not real numbers, the domain of a radical function may have restrictions.

■ EXAMPLE 13 Find the domain of the function given by f1x2 = 2 4 2x - 10.

Solution For f1x2 to be a real number, 2x - 10 cannot be negative. Thus, to find the domain of f, we solve the inequality 2x - 10 Ú 0: 2x - 10 Ú 0 2x Ú 10 x Ú 5.

Quick Check 7

O

Find the domain of the function given by f1x2 = 2x + 3.

Adding 10 to both sides Dividing both sides by 2

The domain of f is 5x | x Ú 56, or, in interval notation, [5, q 2.

O Quick Check 7 Power functions of the form f1x2 = ax k, with k a fraction, occur in many applications.

■ EXAMPLE 14 Life Science: Home Range. The home range of an animal is defined as the region to which the animal confines its movements. It has been shown that for carnivorous (meat-eating) mammals the area of that region can be approximated by the function H1w2 = 0.11w 1.36, where w is the mass of the animal, in grams, and H1w2 is the area of the home range, in hectares. Graph the function. (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. 200 (Reading, MA: Addison-Wesley, 1973), and Harestad, A. S., and Bunnel, F. L., “Home Range and Body Weight—A Reevaluation,” Ecology, Vol. 60, No. 2 (April, 1979), pp. 389–402.)

Solution We can approximate function values using a power key, usually labeled ^ or

A lynx in its territorial area.

yx

100 w 136. . Note that w 1.36 = w 136>100 = 2 w

0

700

1400

H1w2

0

814.2

2089.9

2100

2800

3500

3627.5

5364.5

7266.5

`

R.5

•

63

Nonlinear Functions and Models

The graph is shown below. Note that the function values increase from left to right. As body weight increases, the area over which the animal moves increases.

Home range (in hectares)

H(w) 7500 6000 4500

H(w) = 0.11w1.36

3000 1500

700

1400 2100 2800 3500

w

Mass (in grams)

Supply and Demand Functions Supply and demand in economics are modeled by increasing and decreasing functions.

Demand Functions The table and graph below show the relationship between the price x per bag of sugar and the quantity q of 5-lb bags that consumers will demand at that price. q

Demand Schedule

$5 4 3 2 1

4 5 7 10 15

25 Quantity (in millions)

Price, x, per 5-lb Bag

Quantity, q, of 5-lb Bags (in millions)

20 Demand function: q = D(x) 15 10 5

$1

$2

$3

$4

$5

x

Price

Note that the quantity consumers demand is inversely proportional to the price. As the price goes up, the quantity demanded goes down.

Supply Functions The next table and graph show the relationship between the price x per bag of sugar and the quantity q of 5-lb bags that sellers are willing to supply, or sell, at that price.

64

CHAPTER R

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Functions, Graphs, and Models q

Supply Schedule

$1 2 3 4 5

0 10 16 20 22

25 Quantity (in millions)

Price, x, per 5-lb Bag

Quantity, q, of 5-lb Bags (in millions)

20 15 Supply function: q = S(x) 10 5

$1

$2

$3

$4

$5

x

Price

Note that suppliers are willing to supply greater quantities at higher prices than they are at lower prices. Let’s now look at these curves together. As price increases, supply increases and demand decreases; and as price decreases, demand increases but supply decreases. The point of intersection 1x E, qE2 is called the equilibrium point. The equilibrium price x E (in this case, $2 per bag) corresponds to an equilibrium quantity qE (in this case, 10 million bags). Sellers are willing to sell 10 million bags at $2> bag, and consumers are willing to buy 10 million bags at that price. The situation is analogous to a buyer and seller haggling over the sale of an item. The equilibrium point, or selling price, is what they finally agree on. q 25

S(x)

Quantity (in millions)

20 D(x) 15 Equilibrium point: (xE, qE)

qE = 10 5

$1

xE = 2

$3

$4

$5

x

Price

■ EXAMPLE 15 Economics: Equilibrium Point. Find the equilibrium point for the demand and supply functions for the Ultra-Fine coffee maker. Here q represents the number of coffee makers produced, in hundreds, and x is the price, in dollars. TECHNOLOGY CONNECTION

Demand: q = 50 -

EXERCISE

Supply:

1. Use the INTERSECT feature to

find the equilibrium point for the following demand and supply functions. Demand: q = 1123.6 - 61.4x Supply:

q = 201.8 + 4.6x

1 x 4

q = x - 25

Solution To find the equilibrium point, the quantity demanded must match the quantity produced: 50 -

1 x = x - 25 4

50 + 25 = x +

1 x 4

Adding 25 +

1 x to each side 4

Exercise Set R.5

75 = 75 #

5 x 4

4 = x 5 60 = x.

Quick Check 8

65

Multiplying both sides by

4 5

Economics: Equilibrium Point. Repeat Example 15 for the following functions:

Thus, x E = 60. To find qE, we substitute x E into either function. We select the supply function:

1 Demand: q = 70 - x 5 Supply: q = x - 20

Thus, the equilibrium quantity is 3500 units, and the equilibrium point is 1$60, 35002.

qE = x E - 25 = 60 - 25 = 35.

O Quick Check 8

O

Section Summary • Many types of functions have graphs that are not straight lines; among these are quadratic functions, polynomial functions, power functions, rational functions, absolute-value functions, and square-root functions.

• Demand is modeled by a decreasing function. Supply is modeled by an increasing function. The point of intersection of graphs of demand and supply functions for the same product is called the equilibrium point.

EXERCISE SET

R.5 Graph each pair of equations on one set of axes.

Graph. 17. y = x 2 - 4x + 3

18. y = x 2 - 6x + 5

19. y = - x 2 + 2x - 1

20. y = - x 2 - x + 6

3. y = x 2 and y = x 2 - 1

21. f 1x2 = 2x 2 - 6x + 1

22. f 1x2 = 3x 2 - 6x + 4

4. y = x 2 and y = x 2 - 3

23. g1x2 = - 3x 2 - 4x + 5

1. y =

1 2 2x

and y =

- 12 x 2

2. y = 14 x 2 and y = - 14 x 2

5. y = - 2x 2 and y = - 2x 2 + 1 6. y = - 3x and y = - 3x + 2 2

7. y = ƒ x ƒ and y = ƒ x - 3 ƒ 8. y = ƒ x ƒ and y = ƒ x - 1 ƒ

25. y =

2 x

27. y = -

9. y = x and y = x + 2 3

24. g1x2 = - 2x 2 - 3x + 7

2

3

10. y = x 3 and y = x 3 + 1

29. y =

11. y = 2x and y = 2x - 1 12. y = 2x and y = 2x - 2

2 x

1 x2

13. f 1x2 = x 2 + 4x - 7

14. f 1x2 = x 3 - 2x + 3

15. g1x2 = 2x 4 - 4x 2 - 3

16. g1x2 = 3x 2 - 6x

3 x

28. y =

-3 x

30. y =

1 x - 1

32. y =

31. y = 2 3x

For each of the following, state whether the graph of the function is a parabola. If the graph is a parabola, find the parabola’s vertex.

26. y =

1 ƒxƒ

33. f 1x2 =

x + 5x + 6 x + 3

34. g1x2 =

x 2 + 7x + 10 x + 2

35. f 1x2 =

x2 - 1 x - 1

36. g1x2 =

x 2 - 25 x - 5

2

66

CHAPTER R

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Functions, Graphs, and Models

APPLICATIONS

Solve. 37. x - 2x = 2

38. x - 2x + 1 = 5

39. x + 6x = 1

40. x 2 + 4x = 3

41. 4x = 4x + 1

42. - 4x = 4x - 1

43. 3y + 8y + 2 = 0

44. 2p 2 - 5p = 1

2

2

2

2

2

2

87. Demand: q = 1000 - 10x; Supply: q = 250 + 5x q

9 = 0 (Hint: Multiply both sides by x.) x

1 1 = 2 w w

46. 1 -

Find the equilibrium point for each pair of demand and supply functions.

1000

600 400

Rewrite each of the following as an equivalent expression with rational exponents. 47. 2x 3

48. 2x 5

49. 2 5 a3

50. 2 4 b 2,

51. 2 7t

52. 2 8c

53. 2 4 x 12, x Ú 0

54. 2 3 t6

57.

56.

2 t5 1

200 20 40 60 80 100 x Price (in dollars)

b Ú 0

88. Demand: q = 8800 - 30x; Supply: q = 7000 + 15x q 10,000

1

Quantity

55.

1

2m 4

58. 2x + 4 3

2x 2 + 7

59. x 1>5

60. t 1>7

61. y 2>3

62. t 2>5

63. t -2>5

64. y -2>3

65. b -1>3

66. b -1>5

67. e -17>6

68. m -19>6

69. 1x 2 - 32-1>2

70. 1 y 2 + 72-1>4

71.

72.

t 2>3

4,000

20 40 60 80 x Price (in dollars)

89. Demand: q =

x 5 ; Supply: q = x 5

1

Quantity (in thousands)

q

3

2>3

3>4

5>2

78. 25

Determine the domain of each function. 80. f 1x2 =

81. f 1x2 =

x3

82. f 1x2 =

x - 5x + 6 2

83. f 1x2 = 25x + 4 85. f 1x2 = 2 4 7 - x

x2 - 4 x + 2 x4 + 7 x + 6x + 5 2

84. f 1x2 = 22x - 6 86. f 1x2 = 2 65 - x

90. Demand: q = Quantity (in thousands)

75. 64 2>3

x 2 - 25 x - 5

S

2 1 2 4 6 8 Price (in hundreds of dollars)

74. 165>2

79. f 1x2 =

D

4

w -4>5

73. 93>2

77. 16

S

6,000

Simplify.

76. 8

D

8,000

2,000

Rewrite each of the following as an equivalent expression using radical notation.

1

S

D

800

Quantity

45. x + 7 +

Business and Economics

x

4 x ; Supply: q = x 4 q

D

8 6 4

S

2 2 4 6 8 10 x Price (in hundreds of dollars)

Exercise Set R.5

Quantity (in hundreds)

91. Demand: q = 1x - 322; Supply: q = x 2 + 2x + 1 (assume x … 3) q

96. Demand. The quantity sold x of a plasma television is inversely proportional to the price p. If 85,000 plasma TVs sold for $2900 each, how many will be sold if the price is $850 each?

S

6

Life and Physical Sciences

4

97. Radar range. The function given by

D

R1x2 = 11.74x 0.25

2 1 2 3 4 Price (in dollars)

can be used to approximate the maximum range, R1x2, in miles, of an ARSR-3 surveillance radar with a peak power of x watts. a) Determine the maximum radar range when the peak power is 40,000 watts, 50,000 watts, and 60,000 watts. b) Graph the function.

x

92. Demand: q = 1x - 422; Supply: q = x 2 + 2x + 6 (assume x … 4) q Quantity (in hundreds)

S

98. Home range. Refer to Example 14. The home range, in hectares, of an omnivorous mammal (one that eats both plants and meat) of mass w grams is given by

10 8

H1w2 = 0.059w 0.92.

6

D

(Source: Harestad, A. S., and Bunnel, F. L., “Home Range and Body Weight—A Reevaluation,” Ecology, Vol. 60, No. 2 (April, 1979), pp. 389–402.) Complete the table of approximate function values and graph the function.

4 2 1 2 3 4 Price (in dollars)

x

Quantity (in thousands)

93. Demand: q = 5 - x; Supply: q = 2x + 7 q 5 4 3 2 1

D S

94. Demand: q = 7 - x; Supply: q = 22x + 1 q Quantity (in thousands)

w

0

1000

H1w2

0

34.0

2000

3000

4000

5000

6000

7000

99. Life science: pollution control. Pollution control has become a very important concern in all countries. If controls are not put in place, it has been predicted that the function P = 1000t 5>4 + 14,000

1 2 3 4 5 6 7 x Price (in thousands of dollars)

will describe the average pollution, in particles of pollution per cubic centimeter, in most cities at time t, in years, where t = 0 corresponds to 1970 and t = 35 corresponds to 2005. a) Predict the pollution in 2005, 2008, and 2014. b) Graph the function over the interval [0, 50]. 100. Surface area and mass. The surface area of a person whose mass is 75 kg can be approximated by the function

8 S 6

67

D

4 2 1 2 3 4 5 6 7 8 9x Price (in thousands of dollars)

95. Stock prices and prime rate. It is theorized that the price per share of a stock is inversely proportional to the prime (interest) rate. In January 2010, the price per share S of Apple Inc. stock was $205.93 and the prime rate R was 3.25%. The prime rate rose to 4.75% in March 2010. (Source: finance.yahoo.com and Federal Reserve Board.) What was the price per share in March 2010 if the assumption of inverse proportionality is correct?

f 1h2 = 0.144h1/2,

where f 1h2 is measured in square meters and h is the person’s height in centimeters. (Source: U.S. Oncology.) a) Find the approximate surface area of a person whose mass is 75 kg and whose height is 180 cm. b) Find the approximate surface area of a person whose mass is 75 kg and whose height is 170 cm. c) Graph the function f for 0 … h … 200.

SYNTHESIS 101. Zipf’s Law. According to Zipf’s Law, the number of cities N with a population greater than S is inversely proportional to S. In 2008, there were 52 U.S. cities with a population greater than 350,000. Estimate the number of U.S. cities with a population between 350,000 and 500,000; between 300,000 and 600,000.

68

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102. At most, how many y-intercepts can a function have? Explain. 103. Explain the difference between a rational function and a polynomial function. Is every polynomial function a rational function? Why or why not?

Answers to Quick Checks y

1. (a) −3

Use the ZERO feature or the INTERSECT feature to approximate the zeros of each function to three decimal places. 104. f 1x2 = x 3 - x (Also, use algebra to find the zeros of this function.) 105. f 1x2 = 2x 3 - x 2 - 14x - 10

107. f 1x2 = x 4 + 4x 3 - 36x 2 - 160x + 300 108. f 1x2 = 27 - x 2 - 1

6

−1

4

−2

2 −4

−4 f (x)  x 2  2x  3

2.

−3 −2 −1

x −2

- 1 ; 222 , or 1.230 and - 1.897 3 y

3.

4

− 0.5 −2

f (x)  4  x 2

0.5 1.0 1.5 x g (x)  x 3  1

−4

y 4

4. (a)

y

(b)

4

2

109. f 1x2 = | x + 1 | + | x - 2 | - 5

2 −4

−2

2 −2

110. f 1x2 = | x + 1 | + | x - 2 |

−4

111. f 1x2 = | x + 1 | + | x - 2 | - 3

4

x

x2  9 f (x)  x3

−4

−2

1 f (x) ⴝ ⴚⴚ x 2

4

x

−2 −4

−6

112. f 1x2 = x 8 + 8x 7 - 28x 6 - 56x 5 + 70x 4 + 56x 3 - 28x 2 - 8x + 1

5. Dow Jones Industrial Average increases; D = 11,996.01. y y 6. (a) (b) f(x)  | x  2 |

113. Find the equilibrium point for the following demand and supply functions, where q is the quantity, in thousands of units, and x is the price per unit, in dollars.

8

4

6

2

4

−4

−4 −2

2

4

x

−2

2 −6

−2

f(x) ⴝ √ x ⴙ 4

2

x

−4

7. 5x | x Ú - 36 8. Equilibrium point is ($75, 5500); price is $75 and quantity is 5500.

Demand: q = 83 - x x2 Supply: q = - 1.9 576

• Use curve fitting to find a mathematical model for a set of data and use the model to make predictions.

f (x)  2x 2  10x  5 y

x

2

106. f 1x2 = 12 1| x - 4 | + | x - 7 |2 - 4

OBJECTIVE

1

−3

TECHNOLOGY CONNECTION

R.6

(b)

−2 −1

Mathematical Modeling and Curve Fitting Fitting Functions to Data We have developed a library of functions that can serve as models for many applications. Although others will be introduced later, let’s look at those that we have considered. (Cubic and quartic functions are covered in detail in Chapter 2, but we show them for reference.) We will not consider rational functions in this section. Linear function: f(x) = mx + b

Quadratic function: f(x) = ax2 + bx + c, a > 0

Quadratic function: f(x) = ax2 + bx + c, a < 0

Absolute-value function: f(x) = |x|

Cubic function: f(x) = ax3 + bx2 + cx + d, a > 0

Quartic function: f(x) = ax 4 + bx3 + cx2 + dx + e, a > 0

R.6

•

69

Mathematical Modeling and Curve Fitting

Now let’s consider some real-world data. How can we decide which, if any, type of function might fit the data? One simple way is to examine a graph of the data called a scatterplot. Then we look for a pattern resembling one of the graphs on p. 68. For example, data might be modeled by a linear function if the graph resembles a straight line. The data might be modeled by a quadratic function if the graph rises and then falls, or falls and then rises, in a curved manner resembling a parabola. Let’s now use our library of functions to see which, if any, might fit certain data sets.

■ EXAMPLE 1 Choosing Models. For the scatterplots and graphs below, determine which, if any, of the following functions might be used as a model for the data.

8 4 2

4 6 Year

8

d)

x

c)

8 4 0 0

5 Year

10 x

DRIVER FATALITIES BY AGE

35 30 25 20 15 10 5 0

16–20 21–34 35–44 45–64 65–74 Ages

75+

(Source: National Highway Traffic Administration.)

e)

ACTUAL AND PROJECTED ANNUAL SUIT SALES FOR RAGGS LTD. 600 500 400 300 200 100 0

2008

2009

2010

2011 2012 Year

y 10 5 0 0

4

2 Year

Number of licensed drivers per 100,000 who died in motor vehicle accidents in 2002 Number of driver fatalities per 100,000 drivers

0 0

y

Population (in millions)

b)

Population (in millions)

y

Number of suits sold

a)

Population (in millions)

Linear, f1x2 = mx + b Quadratic, f1x2 = ax 2 + bx + c, a 7 0 Quadratic, f1x2 = ax 2 + bx + c, a 6 0 Polynomial, neither quadratic nor linear

2013

2014

6 x

Functions, Graphs, and Models

a)

Population (in millions)

Solution y 8 4 0 0

2

4 6 Year

8

x

The data rise and then fall in a curved manner fitting a quadratic function, f1x2 = ax 2 + bx + c, a 6 0. b)

Population (in millions)

•

y 8 4 0 0

5 Year

10 x

The data seem to fit a linear function, f1x2 = mx + b. c)

Population (in millions)

CHAPTER R

y 10 5 0 0

4

2

6 x

Year

The data rise in a manner fitting the right-hand side of a quadratic function, f1x2 = ax 2 + bx + c, a 7 0. d)

DRIVER FATALITIES BY AGE Number of licensed drivers per 100,000 who died in motor vehicle accidents in 2002 Number of driver fatalities per 100,000 drivers

70

35 30 25 20 15 10 5 0

16–20 21–34 35–44 45–64 65–74 Ages

75+

(Source: National Highway Traffic Administration.)

The data fall and then rise in a curved manner fitting a quadratic function, f1x2 = ax 2 + bx + c, a 7 0.

R.6

e)

•

71

Mathematical Modeling and Curve Fitting

ACTUAL AND PROJECTED ANNUAL SUIT SALES FOR RAGGS LTD. 600

Number of suits sold

500 400 300 200 100 2008

2009

2010

2011 2012 Year

2013

2014

The data rise and fall more than once, so they do not fit a linear or quadratic function but might fit a polynomial function that is neither quadratic nor linear. It is sometimes possible to find a mathematical model by graphing a set of data as a scatterplot, inspecting the graph to see if a known type of function seems to fit, and then using the data points to derive the equation of a specific function.

■ EXAMPLE 2 Business: Cable TV Subscribers. The following table shows the number of U.S. households with cable television and a scatterplot of the data. It appears that the data can be represented or modeled by a linear function.

1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006,

0 1 2 3 4 5 6 7

76.4 78.6 81.5 87.8 88.4 92.4 94.0 95.0

Scatterplot y Subscribers (in millions)

Year, x

Number of Cable Television Subscribers (in millions), y

Households with Cable Television

100 95 90 85 80 75 1

2

3

4

5

6

7

Year (Source: Nielsen Media Research.)

a) Find a linear function that fits the data. b) Use the model to predict the number of cable TV subscribers in 2014.

Solution a) We can choose any two of the data points to determine an equation. Let’s use 12, 81.52 and 16, 94.02. We first determine the slope of the line: m =

94.0 - 81.5 12.5 = = 3.125. 6 - 2 4

x

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Then we substitude 3.125 for m and either of the points 12, 81.52 or 16, 94.02 for 1x 1, y12 in the point–slope equation. Using 16, 94.02, we get y - 94.0 = 3.1251x - 62, which simplifies to y = 3.125x + 75.25, where x is the number of years after 1999 and y is in millions. We can then graph the linear equation y = 3.125x + 75.25 on the scatterplot to see how it fits the data.

Subscribers (in millions)

y

Households with Cable Television

100 95 90 85 80 75 1

2

3

4

5

6

7

x

Year

Quick Check 1

O

Business. Repeat Example 2 using the data points 11, 78.62 and 17, 95.02. Are the model and the result different from those in Example 2? This illustrates that a model and the predictions it makes are dependent on the pair of data points used.

b) We can predict the number of cable subscribers in 2014 by substituting 15 for x in the model 12014 - 1999 = 152: y = 3.125x + 75.25 = 3.1251152 + 75.25 L 122.1.

Model Substituting Rounding to the nearest tenth

We can then predict that there will be about 122.1 million cable TV subscribers in 2014.

O Quick Check 1

TECHNOLOGY CONNECTION

Linear Regression: Fitting a Linear Function to Data We now consider linear regression, the preferred method for fitting a linear function to a set of data. Although the complete basis for this method is discussed in Section 6.4, we consider it here because we can carry out the procedure easily using technology. One advantage of linear regression is that it uses all data points rather than just two.

EXAMPLE

Business: Cable TV Subscribers. Consider the data in Example 2. a) Find the equation of the regression line for the given data. Then graph the regression line with the graph. b) Use the model to predict the number of cable TV subscribers in 2014. Compare your answer to that found in Example 2. Solution a) To fit a linear function to the data using regression, we select the EDIT option of the STAT menu. We then enter

the data, entering the first coordinate for L1 and the second coordinate for L2. 10, 76.42, 11, 78.62, 12, 81.52, 13, 87.82,

14, 88.42, 15, 92.42, 16, 94.02, 17, 95.02 L1 0 1 2 3 4 5 6 L2(7) = 94

L2 76.4 78.6 81.5 87.8 88.4 92.4 94

L3 ––––––

L1 1 2 3 4 5 6 7 L2(8) = 95

L2 78.6 81.5 87.8 88.4 92.4 94 95

L3 2 ––––––

To view the data points, we turn on PLOT1 by pressing h [ from y1 at the E screen and then pressing D. To set the window, we could select the ZoomStat option of ZOOM. Instead, we select a 3- 1, 20, 70, 1404 window. (continued)

R.6 140

–1

Mathematical Modeling and Curve Fitting

73

140 Y1 = 2.8630952380952X + 76.7

–1 X = 15 70

20 70

To find the line that best fits the data, we press K, select CALC, then LinReg(ax + b), and then press [. LinReg y = ax + b a = 2.863095238 b = 76.74166667 r 2 = .9630983726 r = .9813757551

Y = 119.6881

20

Note that the result found using the regression equation, 119.7 million, is less than the 122.1 million found in Example 2. Statisticians might give more credence to the regression value because more data points were used to construct the equation.

EXERCISE 1. Study time and test scores.

The data in the following table relate study time and test scores.

The equation of the regression is approximately y = 2.86x + 76.7. We enter this as y1 and press D. y1 = 2.86x + 76.7 140

–1

•

Study Time (in hours)

Test Grade (in percent)

7 8 9 10 11

83 85 88 91 ?

20 70

a) Fit a regression line to the data. Then make a scatterplot of the data and graph the regression line with the scatterplot. b) Use the linear model to predict the test score received when one has studied for 11 hr. c) Discuss the appropriateness of a linear model of these data.

b) To find, or predict, the number of U.S. households with cable television in 2014, we press F, m, Y, and enter 15.

■ EXAMPLE 3 Life Science: Hours of Sleep and Death Rate. In a study by Dr. Harold J. Morowitz of Yale University, data were gathered that showed the relationship between the death rate of men and the average number of hours per day that the men slept. These data are listed in the following table. Average Number of Hours of Sleep, x

Death Rate per 100,000 Males, y

5 6 7 8 9

1121 805 626 813 967

(Source: Morowitz, Harold J., “Hiding in the Hammond Report,” Hospital Practice.)

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a) Make a scatterplot of the data, and determine whether the data seem to fit a quadratic function. b) Find a quadratic function that fits the data. c) Use the model to find the death rate for males who sleep 2 hr, 8 hr, and 10 hr.

Solution Death rate per 100,000 males

y

a) The scatterplot is shown to the left. Note that the rate drops and then rises, which suggests that a quadratic function might fit the data. b) We consider the quadratic model,

Possible quadratic function that “fits” data

1200 1000

(1)

y = ax 2 + bx + c.

800 600

To derive the constants (or parameters) a, b, and c, we use the three data points 15, 11212, 17, 6262, and 19, 9672. Since these points are to be solutions of equation (1), it follows that

400 200 1

2

3

4

5

6

7

8

9

1121 = a # 52 + b # 5 + c, or 1121 = 25a + 5b + c, 626 = a # 72 + b # 7 + c, or 626 = 49a + 7b + c, 967 = a # 92 + b # 9 + c, or 967 = 81a + 9b + c.

x

Average number of hours of sleep

We solve this system of three equations in three variables using procedures of algebra and get a = 104.5, b = - 1501.5, and c = 6016. Substituting these values into equation (1), we get the function given by y = 104.5x 2 - 1501.5x + 6016. Quick Check 2

O

Life Science. See the data on live births in the following Technology Connection. Use the data points 116, 342, 127, 113.92, and 137, 35.42 to find a quadratic function that fits the data. Use the model to predict the average number of live births to women age 20.

c) The death rate for males who sleep 2 hr is given by y = 104.51222 - 1501.5122 + 6016 = 3431. The death rate for males who sleep 8 hr is given by y = 104.51822 - 1501.5182 + 6016 = 692. The death rate for males who sleep 10 hr is given by y = 104.511022 - 1501.51102 + 6016 = 1451.

O Quick Check 2

TECHNOLOGY CONNECTION

Mathematical Modeling Using Regression: Fitting Quadratic and Other Polynomial Functions to Data Regression can be extended to quadratic, cubic, and quartic polynomial functions.

EXAMPLE Life Science: Live Births to Women of Age x. The chart to the right relates the average number of live births to women of a particular age. a) Fit a quadratic function to the data using REGRESSION. Then make a scatterplot of the data and graph the quadratic function with the scatterplot. b) Fit a cubic function to the data using REGRESSION. Then make a scatterplot of the data and graph the cubic function with the scatterplot.

c) Which function seems to fit the data better?

Age, x

Average Number of Live Births per 1000 women

16 18.5 22 27 32 37 42

34 86.5 111.1 113.9 84.5 35.4 6.8

(Source: Centers for Disease Control and Prevention.)

(continued)

R.6

•

Mathematical Modeling and Curve Fitting

d) We press F o to leave the graph screen. Pressing O and selecting Y-VARS and then FUNCTION and Y2, we have Y21202 L 99.6 and Y21302 L 97.4 as shown. Y2(20) 99.61232795 Y2(30) 97.38665968

d) Use the function from part (c) to estimate the average number of live births to women of ages 20 and 30.

Solution We proceed as follows. a) To fit a quadratic function using REGRESSION, the procedure is similar to what is outlined in the preceding Technology Connection on linear regression. We enter the data but select QuadReg instead of LinReg(ax + b). For the graph below, we used ZoomStat to set the window. We wrote y1 using approximations of a, b, and c from the QuadReg screen. y1 = –0.49x 2 + 25.95x – 238.49

The TRACE feature can also be used. Thus, the average number of live births is 99.6 per 1000 women age 20 and 97.4 per 1000 women age 30.

EXERCISES 1. Life science: live births.

a) Use the REGRESSION feature to fit a quartic equation to the live-birth data. Make a scatterplot of the data. Then graph the quartic function with the scatterplot. Decide whether the quartic function gives a better fit than either the quadratic or the cubic function. b) Explain why the domain of the cubic live-birth function should probably be restricted to the interval 315, 454. 2. Business: median household income by age.

QuadReg y = ax 2+bx+c a = –.4868465035 b = 25.94985182 c = –238.4892193

b) To fit a cubic function, we select CubicReg and obtain the following: CubicReg y = ax 3+bx 2+cx+d a = .0313582585 b = –3.219568404 c = 101.1751623 d = –886.9296244

A convenient way to copy a regression function, in full detail, onto the Y= screen is to press E, move the cursor to Y2 (or wherever the function is to appear), press O, and select STATISTICS and then EQ and RegEq. y2 = 0.03135825845952x 3 – 3.2195684044498x 2 + 101.17516232219x – 886.92962438781

Age, x

Median Income in 2003

19.5 29.5 39.5 49.5 59.5 65

$27,053 44,779 55,044 60,242 49,215 23,787

(Source: Based on data in the Statistical Abstract of the United States, 2005.)

a) Make a scatterplot of the data and fit a quadratic function to the data using QuadReg. Then graph the quadratic function with the scatterplot. b) Fit a cubic function to the data using CubicReg. Then graph the cubic function with the scatterplot. c) Fit a quartic function to the data using QuartReg. Then graph the quartic function with the scatterplot. d) Which of the quadratic, cubic, or quartic functions seems to best fit the data? e) Use the function from part (d) to estimate the median household income of people age 25; of people age 45. 3. Life science: hours of sleep and death rate.

Repeat Example 3 using quadratic regression to fit a function to the data.

c) The graph of the cubic function seems to fit closer to the data points. Thus we choose it as a model.

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EXERCISE SET

R.6 Choosing models. For the scatterplots and graphs in Exercises 1–9, determine which, if any, of the following functions might be used as a model for the data:

5. Player salaries of NBA players for the ending year of season (in millions of dollars)

Linear, f 1x2  mx  b Quadratic, f 1x2  ax 2  bx  c, a>0 Quadratic, f 1x2  ax 2  bx  c, ah for every 200 million km driven.

APPLICATIONS Business and Economics 10. Prime interest rate. a) For the prime interest rate data in Exercise 9, find a linear function that fits the data using the values given for January 2008 and December 2008. b) Use the linear function to estimate the prime rate in June 2009. 11. Average salary in NBA. Use the data from the bar graph in Exercise 5. a) Find a linear function that fits the data using the average salaries given for the years 2000 and 2009. Use 0 for 2000 and 9 for 2009. b) Use the linear function to predict average salaries in 2012 and 2020. c) In what year will the average salary reach 9.0 million?

Life and Physical Sciences 12. Absorption of an asthma medication. Use the data from Exercise 3. a) Find a quadratic function that fits the data using the data points 10, 02, 12, 2002, and 13, 1672. b) Use the function to estimate the amount of albuterol in the bloodstream after 4 hr. c) Does it make sense to use this function for t = 6? Why or why not? 13. Braking distance. a) Find a quadratic function that fits the following data. Travel Speed (mph)

Braking Distance (ft)

20 40 60

25 105 300

(Source: New Jersey Department of Law and Public Safety.)

15. High blood pressure in women. a) Choose two points from the following data and find a linear function that fits the data.

Age of Female

Percentage of Females with High Blood Pressure

30 40 50 60 70

1.4 8.5 19.1 31.9 53.0

(Source: Based on data from Health United States 2005, CDC/NCHS.)

b) Graph the scatterplot and the function on the same set of axes. c) Use the function to estimate the percent of 55-yr-old women with high blood pressure. 16. High blood pressure in men. a) Choose two points from the following data and find a linear function that fits the data.

Age of Male

Percentage of Males with High Blood Pressure

30 40 50 60 70

7.3 12.1 20.4 24.8 34.9

(Source: Based on data from Health United States 2005, CDC/NCHS.)

b) Graph the scatterplot and the function on the same set of axes. c) Use the function to estimate the percent of 55-yr-old men with high blood pressure.

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SYNTHESIS 17. Suppose that you have just 3 or 4 data points. Why might it make better sense to use a linear function rather than a quadratic or cubic function that fits these data more closely? 18. When modeling the number of hours of daylight for the dates April 22 to August 22, which would be a better choice: a linear function or a quadratic function? Explain. 19. Explain the restrictions that should be placed on the domain of the quadratic function found in Exercise 12 and why such restrictions are needed. 20. Explain the restrictions that should be placed on the domain of the quadratic function found in Exercise 13 and why such restrictions are needed.

TECHNOLOGY CONNECTION 21. Business: prime interest rate. a) Use regression to fit a linear function to the data in Exercise 9. b) Use the function to estimate the prime rate in June 2009. c) Compare your answers to those found in Exercise 10. Which is more accurate?

d) Fit a cubic function to the data and use it to estimate the prime rate in June 2009. e) Is a linear or cubic model more appropriate for this set of data? Explain. 22. Business: trade deficit with Japan. a) Use regression to fit a cubic function to the data in Exercise 4. Let x be the number of years after 1996. b) Use the function to estimate the trade deficit with Japan in 2012. c) Why might a linear function be a more logical choice than a cubic function for modeling this set of data?

Answers to Quick Checks 1. y = 2.733x + 75.869; 116.864 million cable TV subscribers. The model is different, and the prediction is lower. 2. y = - 0.7197x 2 + 38.2106x - 393.1272; 83.2 live births per 1000 women age 20.

CHAPTER R SUMMARY KEY TERMS AND CONCEPTS

EXAMPLES

SECTION R.1

To graph an ordered pair 1x, y2, move x units horizontally, then y units vertically, depending on whether the coordinates are positive, negative, or zero.

y 5 4 3 2 1

(−4, 2)

−5 −4 −3 −2 −1 −1 −2 −3 −4 (−2, −5) −5

A solution of an equation in two variables is an ordered pair of numbers that, when substituted for the variables, forms a true equation. The graph of an equation is a drawing that represents all ordered pairs that are solutions of the equation.

(4, 3)

1 2 3 4 5

x

(4, −3)

y

This is the graph of the equation y = x 4 - 2x 2.

2 1 −2

−1

1

2

x

−1

SECTION R.2 The mathematics used to represent the essential features of an applied problem comprise a mathematical model. A function is a correspondence between a first set, called a domain, and a second set, called a range, such that each member of the domain corresponds to exactly one member of the range.

i 7*4 can be used to estimate b 4 the amount A, in dollars, to which $1000 will grow in 7 years, at interest rate i, compounded quarterly. Business. The equation A = $1000a1 +

A graph represents a function if it is impossible to draw a vertical line that crosses the graph more than once. y

y

x

Not a function. Function notation permits us to easily determine what member of the range, output, is paired with a member of the domain, input.

x

A function.

The function given by f 1x2 = x 2 - 5x - 8 allows us to determine what member of the range is paired with the number 2 of the domain: f 1x2 = x 2 - 5x - 8, f 122 = 2 2 - 5122 - 8 = 4 - 10 - 8 = - 14

(continued)

Chapter R Summary

79

80

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KEY TERMS AND CONCEPTS SECTION R.2 (continued ) In calculus, it is important to be able to simplify an expression like f1x + h2 - f1x2 h

EXAMPLES For f 1x2 = x 2 - 5x - 8, f 1x + h2 - f 1x2 h

.

=

31x + h22 - 51x + h2 - 84 - f 1x2

h x + 2xh + h - 5x - 5h - 8 - 3x 2 - 5x - 84 2

=

2

h x + 2xh + h - 5x - 5h - 8 - x 2 + 5x + 8 = h 2xh + h2 - 5h = h h12x + h - 52 = h = 2x + h - 5, x Z 0 2

A function that is defined piecewise specifies different rules for parts of the domain.

2

The graph of 1 x + 3, g1x2 = c 3 - x, is

for x 6 3, for x Ú 3, 1 3

g(x) = − x + 3, for x < 3 y 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5

x

g(x) = −x, for x ≥ 3

1 We graph x + 3 only for inputs x less than 3. We graph - x only for 3 inputs x greater than or equal to 3.

SECTION R.3 The domain of any function is a set, or collection of objects. In this book, the domain is usually a set of real numbers.

There are three ways of naming sets: Roster Notation: 5- 2, 5, 9, p6 Set-Builder Notation: 5x ƒ x is a real number and x Ú 36 Interval Notation: Closed 3

Open 5

3

5

[3, 5]

(3, 5)

Half-open

Half-open

3

5

3

[3, 5)

5

(−∞, 5]

5

(3, 5]

3

(3, ∞)

Chapter R Summary

KEY TERMS AND CONCEPTS

EXAMPLES

For a function given by an equation or formula, the domain is the largest set of real numbers (inputs) for which function values (outputs) can be calculated.

Function f 1x2 =

81

Domain 8x

x2 - 9

=

8x 1x - 321x + 32

All real numbers except - 3 and 3.

g1x2 = 210 - 5x

All real numbers x for which 10 - 5x Ú 0 or 10 Ú 5x or 2 Ú x. The domain is 1- q , 24.

h1x2 = ƒ x - 3 ƒ + 5x 2

All real numbers, .

SECTION R.4 The slope m of the line containing the points 1x 1, y12 and 1x 2, y22 is given by y2 - y1 and can be regarded as m = x2 - x1 average rate of change. The graph of an equation f1x2 = mx + b, called slope–intercept form, is a line with slope m and y-intercept 10, b2. The graph of an equation y - y1 = m1x - x 12, called point–slope form, is a line with slope m passing through the point 1x 1, y12.

y (x2, y2) y2 − y1

(x1, y1)

y = mx + b y − y1 = m(x − x1)

x2 − x1

b

y2 − y1 =m x2 − x1

x

y

The graph of a constant function, given by f 1x2 = c, is a horizontal line.

c

f(x) = c x

The graph of an equation of the form x = a is a vertical line and is not a function.

y x=a a

x

SECTION R.5 The graph of a quadratic function f 1x2 = ax 2 + bx + c is a parabola with b b , f a - b b. The graph a vertex at a 2a 2a opens upward if a 7 0 and downward if a 6 0. It is an example of a nonlinear function or model.

y a>0 Vertex a12a2. [R.5] 12. A scatterplot is a random collection of points near a line. [R.6]

3. y = - 2x - 1

c)

y

13. Unless stated otherwise, the domain of a polynomial function is the set of all real numbers. [R.5] 14. The graph of a constant function has a slope of 0. [R.4] x

REVIEW EXERCISES 4. y = x

d)

15. Life science: babies born to women of age x. The following graph relates the number of babies born per 1000 women to the women’s age. [R.1, R.3]

y

y 120

5. g1x2 = 2x

e)

Number of babies born per 1000 women

x

y

x

100 80 60 40 20 0

10

20 30 40 Women's age

50

60 x

Chapter R Review Exercises

85

86

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Use the graph to answer the following. a) What is the incidence of babies born to women of age 35? b) For what ages are approximately 100 babies born per every 1000 women? c) Make an estimate of the domain of the function, and explain why it should be so.

29. Consider the function given by - x 2 + 2, for x 6 1, for 1 … x 6 2, f 1x2 = c 4, 1 x, for x Ú 2. 2

a) Find f 1- 12, f 11.52, and f 162. [R.2] b) Graph the function. [R.2]

16. Business: compound interest. Suppose that $1100 is invested at 5%, compounded semiannually. How much is in the account at the end of 4 yr? [R.1]

30. Write interval notation for each graph. [R.3] a) −2

17. Finance: compound interest. Suppose that $4000 is borrowed at 12%, compounded annually. How much is owed at the end of 2 yr? [R.1]

b)

32. For the function graphed below, determine (a) f 1 - 32; (b) the domain; (c) all x-values for which f 1x2 = 4; (d) the range. [R.3] y

19. A function is given by f 1x2 = - x 2 + x. Find each of the following. [R.2] a) f 132 b) f 1- 52 c) f 1a2 d) f 1x + h2

22. f 1x2 =

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

21. f 1x2 = 1x - 222

20. y = ƒ x + 1 ƒ x 2 - 16 x + 4

1 2 3 4 5

x

23. g1x2 = 2x + 1 33. Find the domain of f. [R.3, R.5] 7 a) f 1x2 = 2x - 10

Use the vertical-line test to determine whether each of the following is the graph of a function. [R.2] 24.

3

31. Write interval notation for each of the following. Then graph the interval on a number line. [R.3] a) 5x ƒ - 4 … x 6 56 b) 5x ƒ x 7 26

Becky Rick Rich Dick

Graph. [R.5]

0

a

Pat

Richard

−1

5

c)

18. Is the following correspondence a function? Why or why not? [R.2] Patricia Patrick Rebecca

0

y

25.

b) f 1x2 = 2x + 6

y

34. What are the slope and the y-intercept of y = - 3x + 2? [R.4]

x

x

35. Find an equation of the line with slope 14 , containing the point 18, - 52. [R.4] y

27.

36. Find the slope of the line containing the points 12, - 52 and 1- 3, 102. [R.4]

y

Find the average rate of change. [R.4]

x

37.

x

28. For the graph of function f shown to the right, determine (a) f 122; (b) the domain; (c) all x-values such that f 1x2 = 2; and (d) the range. [R.3]

y 5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

f 1 2 3 4 5

Value of computer (in hundreds of dollars)

26.

18 15 12 9 6 3 0

x

1

2

3

Number of years of use

87

Chapter R Review Exercises

47. Economics: equilibrium point. Find the equilibrium point for the given demand and supply functions. [R.5]

400

Demand: q = 1x - 722 Supply: q = x 2 + x + 4 1assume x … 72

300 200

Quantity (in hundreds)

Number of pages read

38.

100 0 1 2 3 4 5 6 Number of days spent reading

39. Business: shipping charges. The amount A that PetTreats-to-U charges for shipping is directly proportional to the value V of the item(s) being shipped. If the business charges $2.10 to ship a $60 gift basket, find an equation of variation expressing A as a function of V. [R.4] 40. Business: profit-and-loss analysis. The band Soul Purpose has fixed costs of $4000 for producing a new CD. Thereafter, the variable costs are $0.50 per CD, and the CD will sell for $10. [R.4] a) Find and graph C1x2, the total cost of producing x CDs. b) Find and graph R1x2, the total revenue from the sale of x CDs. Use the same axes as in part (a). c) Find and graph P1x2, the total profit from the production and sale of x CDs. Use the same axes as in part (b). d) How many CDs must the band sell in order to break even?

43. Solve each of the following. [R.5] a) 5 + x 2 = 4x + 2 b) 2x 2 = 4x + 3 44. Rewrite each of the following as an equivalent expression with rational exponents. [R.5] 5 a) 2x 4 b) 2t 8 1 1 c) 3 d) 2 2m 2x 2 - 9 45. Rewrite each of the following as an equivalent expression using radical notation. [R.5] a) x 2>5 b) m -3>5 1 c) 1x 2 - 521>2 d) -1>3 t 46. Determine the domain of the function given by 8 f 1x2 = 22x - 9. [R.5]

28 24 20 16 12 8 4

D

S

x

1 2 3 4 5 6 7

Price (in dollars)

48. Trail maintenance. The amount of time required to maintain a section of the Appalachian Trail varies inversely as the number of volunteers working. If a particular section of trail can be cleared in 4 hr by 9 volunteers, how long would it take 11 volunteers to clear the same section? [R.5] 49. Life science: maximum heart rate. A person exercising should not exceed a maximum heart rate, which depends on his or her gender, age, and resting heart rate. The following table shows data relating resting heart rate and maximum heart rate for a 20-yr-old woman. [R.6]

41. Graph each pair of equations on one set of axes. [R.5] a) y = 2x and y = 2x - 3 b) y = x 3 and y = 1x - 123 42. Graph each of the following. If the graph is a parabola, identify the vertex. [R.5] a) f 1x2 = x 2 - 6x + 8 3 b) g1x2 = 2x + 2 1 c) y = x x2 + x - 6 d) y = x - 2

q

Resting Heart Rate, r (in beats per minute)

Maximum Heart Rate, M (in beats per minute)

50 60 70 80

170 172 174 176

(Source: American Heart Association.)

a) Using the data points 150, 1702 and 180, 1762, find a linear function that fits the data. b) Graph the scatterplot and the function on the same set of axes. c) Use the function to predict the maximum heart rate of a woman whose resting heart rate is 67. 50. Business: ticket profits. The Spring Valley Drama Troupe is performing a new play. Data relating the daily profit P to the number of days after opening night are given below. [R.6] Days, x Profit, P (in dollars)

0

9

18

27

36

45

870

548

- 100

- 100

510

872

a) Make a scatterplot of the data. b) Decide whether the data seem to fit a quadratic function. c) Using the data points 10, 8702, 118, - 1002, and 145, 8722, find a quadratic function that fits the data.

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d) Use the function to estimate the profit made on the 30th day. e) Make an estimate of the domain of this function. Explain its restrictions.

SYNTHESIS 51. Economics: demand. The demand function for Clifton Cheddar Cheese is given by Demand: q = 800 - x 3, 0 … x … 9.28, where x is the price per pound and q is in thousands of pounds. a) Find the number of pounds sold when the price per pound is $6.50. b) Find the price per pound when 720,000 lb are sold.

TECHNOLOGY CONNECTION

56. Business: ticket profits. Use the data in Exercise 50. [R.6] a) Use regression to fit a quadratic function to the data. b) Use the function to estimate the profit made on the 30th day. c) What factors might cause the Spring Valley Drama Troupe’s profit to drop and then rise? 57. Social sciences: time spent on home computer. The data in the table below relate the average number of minutes spent per month on a home computer, A, to a person’s age, x. [R.6] a) Use regression to fit linear, quadratic, cubic, and quartic functions to the data. b) Make a scatterplot of the data and graph each function on the scatterplot. c) Which function fits the data best? Why?

Graph the function and find the zeros, the domain, and the range. [R.5] 52. f 1x2 = x 3 - 9x 2 + 27x + 50

Age (in years)

Average Use (in minutes per month)

6.5 14.5 21 29.5 39.5 49.5 55

363 645 1377 1727 1696 2052 2299

53. f 1x2 = 2 ƒ 4 - x 2 ƒ + 1 3

54. Approximate the point(s) of intersection of the graphs of the two functions in Exercises 52 and 53. [R.5] 55. Life science: maximum heart rate. Use the data in Exercise 49. [R.6] a) Use regression to fit a linear function to the data. b) Use the linear function to predict the maximum heart rate of a woman whose resting heart rate is 67. c) Compare your answer to that found in Exercise 49. Are the answers equally reliable? Why or why not?

(Source: Media Matrix; The PC Meter Company.)

CHAPTER R TEST

3. What are the slope and the y-intercept of y = 45 x - 23 ? 4. Find an equation of the line with slope 14 , containing the point 1- 3, 72. 5. Find the slope of the line containing the points 1- 9, 22 and 13, - 42.

6.

7. 36 30 24 18 12 6 0

1

2

3

Number of years of use

Weight gained (in pounds)

2. A function is given by f 1x2 = - x 2 + 5. Find: a) f 1- 32 b) f 1a + h2.

Find the average rate of change.

Value of color copier (in hundreds of dollars)

1. Business: compound interest. A person made an investment at 6.5% compounded annually. It has grown to $798.75 in 1 yr. How much was originally invested?

6 5 4 3 2 1 0 1 2 3 4 5 6

Number of bags of feed used

Chapter R Test

8. Life science: body fluids. The weight F of fluids in a human is directly proportional to body weight W. It is known that a person who weighs 180 lb has 120 lb of fluids. Find an equation of variation expressing F as a function of W. 9. Business: profit-and-loss analysis. A printing shop has fixed costs of $8000 for producing a newly designed note card. Thereafter, the variable costs are $0.08 per card. The revenue from each card is expected to be $0.50. a) Formulate a function C1x2 for the total cost of producing x cards. b) Formulate a function R1x2 for the total revenue from the sale of x cards. c) Formulate a function P1x2 for the total profit from the production and sale of x cards. d) How many cards must the company sell in order to break even?

20. Write interval notation for the following graph. c

f 1x2 = e

Age

Number of Calories Needed Daily

6 11 16 24 41

1800 2200 2400 2400 2200

(Source: Based on data from U.S. Department of Agriculture.)

given that x is the unit price, in dollars, and q is the quantity demanded or supplied, in thousands. Use the vertical-line test to determine whether each of the following is the graph of a function. y

12.

x 2 + 2, for x Ú 0, x 2 - 2, for x 6 0.

22. Nutrition. As people age, their daily caloric needs change. The following table shows data for physically active females, relating age to number of calories needed daily.

Demand: q = 1x - 822, 0 … x … 8, Supply: q = x 2 + x + 13,

y

d

21. Graph:

10. Economics: equilibrium point. Find the equilibrium point for these demand and supply functions:

11.

89

x x

a) Make a scatterplot of the data. b) Do the data appear to fit a quadratic function? c) Using the data points 16, 18002, 116, 24002, and 141, 22002, find a quadratic function that fits the data. d) Use the function from part (c) to predict the number of calories needed daily by a physically active 30-yr-old woman. e) Estimate the domain of the function from part (a). Explain its restrictions.

SYNTHESIS 13. For the following graph of a quadratic function f, determine (a) f 112; (b) the domain; (c) all x-values such that f 1x2 = 4; and (d) the range. y

–1 –1 –2 –3 –4

1

3 4 5 f

TECHNOLOGY CONNECTION 27. Graph the function and find the zeros and the domain and the range: f 1x2 = 2 ƒ 9 - x 2 ƒ - 1. 3

15. Convert to rational exponents: 1> 2t. 16. Convert to radical notation: t -3>5. x2 - 1 . x + 1

Determine the domain of each function. 18. f 1x2 =

x 2 + 20 x + 5x - 14 2

f 1x2 = 15 - 3x21>4 - 7.

26. A function’s average rate of change over the interval [1, 5] is - 73 . If f 112 = 9, find f 152.

x

14. Graph: f 1x2 = 8>x.

17. Graph: f 1x2 =

24. Find the domain and the zeros of the function given by 25. Write an equation that has exactly three solutions: - 3, 1, and 4. Answers will vary.

5 4 3 2 1 –5–4 –3

23. Simplify: 164 4>3 2-1>2.

19. f 1x2 =

x 23x + 6

28. Nutrition. Use the data in Exercise 22. a) Use REGRESSION to fit a quadratic function to the data. b) Use the function from part (a) to predict the number of calories needed daily by a physically active 30-yr-old woman. c) Compare your answer from part (b) with that from part (d) of Exercise 22. Which answer do you feel is more accurate? Why?

Extended Technology Application Average Price of a Movie Ticket Extended Technology Applications occur at the end of each chapter. They are designed to consider certain applications in greater depth, make use of calculator skills, and allow for possible group or collaborative learning.

the average price of a movie ticket for the years 1950 to 2008. How much did you pay the last time you went to the movies in the evening (not a matinee)? The average prices in the table may seem low, but they reflect discounts for matinees and children’s and senior citizens’ tickets. Let’s use our skills with REGRESSION to analyze the data.

Have you noticed that the price of a movie ticket seems to increase? The table and graph that follow show

AVERAGE TICKET PRICE, P(t)

1950, 0 1955, 5 1960, 10 1965, 15 1970, 20 1975, 25 1980, 30 1985, 35 1990, 40 1995, 45 1999, 49 2000, 50 2001, 51 2002, 52 2003, 53 2004, 54 2005, 55 2006, 56 2007, 57 2008, 58

$ 0.46 0.58 0.76 1.01 1.55 2.05 2.69 3.55 4.23 4.35 5.08 5.39 5.66 5.81 6.03 6.21 6.41 6.55 6.88 7.18

P(t) $8 7

Average ticket price

YEAR, t

6 5 4 3 2 1 0

5

10

15

20

(Source: Motion Picture Association of America.)

90

CHAPTER R

•

Functions, Graphs, and Models

25

30

Year

35

40

45

50

55

60 t

Extended Technology Application

EXERCISES 1. a) Using REGRESSION, find a linear function that fits the data. b) Graph the linear function. c) Use the linear function to predict the average price of a movie ticket in 2012 and in 2020. Do these estimates appear reasonable? d) Use the function to predict when the average price of a ticket will reach $20. Does this estimate seem reasonable? 2. a) Using REGRESSION, find a quadratic function, y = ax 2 + bx + c, that fits the data. b) Graph the quadratic function. c) Use the quadratic function to predict the average price of a movie ticket in 2012 and in 2020. Do these estimates appear reasonable? d) Use the function to predict when the average price of a ticket will reach $20. Does this estimate seem reasonable? 3. a) Using REGRESSION, find a cubic function, y = ax 3 + bx 2 + cx + d, that fits the data. b) Graph the cubic function. c) Use the cubic function to predict the average price of a movie ticket in 2012 and in 2020. Do these estimates appear reasonable? d) Use the function to predict when the average price of a ticket will reach $20. Does this estimate seem reasonable?

4. a) Using REGRESSION, find a quartic function, y = ax 4 + bx 3 + cx 2 + dx + e, that fits the data. b) Graph the quartic function. c) Use the quartic function to predict the average price of a movie ticket in 2012 and in 2020. Do these estimates appear reasonable? d) Use the function to predict when the average price of a ticket will reach $20. Does this estimate seem reasonable? 5. You are a research statistician assigned the task of making an accurate prediction of movie ticket prices. a) Why might you not use the linear function? b) Why might you use the quadratic function rather than the linear function? c) Examine the graphs and equations of the four functions and the estimates they provided. One choice the research statistician makes is not to use a higher-order polynomial function when a lower-order one works just as well. Under this criterion, why would a quadratic work just as well as a cubic? Look at the coefficients of the leading terms. There are yet other procedures a statistician uses to choose predicting functions but they are beyond the scope of this text.

91

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Differentiation

1

Chapter Snapshot What You’ll Learn

Why It’s Important

1.1 1.2 1.3 1.4

With this chapter, we begin our study of calculus. The first concepts we consider are limits and continuity. We apply those concepts to establishing the first of the two main building blocks of calculus: differentiation. Differentiation is a process that takes a formula for a function and derives a formula for another function, called a derivative, that allows us to find the slope of a tangent line to a curve at a point. A derivative also represents an instantaneous rate of change. Throughout the chapter, we will learn various techniques for finding derivatives.

Limits: A Numerical and Graphical Approach Algebraic Limits and Continuity Average Rates of Change Differentiation Using Limits of Difference Quotients 1.5 Differentiation Techniques: The Power and Sum–Difference Rules 1.6 Differentiation Techniques: The Product and Quotient Rules 1.7 The Chain Rule 1.8 Higher-Order Derivatives

Where It’s Used MARKET SATURATION

A new product is placed on the market and becomes very popular. How can the derivative help us understand its rate of sales and the phenomenon of market saturation? This problem appears as Example 8 in Section 1.7.

SALES LEVELING OFF

Number of units sold

N(t)

N(t) =

50

250,000 t 2 (2t + 1) 2

100

150

Week

200

t

93

94

CHAPTER 1

• Differentiation

Limits: A Numerical and Graphical Approach

1.1

In this section, we discuss the concept of a limit. The discussion is intuitive—that is, relying on prior experience and lacking formal proof. Suppose a football team has the ball on its own 10-yard line. Then, because of a penalty, the referee moves the ball back half the distance to the goal line; the ball is now on the 5-yard line. If the team commits the same infraction again, the ball will again be moved half the distance to the goal line; now it’s on the 2.5-yard line. If this kind of penalty were repeated over and over again, the ball would move steadily closer to the goal line but never actually be placed on the goal line. We would say that the limit of the distance between the ball and the goal line is zero.

OBJECTIVE • Find limits of functions, if they exist, using numerical or graphical methods.

Limits One important aspect of the study of calculus is the analysis of how function values (outputs) change as input values change. Basic to this study is the notion of a limit. Suppose a function f is given and suppose the x-values (the inputs) get closer and closer to some number a. If the corresponding outputs—the values of f1x2—get closer and closer to another number, then that number is called the limit of f1x2 as x approaches a. For example, let f1x2 = 2x + 3 and select x-values that get closer and closer to 4. In the table and graph below, we see that as the input values approach 4 from the left (that is, are less than 4), the output values approach 11, and as the input values approach 4 from the right (that is, are greater than 4), the output values also approach 11. Thus, we say: As x approaches 4 from either side, the function f1x2 = 2x + 3 approaches 11. Limit Numerically

Limit Graphically y

x 2

These inputs approach 4 from the left. 4 These inputs approach 4 from the right.

f1x2 11

7

3.6

10.2

3.9

10.8

3.99

10.98

3.999

10.998

...

...

4.001

11.002

4.01

11.02

4.1

11.2

4.8

12.6

5

13

12 10 8 6 4

–12 –10 –8 –6 –4

These outputs approach 11. 11 These outputs approach 11.

–2 –4

f(x) = 2x + 3

–6 –8 –10 –12

2

6

4

8 10 12

x

1.1

• Limits: A Numerical and Graphical Approach

95

An arrow, :, is often used to stand for the words “approaches from either side.” Thus, the statement above can be written: As x : 4, 2x + 3 : 11. The number 11 is said to be the limit of 2x + 3 as x approaches 4 from either side. We can abbreviate this statement as follows: lim 12x + 32 = 11.

x:4

This is read: “The limit, as x approaches 4, of 2x + 3 is 11.”

DEFINITION As x approaches a, the limit of f1x2 is L, written lim f1x2 = L,

x:a

if all values of f1x2 are close to L for values of x that are sufficiently close, but not equal, to a. The limit L must be a unique real number.

When we write lim f1x2, we are indicating that x is approaching a from both sides. x:a If we want to be specific about the side from which the x-values approach the value a, we use the notation lim f1x2 to indicate the limit from the left 1that is, where x 6 a2,

x : a-

or

lim f1x2 to indicate the limit from the right 1that is, where x 7 a2.

x : a+

These are called left-hand and right-hand limits, respectively. In order for a limit to exist, both the left-hand and right-hand limits must exist and be the same. This leads to the following theorem.

THEOREM As x approaches a, the limit of f1x2 is L if the limit from the left exists and the limit from the right exists and both limits are L. That is, if

lim f1x2 = lim- f1x2 = L,

x : a+

x:a

then

lim f1x2 = L.

x:a

The converse of this theorem is also true: if lim f1x2 = L, then it is assumed x:a that both the left-hand limit, lim- f1x2, and the right-hand limit, lim+ f1x2, exist and x:a x:a are equal to L. TECHNOLOGY CONNECTION

Finding Limits Using the TABLE and TRACE Features Consider the function given by f 1x2 = 3x - 1. Let’s use the TABLE feature to complete the following table. Note that the

inputs do not have the same increment from one to the next, but do approach 6 from either the left or the right. We use TblSet and select Indpnt and Ask mode. Then we enter the inputs shown and use the corresponding outputs to complete the table. (continued)

96

• Differentiation

CHAPTER 1

Finding Limits Using the TABLE and TRACE Features (continued) f 1x2 = 3x - 1 x

lim f 1x2 = 17 and lim- f 1x2 = 17 .

f 1x2

5

14

5.8

16.4

5.9

16.7

5.99

16.97

5.999

16.997

to visualize the limit process in this early stage of your learning. Using the TABLE and TRACE features, let’s complete the following: x : 6+

x:6

Thus, lim f 1x2 = 17 .

x:6

EXERCISES

Consider f 1x2 = 3x - 1. Use the TABLE and TRACE features, making up your own tables, to find each of the following. 1. lim f 1x2

2. lim f 1x2

x:2

6

?

x : -1

Consider g1x2 = x - 2x - 2 for Exercises 3–5. 3

3. Complete the following table.

6.001

17.003

6.01

17.03

6.1

17.3

7

6.4

18.2

7.8

7

20

7.9

x

g1x2

7.99 Now we set the table in Auto mode and starting (TblStart) with a number near 6, we make tables for some increments ( ¢ Tbl) like 0.1, 0.01, - 0.1, - 0.01, and so on, to determine lim f 1x2.

7.999 8

x:6

TABLE SETUP TblStart = 5.97 Tbl = .01 Indpnt: Auto Ask Depend: Auto Ask

X 5.97 5.98 5.99 6 6.01 6.02 6.03

?

8.001

Y1 16.91 16.94 16.97 17 17.03 17.06 17.09

8.01 8.1 8.4

As an alternative, graphical approach, let’s use the

8.9

TRACE feature with the graph of f. We move the cursor

from left to right so that the x-coordinate approaches 6 from the left, changing the window as needed, to see what happens. For example, let’s use 35.3, 6.4, 14, 184 and move the cursor from right to left so that the x-coordinate approaches 6 from the right. In general, the TRACE feature is not an efficient way to find limits, but it will help you

Use the TABLE and TRACE features to find each of the following. 4. lim g1x2 x:8

5. lim g1x2 x : -1

Refer again to the function f1x2 = 2x + 3. We showed that as x approaches 4, the function values approach 11, summarized as lim 12x + 32 = 11. You may have thought x:4 “Why not just substitute 4 into the function to get 11?” You are partially correct to make this observation (and in the next section, we see that we can use such shortcuts in certain cases), but keep in mind that we are curious about the behavior of the function

1.1

97

• Limits: A Numerical and Graphical Approach

f1x2 = 2x + 3 for values of x close to 4, not necessarily at 4 itself. It may help to summarize what we know already:

• At x = 4, the function value is 11, and this is visualized as the point 14, 112 on the graph of f. • For values of x close to 4, the values of f1x2 are correspondingly near 11; this is the limit we have been discussing.

The limit can help us understand the behavior of some functions a little more clearly. Consider the following example.

■ EXAMPLE 1 Let f1x2 =

x2 - 1 . x - 1

a) What is f112? b) What is the limit of f1x2 as x approaches 1?

Solution a) There is no answer, since we get a 0 in the denominator: f112 =

1122 - 1 112 - 1

=

0 0

Thus, f112 does not exist. There is no point on the graph of f1x2 shown below for x = 1. Limit Numerically x : 1-

1x 6 12

Limit Graphically y

f1x2 4

0.9

1.9

0.99

1.99

0.999

1.999

x2 − 1 f(x) = − x−1

2

4 3 2 1 1

1

3

4

x

2

x : 1+ 1x 7 12

Quick Check 1

f1x2

1.1

2.1

1.01

2.01

1.001

2.001

3 4

Let f1x2 =

x2 - 9 . x - 3

(See Example 5 in Section R.5.) a) What is f132? b) What is the limit of f as x approaches 3?

b) We select x-values close to 1 on either side (see the table above), and we see that the function values get closer and closer to 2. Thus, the limit of f1x2 as x approaches 1 is 2: lim f1x2 = 2. x:1

The graph has a “hole” at the point 11, 22. Thus, even though the function is not defined at x = 1, the limit does exist as x : 1.

O Now try Quick Check 1 Limits are also useful when discussing the behavior of piecewise-defined functions, as the following example illustrates.

O

98

CHAPTER 1

• Differentiation

■ EXAMPLE 2 Consider the function H given by H1x2 = e

2x + 2, for x 6 1, 2x - 4, for x Ú 1.

Graph the function and find each of the following limits, if they exist. When necessary, state that the limit does not exist. b) lim H1x2

a) lim H1x2 x:1

x : -3

Solution We check the limits from the left and from the right both numerically, with an input–output table, and graphically. a)

Limit Numerically x : 1-

1x 6 12

Limit Graphically

H 1x2

y 5

lim −H(x) = 4

0

2

0.5

3

0.8

3.6

1

0.9

3.8

5 4 3 2 1 1

0.99

3.98

x→1

H

4 3 2

0.999 These choices can vary.

3.998

2 3

1

2

3

4

5

x

lim +H(x) = –2 x→1

4 5

x : 1+ 1x 7 12

H 1x2 0

2 1.8

- 0.4

1.1

- 1.8

1.01

- 1.98

1.001

- 1.998

1.0001

- 1.9998

As inputs x approach 1 from the left, outputs H1x2 approach 4. Thus, the limit from the left is 4. That is, lim H1x2 = 4.

x : 1-

But as inputs x approach 1 from the right, outputs H1x2 approach - 2. Thus, the limit from the right is - 2. That is, lim H1x2 = - 2.

x : 1+

Since the limit from the left, 4, is not the same as the limit from the right, 2, we say that lim H1x2 does not exist.

x:1

We note in passing that H112 = - 2. In this example, the function value exists for x = 1, but the limit as x approaches 1 does not exist.

1.1

TECHNOLOGY CONNECTION

Exploratory Check the results of Examples 1 and 2 using the TABLE feature. See Section R.2 to recall how to graph functions defined piecewise. Note: Each Technology Connection labeled “Exploratory” is designed to lead you through a discovery process. Because of this, answers are not provided.

b)

Limit Numerically x : - 3-

1x 6 - 32

99

• Limits: A Numerical and Graphical Approach

Limit Graphically H 1x2

-4

-6

- 3.5

-5

- 3.1

- 4.2

- 3.01

- 4.02

- 3.001

- 4.002

y 5

H

4 3 2 1 5 4 3 2 1 1

1

2

3

4

5

x

2 3 4

x : - 3 + 1x 7 - 32

H 1x2

-2

-2

- 2.5

-3

- 2.9

- 3.8

- 2.99

- 3.98

- 2.999

- 3.998

5

As inputs x approach - 3 from the left, outputs H1x2 approach - 4, so the limit from the left is - 4. That is, Quick Check 2 Let - x + 4, for x … 3, k1x2 = e 2x + 1, for x 7 3. Find these limits: a) lim- k1x2, lim+ k1x2, x:3

x:3

and lim k1x2; x:3

b) lim k1x2. x:1

lim H1x2 = - 4.

x : -3 -

As inputs x approach - 3 from the right, outputs H1x2 approach - 4, so the limit from the right is - 4. That is, lim H1x2 = - 4.

x : -3 +

Since the limits from the left and from the right exist and are the same, we have lim H1x2 = - 4.

x : -3

O Now try Quick Check 2 The limit at a number a does not depend on the function value at a or even on whether that function value, f1a2, exists. That is, whether or not a limit exists at a has nothing to do with the function value f1a2.

The “Wall” Method As an alternative approach for Example 2, we can draw a “wall” at x = 1, as shown in blue on the graph to the left on the next page. We then follow the curve from left to right with a pencil until we hit the wall and mark the location with an *, assuming it can be determined. Then we follow the curve from right to left until we hit the wall and

O

100

CHAPTER 1

• Differentiation

mark that location with an *. If the locations are the same, as in the graph to the right below, a limit exists. Thus, for Example 2, lim H1x2 = - 4.

lim H1x2 does not exist, and

x:1

x : -3

y

y

5

5

H

4 3

3

2

2

1

1

5 4 3 2 1 1

1

2

H

4

3

4

5

5 4 3 2 1 1

x

2

2

3

3

4

4

5

5

1

2

3

4

5

x

■ EXAMPLE 3 Consider the function defined as follows: G1x2 = e

5, for x = 1, x + 1, for x Z 1.

Graph the function, and find each of the following limits, if they exist. If necessary, state that the limit does not exist. a) lim G1x2

b) lim G1x2

x:1

x : -2

Solution The graph of G follows. y 5 4

G

3 2 1 –5 – 4 –3 –2 –1

1

2

3

4

5

x

–1 –2 –3 –4 –5

a) As inputs x approach 1 from the left, outputs G1x2 approach 2, so the limit from the left is 2. As inputs x approach 1 from the right, outputs G1x2 also approach 2, so the limit from the right is 2. Since the limit from the left, 2, is the same as the limit from the right, 2, we have lim G1x2 = 2.

x:1

Note that the limit, 2, is not the same as the function value at 1, which is G112 = 5.

1.1

Limit Numerically x : 1- 1x 6 12

101

• Limits: A Numerical and Graphical Approach

Limit Graphically y

G1x2

0

1

0.5

1.5

0.9

1.9

0.99

1.99

5

G

4 3 2 1 5 4 3 2 1 1

2

3

4

5

x

5

x

2

lim G1x2 = 2

x : 1+ 1x 7 12

1

x:1

3

G1x2

4 5

1.5

2.5

1.1

2.1

1.01

2.01

1.001

2.001

b) Using the same approach as in part (a), we have lim G1x2 = - 1.

x : -2

Note that in this case, the limit, - 1, is the same as the function value at - 2, which is G1- 22 = - 1. Limit Numerically x : - 2 - 1x 6 - 22

Limit Graphically y

G1x2

-3

-2

- 2.5

- 1.5

- 2.1

- 1.1

- 2.01

- 1.01

Calculate the following limits based on the graph of f. y

lim G1x2 = - 1

x : -2

4

f

x : - 2 + 1x 7 - 22

3 2 1

a) lim- f1x2 x:2

b) lim+ f1x2

2

3

4

x

x:2

G1x2

- 1.5

- 0.5

- 1.9

- 0.9

- 1.99

- 0.99

- 1.999

- 0.999

x:2

c) lim f1x2

G

4 3 2

Quick Check 3

1

5

O Now try Quick Check 3

1 5 4 3 2 1 1 2 3 4 5

1

2

3

4

O

102

CHAPTER 1

• Differentiation

Limits Involving Infinity Limits also help us understand the role of infinity with respect to some functions. Consider the following example. 1

■ EXAMPLE 4 Let f1x2 = x . a) Find lim- f1x2.

b) Find lim+ f1x2.

x:0

x:0

c) Use the information from parts (a) and (b) to form a conclusion about lim f1x2. x:0

Solution We note first that f102 does not exist: there is no point on the graph that

corresponds to x = 0.

Limit Numerically x : 0-

1x 6 02

Limit Graphically y

f1x2

- 0.1

- 10

- 0.01

- 100

5

lim f(x) = ∞ 0+

x

4

1 f(x) = − x

3 2

- 0.001

- 1,000

1

- 0.0001

- 10,000

5 4 3 2 1 1

1

2

3

4

5

x

2 3

f 1x2

x : 0+ 1x 7 02

4

lim f(x) = − ∞

0.1

10

0.01

100

0.001

1,000

0.0001

10,000

x

0−

5

a) The table and graph show that as x approaches 0 from the left, the corresponding f1x2 values are decreasing without bound. We conclude that the left-hand limit is negative infinity; that is, lim- f1x2 = - q . We describe the notion of “infinity” by x:0 the symbol q . The symbol q does not represent a real number. b) The table and graph show that as x approaches 0 from the right, the f1x2 values increase without bound toward positive infinity. The right-hand limit is positive infinity; that is, lim+ f1x2 = q . x:0

c) Since the left-hand and right-hand limits do not match (and are not finite), lim f1x2 does not exist. x:0

Sometimes we need to determine limits when the inputs get larger and larger without bound, that is, as the inputs approach infinity. In such cases, we are finding limits at infinity. Such a limit is expressed as lim f1x2

x: q

or

lim f1x2.

x: -q

These limits are approached from one side only: from the left if approaching positive infinity or from the right if approaching negative infinity.

1.1

103

• Limits: A Numerical and Graphical Approach

1

■ EXAMPLE 5 Let f1x2 = x . Find limq f1x2 and limq f1x2. x: x: Solution The table shows that as x gets larger and larger in the positive direction,

the values for f1x2 approach 0. Thus, lim f1x2 = 0. As x decreases in the negative x: q

direction, we get the same value for the limit: lim f1x2 = 0. x: -q

Limit Numerically x: q 10 100

Limit Graphically

f 1x2

y 5

0.1

4

1 f(x) = − x

0.01

3 2

1,000

0.001

10,000

0.0001

lim f 1x2 = 0

lim f(x) = 0 x

∞

x

5 4 3 2 1 1

1

2

3

4

5

x

2

−∞

x: q

x: -q

lim f(x) = 0

1

3 4

f1x2

- 10

- 0.1

- 100

- 0.01

- 1,000

- 0.001

- 10,000

- 0.0001

5

lim f 1x2 = 0

x: -q

■ EXAMPLE 6 Consider the function f given by f1x2 =

1 + 3. x - 2

Graph the function, and find each of the following limits, if they exist. If necessary, state that the limit does not exist. a) lim f1x2 x:3

Solution The graph of f1x2 is shown to the right. Note that it is the same as the 1 graph of f1x2 = but shifted 2 units to the x right and 3 units up.

b) lim f1x2 x:2

y 8 7 6

f(x) 

1 3 x2

3

5

5 4 3 2 1 3 2 1 1 2

1

2

4

6

7

x

104

CHAPTER 1

• Differentiation

a) As inputs x approach 3 from the left, outputs f1x2 approach 4, so the limit from the left is 4. As inputs x approach 3 from the right, outputs f1x2 also approach 4. Since the limit from the left, 4, is the same as the limit from the right, we have lim f1x2 = 4.

x:3

Limit Numerically 1x 6 32

x : 3-

Limit Graphically f 1x2

2.1

13

2.5

5

2.9

4.1

2.99

4.01

y 8 7

f(x) 

6

1 3 x2

5 4 3 2

lim f 1x2 = 4

x:3

x : 3 + 1x 7 32

f 1x2

3.5

3.6

3.2

3.83

3.1

3.90

3.01

3.9900

1 3 2 1 1

1

2

3

4

5

6

7

x

2

b) As inputs x approach 2 from the left, outputs f1x2 become more and more negative, without bound. These numbers do not approach any real number, although it might be said that the limit from the left is negative infinity, - q . That is, lim f1x2 = - q .

x : 2-

As inputs x approach 2 from the right, outputs f1x2 become larger and larger, without bound. These numbers do not approach any real number, although it might be said that the limit from the right is infinity, q . That is, lim f1x2 = q .

x:2 +

Because the left-sided limit differs from the right-sided limit, lim f1x2 does not exist.

x:2

1.1

Limit Numerically 1x 6 22

x : 2-

105

• Limits: A Numerical and Graphical Approach

Limit Graphically f 1x2

1.5

1

1.9

-7

1.99

- 97

1.999

- 997

y 8 7 6

f(x) 

1 3 x2

3

5

5 4 3

lim f 1x2 does

x:2

x:2

f 1x2

1x 7 22

+

2.5

5

2.1

13

2.01

103

2.001

1003

not exist.

2 1 3 2 1 1

1

2

4

6

7

x

2

■ EXAMPLE 7 Consider again the function in Example 6, given by f1x2 =

1 + 3. x - 2

Find lim f1x2. x: q

Solution As inputs x get larger and larger, outputs f1x2 get closer and closer to 3. We have lim f1x2 = 3.

x: q

Limit Numerically x: q 5

Limit Graphically

f 1x2

y 8

3.3

10

3.125

100

3.0102

1000

3.0010

7

f(x) 

6

1 3 x2

5

Quick Check 4

c) lim h1x2. x: q

lim f 1x2 = 3

x: q

3

lim f(x) = 3

2

x→

1

1 + 6. Find 1 - x these limits: a) lim h1x2; b) lim h1x2; Let h1x2 =

x:1

4

3 2 1 1 2

x:2

O Now try Quick Check 4

1

2

3

4

5

6

7

x

O

106

CHAPTER 1

• Differentiation

Section Summary • The limit of a function f, as x approaches a, is written lim f1x2 = L. This means that as the values of x approach a, x:a

the corresponding values of f1x2 approach L. The value L must be a unique, finite number. • A left-hand limit is written lim- f1x2. The values of x are x:a

approaching a from the left, that is, x 6 a. • A right-hand limit is written lim+ f1x2. The values of x are x:a

approaching a from the right, that is, x 7 a.

• If the left-hand and right-hand limits (as x approaches a) are not equal, the limit does not exist. On the other hand, if the left-hand and right-hand limits are equal, the limit does exist. • A limit lim f1x2 may exist even though the function value x:a

f1a2 does not. (See Example 1.) • A limit lim f1x2 may exist and be different from the funcx:a

tion value f1a2. (See Example 3b.) • Graphs and tables are useful tools in determining limits.

EXERCISE SET

1.1 Complete each of the following statements.

When necessary, state that the limit does not exist.

1. As x approaches 3, the value of 2x + 5 approaches .

3. As x approaches approaches 6.

, the value of -3x

4. As x approaches approaches 5.

, the value of x - 2

5. The notation lim f 1x2 is read

.

6. The notation lim g1x2 is read

.

x:1

14. Find lim + f 1x2.

15. Find lim f 1x2.

16. Find lim f 1x2.

17. Find lim f 1x2.

18. Find lim f 1x2.

x : -1

8. The notation lim+ G1x2 is read

.

y 5 4 3 2 1 5 4321 1 2 3 4 5

For Exercises 11–18, consider the function f given by x  2, for x ◊ 3, x  1, for x>3.

g

1 2 3 4 5

x

If a limit does not exist, state that fact.

y

5 4321 1 2 3 4 5

x:2

for x2.

10. The notation is read “the limit, as x approaches 3 from the left.”

5 4 3 2 1

x : -1

x:4

9. The notation is read “the limit, as x approaches 2 from the right.”

f1x2  e

x : -1

x:3

g1x2  e .

x:4

13. Find lim - f 1x2.

x:3

For Exercises 19–26, consider the function g given by

7. The notation lim- F1x2 is read x:5

12. Find lim- f 1x2.

x:3

2. As x approaches - 4, the value of 3x + 7 approaches .

x:4

11. Find lim+ f 1x2.

19. f

lim g1x2

x : -2 -

x

lim g1x2

x : -2 +

21. lim+ g1x2

22. lim- g1x2

23. lim g1x2

24. lim g1x2

25. lim g1x2

26. lim g1x2

x:4

1 2 3 4 5

20.

x:4

x:2

x: 4

x : -2

x : -4

Exercise Set 1.1

For Exercises 27–34, use the following graph of F to find each limit. When necessary, state that the limit does not exist.

y

6 5 4 3 2 1

6 5 4 3 2 1

F

1 2 3 4 5 6 7

28. lim F1x2

29. lim F1x2

30. lim F1x2

31. lim F1x2

32. lim F1x2

33.

34.

x : -3

x:2

x : -2

x : -5

x:4

x:6

lim F1x2

x : -2 +

lim F1x2

x : -2 -

For Exercises 35–42, use the following graph of G to find each limit. When necessary, state that the limit does not exist.

5 4 3 2 1 1 2 3 4 5

x

39. lim G1x2

40. lim- G1x2

41. lim+ G1x2

42. lim G1x2

x:0 x:1 x:3 x:3

For Exercises 43–52, use the following graph of H to find each limit. When necessary, state that the limit does not exist.

5 4 3 2 1

x : -3

1 2 3 4 5

x

x:3

45.

46. lim H1x2

lim H1x2

lim H1x2

x : -2 x : -2

47. lim- H1x2

48. lim+ H1x2

49. lim H1x2

50. lim- H1x2

51. lim+ H1x2

52. lim H1x2

x:1 x:1 x:3

x:1

60. lim f 1x2

x : -4

x : -2

61. lim f 1x2

62.

x: q

lim f 1x2

x: -q

find lim f 1x2 and lim f 1x2. x:0

x : -2

x : -1

x:0

64. f 1x2 = x 2; find lim f 1x2 and lim f 1x2.

65. g1x2 = x 2 - 5; find lim g1x2 and lim g1x2. x : -1

x:1 x:3 x:3

x:0

1 67. F1x2 = ; find lim F1x2 and lim F1x2. x:3 x:4 x - 3 68. G1x2 =

1 ; find lim G1x2 and lim G1x2. x : -1 x : -2 x + 2

69. f 1x2 =

1 - 2; find lim f 1x2 and lim f 1x2. x x: q x:0

70. f 1x2 =

1 + 3; find lim f 1x2 and lim f 1x2. x x: q x:0

71. g1x2 =

1 + 4; find lim g1x2 and lim g1x2. x: q x : -2 x + 2

72. g1x2 =

1 + 2; find lim g1x2 and lim g1x2. x: q x:3 x - 3

2x + 1, for x 6 1, x, for x Ú 1. Find lim- F1x2, lim+ F1x2, and lim F1x2.

73. F1x2 = e x:1

44.

x : -2 +

x:0

58. lim f 1x2

59. lim f 1x2

H

43. lim H1x2 x : -3

56. lim f 1x2

57. lim f 1x2

y

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x:2

55. lim f 1x2

x : -3

38. lim+ G1x2

x:3

54. lim f 1x2

x : -1

x:0

37. lim- G1x2 x:1

x

66. g1x2 = ƒ x ƒ + 1; find lim g1x2 and lim g1x2. 36. lim G1x2

x:1

1 2 3 4 5 6 7

53. lim f 1x2

63. f 1x2 = ƒ x ƒ ;

35. lim G1x2 x : -2

f

For Exercises 63–80, graph each function and then find the specified limits. When necessary, state that the limit does not exist.

y

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

76 5 4 321 1 2 3 4 5 6

x

27. lim F1x2

G

For Exercises 53–62, use the following graph of f to find each limit. When necessary, state that the limit does not exist.

y

5 4321 1 2 3 4 5

107

x:1

x:1

- x + 3, for x 6 2, x + 1, for x Ú 2. Find lim- G1x2, lim+ G1x2, and lim G1x2.

74. G1x2 = e x:2

x:2

x:2

- x + 4, for x 6 3, x - 3, for x 7 3. Find lim- g1x2, lim+ g1x2, and lim g1x2.

75. g1x2 = e x:3

x:3

x:3

108

• Differentiation

CHAPTER 1

3x - 4, for x 6 1, x - 2, for x 7 1. Find lim- f 1x2, lim+ f 1x2, and lim f 1x2.

76. f 1x2 = e x:1

x:1

x:1

77. G1x2 = e

x , for x 6 - 1, Find lim G1x2. x : -1 x + 2, for x 7 - 1.

78. F1x2 = e

- 2x - 3, for x 6 - 1, Find lim F1x2. x : -1 x 3, for x 7 - 1.

2

x:0

for x 6 0, for 0 … x 6 1, for x Ú 1. lim H1x2.

p1x2  $0.88, if 0 500

250

100

200

300

400

500

600

700

x

Quantity (in pounds)

3t, for 0 … t … 30, k - 4t, for t 7 30.

Determine k such that the volume function v is continuous at t = 30. Explain why this must be true.

1250

This is a fair compromise: the customers still get a cheaper rate per pound once x is above 500 pounds, but Rick is no longer losing money as he was with his previous pricing function.

O Quick Check 5

Section Summary • For a rational function for which a is in the domain, the limit as x approaches a can be found by direct evaluation of the function at a. • If direct evaluation leads to the indeterminate form 0>0, the limit may still exist: algebraic simplification and/or a table and graph are used to find the limit. • Informally, a function is continuous if its graph can be sketched without lifting the pencil off the paper.

• Formally, a function is continuous at x = a if (1) the function value f1a2 exists, (2) the limit as x approaches a exists, and (3) the function value and the limit are equal. This can be summarized as lim f1x2 = f1a2. x:a • If any part of the continuity definition fails, then the function is discontinuous at x = a.

EXERCISE SET

1.2 Classify each statement as either true or false. 1. lim 7 = 7 x:3

2. If lim f 1x2 = 9, then lim 2f 1x2 = 3.

5. If f is continuous at x = 2, then f 122 must exist. 6. If g is discontinuous at x = 3, then g132 must not exist.

x:2

x:2

7. If lim F1x2 exists, then F must be continuous at x = 4.

x:1

x:1

8. If lim G1x2 equals G172, then G must be continuous at

x:4

x:4

3. If lim g 1x2 = 5, then lim 3g 1x242 = 10.

4. If lim F1x2 = 7, then lim 3c # F1x24 = 7c.

x:4 x:7

x = 7.

118

CHAPTER 1

• Differentiation

39.

Use the Theorem on Limits of Rational Functions to find the following limits. When necessary, state that the limit does not exist. 9. lim 13x + 22

5 4 3 2 1

10. lim 14x - 52

x:1

x:2

11. lim 1x 2 - 42

12. lim 1x 2 + 32

x : -1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x : -2

13. lim 1x - 4x + 72 2

14. lim 1x - 6x + 92 2

x:3

15. lim 12x 4 - 3x 3 + 4x - 12

x:5

x:2

16. lim 13x 5 + 4x 4 - 3x + 62

41.

x : -1

x - 8 x - 2 2

17. lim

x:3

x - 25 2

18. lim

x2 - 5

x:3

For Exercises 19–30, the initial substitution of x  a yields the form 0/0. Look for ways to simplify the function algebraically, or use a table and/or a graph to determine the limit. When necessary, state that the limit does not exist. x2 - 9 19. lim x:3 x - 3

x 2 - 25 20. lim x:5 x - 5

x + 5x - 6

21. lim

23. lim

22. lim

3x 2 + x - 14

24. lim

x - 4 2

x:2

x3 - 1 x:1 x - 1 x : 25

29. lim

x : -1

9 - x

28. lim

x:9

x 2 + 5x + 4

30. lim

x 2 + 2x + 1

x:2

x

–5 – 4 –3 –2 –1 –1 –2

1 2 3 4 5

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2

y = t(x) 1 2 3 4 5

x

Use the graphs and functions in Exercises 37–41 to answer each of the following. If an expression does not exist, state that fact. x:1

x:1

43. a) Find lim+ g 1x2, lim- g 1x2, and lim g 1x2. x:1

x:1

x:1

b) Find g112. c) Is g continuous at x = 1? Why or why not? d) Find lim g 1x2.

9 - x 2x - 3

x : -2

x 2 + 3x - 10

e) Find g1- 22. f) Is g continuous at x = - 2? Why or why not?

x 2 - 4x + 4

44. a) Find lim k1x2. Use the Limit Properties to find the following limits. If a limit does not exist, state that fact. 31. lim 2x 2 - 9

32. lim 2x 2 - 16

33. lim 2x - 9

34. lim 2x - 16

35. lim+ 2x - 9

36.

x:4

e) Find k132. f) Is k continuous at x = 3? Why or why not?

2

x:2

x:3

2

x:3

45. a) Find lim h 1x2.

lim 2x 2 - 16

x : -4 -

x:1

b) Find h112. c) Is h continuous at x = 1? Why or why not? d) Find lim h 1x2.

Determine whether each of the functions shown in Exercises 37–41 is continuous over the interval 16, 62. 37.

y

38.

5 4 3 2 1 –5 – 4 –3 –2 –1 –1 –2 –3 –4 –5

x : -2

y

e) Find h1- 22. f) Is h continuous at x = - 2? Why or why not?

5 4 3 2 1 1 2 3 4 5

y = f(x)

x

x : -1

b) Find k1- 12. c) Is k continuous at x = - 1? Why or why not? d) Find lim k1x2. x:3

x:5

2

–5 – 4 –3 –2 –1 –1 –2 –3 –4 –5

46. a) Find lim t1x2. x:1

1 2 3 4 5

y = g(x)

x

x

y

x : -2

x3 - 8 x:2 2 - x

2x - 5 x - 25

1 2 3 4 5

y = h(x)

e) Find f 1- 22. f) Is f continuous at x = - 2? Why or why not?

2

26. lim

25. lim

27. lim

2x 2 - x - 21

x : -3

y = k(x)

x:1

x2 - 4

x : -2

5 4 3 2 1

b) Find f 112. c) Is f continuous at x = 1? Why or why not? d) Find lim f 1x2.

2

x2 - 1

x:1

y

40.

42. a) Find lim+ f 1x2, lim- f 1x2, and lim f 1x2.

x - 2x - 8

2

y

b) Find t112. c) Is t continuous at x = 1? Why or why not? d) Find lim t1x2. x : -2

e) Find t1- 22. f) Is t continuous at x = - 2? Why or why not?

Exercise Set 1.2

In Exercises 47 and 48, use the graphs to find the limits and answer the related questions. y

47.

5

1 x + 4, for x … 3, F1x2 = b 3 2x - 5, for x 7 3,

2 1

continuous at x = 3? Why or why not?

–2 –1 0

1

2

3

4

x

5

56. Is the function given by 1 x + 1, for x 6 4, G1x2 = b 2 - x + 5, for x 7 4,

a) Find lim+ G1x2. x:3

b) Find lim- G1x2.

continuous at x = 4? Why or why not?

x:3

c) Find lim G1x2. x:3

Find G132. Is G continuous at x = 3? Why or why not? Is G continuous at x = 0? Why or why not? Is G continuous at x = 2.9? Why or why not?

48. Consider the function

1 x + 4, for x 6 3, f 1x2 = b 3 2x - 1, for x Ú 3,

continuous at x = 3? Why or why not? 1 x + 1, for x 6 4, g 1x2 = b 2 - x + 7, for x 7 4,

continuous at x = 4? Why or why not?

y

59. Is the function given by

1 y = C(x)

0.5 – 2 –1 0

57. Is the function given by

58. Is the function given by

- 1, for x 6 2, C1x2 = e 1, for x Ú 2.

1

2

3

4

x

60. Is the function given by

–1

x2 - 1 , for x Z 1, F1x2 = c x - 1 4, for x = 1,

a) Find lim+ C1x2. x:2

b) Find lim- C1x2.

continuous at x = 1? Why or why not?

x:2

c) Find lim C1x2. x:2

d) Find C122. e) Is C continuous at x = 2? Why or why not? f) Is C continuous at x = 1.95? Why or why not?

49. Is the function given by f 1x2 = 3x - 2 continuous at x = 5? Why or why not?

50. Is the function given by g 1x2 = x - 3x continuous at x = 4? Why or why not? 2

51. Is the function given by G1x2 =

x2 - 4 , for x Z 2, G1x2 = c x - 2 5, for x = 2, continuous at x = 2? Why or why not?

– 0.5

1 continuous at x = 0? x

Why or why not? 52. Is the function given by F1x2 = 2x continuous at x = - 1? Why or why not? 53. Is the function given by g 1x2 = b

1 x + 1, for x 6 4, f 1x2 = b 2 - x + 7, for x Ú 4,

55. Is the function given by

3

d) e) f) g)

54. Is the function given by

continuous at x = 4? Why or why not?

y = G(x)

4

119

1 3x

+ 4, for x … 3, 2x - 1, for x 7 3,

continuous at x = 3? Why or why not?

61. Is the function given by x 2 - 4x - 5 , for x 6 5, x - 5 f 1x2 = c x + 1, for x Ú 5, continuous at x = 5? Why or why not? 62. Is the function given by x 2 - 3x - 4 , for x 6 4, x - 4 G1x2 = c 2x - 3, for x Ú 4, continuous at x = 4? Why or why not? 63. Is the function given by g 1x2 = at x = 5? Why or why not? 64. Is the function given by f 1x2 = at x = 3? Why or why not?

1 x 2 - 7x + 10 1 x 2 - 6x + 8

continuous

continuous

120

CHAPTER 1

• Differentiation

65. Is the function given by F1x2 = at x = 4? Why or why not?

1 x 2 - 7x + 10

66. Is the function given by G1x2 = at x = 2? Why or why not?

1 x - 6x + 8 2

continuous

Find each limit, if it exists. If a limit does not exist, state that fact. continuous

67. Is the function given by g 1x2 = x 2 - 3x + 2 continuous over the interval 1- 4, 42? Why or why not? 68. Is the function given by F1x2 = x 2 - 5x + 6 continuous over the interval 1- 5, 52? Why or why not? 1 69. Is the function given by f 1x2 = + 3 continuous over x the interval 1- 7, 72? Why or why not? 1 continuous over x - 1 the interval 10, q 2? Why or why not?

70. Is the function given by G1x2 =

71. Is the function given by g 1x2 = 4x 3 - 6x continuous on ? 72. Is the function given by F1x2 =

SYNTHESIS

3 continuous on ? x - 5

APPLICATIONS

75. lim

x:0

ƒxƒ x

76. lim

x : -2

x3 + 8 x2 - 4

TECHNOLOGY CONNECTION In Section 1.1, we discussed how to use the TABLE feature to find limits. Consider lim

x:0

21 + x - 1 . x

Input–output tables for this function are shown below. The table on the left uses TblStart = - 1 and ¢ Tbl = 0.5. By using smaller and smaller step values and beginning closer to 0, we can refine the table and obtain a better estimate of the limit. On the right is an input–output table with TblStart = - 0.03 and ¢ Tbl = 0.01. X –1 –0.5 0 0.5 1 1.5 2 X = –1

Y1 1 0.58579 ERROR 0.44949 0.41421 0.38743 0.36603

X –0.03 –0.02 –0.01 0 0.01 0.02 0.03 X = –0.03

Y1 0.50381 0.50253 0.50126 ERROR 0.49876 0.49752 0.49631

Business and Economics 73. The Candy Factory sells candy by the pound, charging $1.50 per pound for quantities up to and including 20 pounds. Above 20 pounds, the Candy Factory charges $1.25 per pound for the entire quantity, plus a quantity surcharge k. If x represents the number of pounds, the price function is p1x2 = e

It appears that the limit is 0.5. We can verify this by graphing y =

21 + x - 1 x

and tracing the curve near x = 0, zooming in on that portion of the curve.

1.50x, for x … 20, 1.25x + k, for x 7 20.

y= 1+ x – 1 x 2

a) Find k such that the price function p is continuous at x = 20. b) Explain why it is preferable to have continuity at x = 20.

Limit = 0.5 –2

Life and Physical Sciences 74. A lab technician controls the temperature T inside a kiln. From an initial temperature of 0 degrees Celsius (°C), he allows the kiln to increase by 2°C per minute for the next 60 min. After the 60th minute, he allows the kiln to cool at the rate of 3°C per minute. The temperature function T is defined by T1t2 = e

2t, for t … 60, k - 3t, for t 7 60.

a) Find k such that T is continuous at t = 60. b) Explain why T must be continuous at t = 60 min.

5

–1

We see that lim

x:0

21 + x - 1 = 0.5. x

To verify this algebraically, multiply 21 + x + 1 21 + x + 1

21 + x - 1 by 1, using x

. Then simplify the result and find the limit.

1.3

In Exercises 77–84, find each limit. Use the TABLE feature and start with ¢ Tbl  0.1. Then use 0.01, 0.001, and 0.0001. When you think you know the limit, graph and use the TRACE feature to verify your assertion. Then try to verify it algebraically. a2 - 4

2x - 1 x - 1

80. lim

24 + x - 24 - x x

x - 2x x:1 x - 1

82. lim

27 + 2x - 27 x

2 - 2x x:4 4 - x

84. lim

a : -2

79. lim

x:0

2a 2 + 5 - 3

23 - x - 23 x

x:1

x:0

4

81. lim

x:0

121

Answers to Quick Checks

78. lim

77. lim

• Average Rates of Change

1. (a) - 1; L1, L2, L3, L6 51 (b) , or 5.1; L1, L2, L3, L5, L6 10 (c) 213; L1, L2, L3, L6 2. 2 3. No, the limit as x : 2 does not exist. 4. (a) No, the limit as x : 3 is 6, but h132 = 7, so the limit does not equal the function value. (b) c = 10 5. k = 210; the amount of water (in gallons) is continuous as a function of time.

7 - 249 - x 2 x x:0

1.3 OBJECTIVES • Compute an average rate of change. • Find a simplified difference quotient.

Average Rates of Change Let’s say that a car travels 110 mi in 2 hr. Its average rate of change (speed) is 110 mi>2 hr, or 55 mi>hr (55 mph). Suppose that you are on a freeway and you begin accelerating. Glancing at the speedometer, you see that at that instant your instantaneous rate of change is 55 mph. These are two quite different concepts. The first you are probably familiar with. The second involves ideas of limits and calculus. To understand instantaneous rate of change, we first need to develop a solid understanding of average rate of change. The following graph shows the total production of suits by Raggs, Ltd., during one morning of work. Industrial psychologists have found curves like this typical of the production of factory workers. N S(4, 100)

100 Total number of suits produced

83. lim

75 R(3, 64) Q(2, 55)

50 P(1, 20) 25

(0, 0) 0 (8 A.M.)

1 (9 A.M.)

2 Time (10 A.M.)

3

4

(11 A.M.)

(12 P.M.)

t

■ EXAMPLE 1 Business: Production. What was the number of suits produced at Raggs, Ltd., from 9 A.M. to 10 A.M.?

Solution At 9 A.M., 20 suits had been produced. At 10 A.M., 55 suits had been produced. In the hour from 9 A.M. to 10 A.M., the number of suits produced was 55 suits - 20 suits, or 35 suits. Note that 35 is the slope of the line from P to Q.

122

CHAPTER 1

• Differentiation

■ EXAMPLE 2 Business: Average Rate of Change. What was the average number of suits produced per hour from 9 A.M. to 11 A.M.?

Solution

We have

44 suits 64 suits - 20 suits = 11 A.M. - 9 A.M. 2 hr = 22

suits . hr

Note that 22 is the slope of the line from P to R. N S(4, 100)

Total number of suits produced

100

Quick Check 1 State the average rate of change for each situation in a short sentence. Be sure to include units. a) It rained 4 inches over a period of 8 hours. b) Your car travels 250 miles on 20 gallons of gas. c) At 2 P.M., the temperature was 82 degrees. At 5 P.M., the temperature was 76 degrees.

75

R(3, 64) Q(2, 55) 50

64 − 20 P(1, 20) 25

3−1 (0, 0) 0

1

2

3

4

(11 A.M.)

(12 P.M.)

t

Time (8 A.M.)

(9 A.M.)

(10 A.M.)

O Quick Check 1 Let’s consider a function y = f1x2 and two inputs x 1 and x 2. The change in input, or the change in x, is x 2 - x 1. The change in output, or the change in y, is y2 - y1,

where y1 = f1x 12 and y2 = f1x 22.

DEFINITION The average rate of change of y with respect to x, as x changes from x 1 to x 2, is the ratio of the change in output to the change in input: y2 - y1 , where x 2 Z x 1. x2 - x1

If we look at a graph of the function, we see that y2 - y1 f1x 22 - f1x 12 = , x2 - x1 x2 - x1

O

1.3

• Average Rates of Change

123

which is both the average rate of change and the slope ! the line from P1x 1, y12 to Í of Q1x 2, y22.* The line passing through P and Q , denoted PQ, is called a secant line. y

Secant line

P

y1 = f(x1)

f(x 2) – f(x1)

x

x2

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

x1

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

Q

y2 = f(x 2)

x 2 – x1

The slope of the secant line is interpreted as the average rate of change of f from x 1 to x 2.

■ EXAMPLE 3 For y = f1x2 = x 2, find the average rate of change as: a) x changes from 1 to 3. b) x changes from 1 to 2. c) x changes from 2 to 3.

Solution The following graph is not necessary to the computations but gives us a look at two of the secant lines whose slopes are being computed. y 9 8 7 6 5 4 3 2 1

R(3, 9) y = f(x) = x 2

Q(2, 4)

P(1, 1) 1

2

3

4

a) When x 1 = 1,

y1 = f1x 12 = f112 = 12 = 1;

and when x 2 = 3,

y2 = f1x 22 = f132 = 32 = 9.

The average rate of change is

y2 - y1 f1x 22 - f1x 12 = x2 - x1 x2 - x1 9 - 1 3 - 1 8 = = 4. 2 =

*The notation P1x 1, y12 simply means that point P has coordinates 1x 1, y12.

x

124

CHAPTER 1

• Differentiation

b) When x 1 = 1,

y1 = f1x 12 = f112 = 12 = 1;

and when x 2 = 2,

y2 = f1x 22 = f122 = 2 2 = 4.

The average rate of change is 4 - 1 3 = = 3. 2 - 1 1 c) When x 1 = 2,

y1 = f1x 12 = f122 = 2 2 = 4;

and when x 2 = 3,

The average rate of change is 9 - 4 5 = = 5. 3 - 2 1

O Quick Check 2 For a linear function, the average rate of change is the same for any choice of x 1 and x 2. As we saw in Example 3, a function that is not linear has average rates of change that vary with the choice of x 1 and x 2.

Difference Quotients as Average Rates of Change We now develop a notation for average rates of change that does not require subscripts. Instead of x 1, we will write simply x; in place of x 2, we will write x + h. y

(x + h, f(x + h))

y2 = f(x + h)

(x, f(x)) y1 = f(x)

f(x + h) – f(x)

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

For f1x2 = x 3, find the average rate of change between: a) x = 1 and x = 4; b) x = 1 and x = 2; c) x = 1 and x = 1.2.

y2 = f1x 22 = f132 = 32 = 9.

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

Quick Check 2

h x1 = x

x2 = x + h

x

It may help to think of the h as the horizontal distance between the inputs x 1 and x 2. That is, to get from x 1, or x, to x 2, we move a distance h. Thus, x 2 = x + h. Then the average rate of change, also called a difference quotient, is given by y2 - y1 f1x 22 - f1x 12 f1x + h2 - f1x2 f1x + h2 - f1x2 = = = . x2 - x1 x2 - x1 1x + h2 - x h

O

• Average Rates of Change

1.3

125

DEFINITION The average rate of change of f with respect to x is also called the difference quotient. It is given by

TECHNOLOGY CONNECTION

EXERCISES

f1x + h2 - f1x2

Use a calculator to show that f 1x + h2 Z f 1x2 + f 1h2 for each of the following functions.

h

, where h Z 0.

The difference quotient is equal to the slope of the secant line from 1x, f1x22 to 1x + h, f1x + h22.

1. f 1x2 = x 4 + x 2; let x = 6 and

h = 0.02.

2. f 1x2 = x 3 - 2x 2 + 4; let

Keep in mind that, in general, f1x + h2 Z f1x2 + f1h2. (You can check this using f1x2 = x 2, as in the Technology Connection at the left.)

x = 6 and h = 0.1.

EXAMPLE 4 For f1x2 = x 2, find the difference quotient when: a) x = 5 and h = 3. b) x = 5 and h = 0.1. c) x = 5 and h = 0.01.

■

Solution a) We substitute x = 5 and h = 3 into the formula: f1x + h2 - f1x2

=

h

f15 + 32 - f152 3

f182 - f152

=

3

.

Now f182 = 82 = 64 and f152 = 52 = 25, and we have f182 - f152 3

64 - 25 39 = = 13. 3 3

=

The difference quotient is 13. It is also the slope of the line from 15, 252 to 18, 642. b) We substitute x = 5 and h = 0.1 into the formula: f1x + h2 - f1x2 h

=

f15 + 0.12 - f152 0.1

=

f15.12 - f152 0.1

.

Now f15.12 = 15.122 = 26.01 and f152 = 25, and we have f15.12 - f152 0.1

=

26.01 - 25 1.01 = = 10.1. 0.1 0.1

c) We substitute x = 5 and h = 0.01 into the formula: f1x + h2 - f1x2 h

=

f15 + 0.012 - f152 0.01

=

f15.012 - f152 0.01

.

Now f15.012 = 15.0122 = 25.1001 and f152 = 25, and we have f15.012 - f152 0.01

=

25.1001 - 25 0.1001 = = 10.01. 0.01 0.01

Note the trend in the average rate of change as h gets closer to 0. For the function in Example 4, let’s find a form of the difference quotient that will allow for more efficient computations.

126

CHAPTER 1

• Differentiation

■ EXAMPLE 5 For f1x2 = x 2, find a simplified form of the difference quotient. Then

find the value of the difference quotient when x = 5 and h = 0.1 and when x = 5 and h = 0.01.

Solution We have f1x2 = x 2, so

Then

f1x + h2 = 1x + h22 = x 2 + 2xh + h2.

Multiplying 1x + h21x + h2

f1x + h2 - f1x2 = 1x 2 + 2xh + h22 - x 2 = 2xh + h2.

The x 2 terms sum to 0.

Thus, f1x + h2 - f1x2 h

=

h12x + h2 2xh + h2 = = 2x + h, h Z 0. h h

h>h = 1

This is a simplified form of this difference quotient. It is important to note that any difference quotient is defined only when h Z 0. The above simplification is valid only for nonzero values of h. When x = 5 and h = 0.1, f1x + h2 - f1x2 h

= 2x + h = 2 # 5 + 0.1 = 10 + 0.1 = 10.1.

When x = 5 and h = 0.01, f1x + h2 - f1x2 h

= 2x + h = 2 # 5 + 0.01 = 10.01.

Although the expression 2x + h is valid only when h Z 0, there is nothing stopping us from allowing h to get closer and closer to 0. Perhaps you can sense what the value of the difference quotient would be in this example if you allowed h to get close to 0 as a limit. Compare the results of Example 4(b) and 4(c) and Example 5. In general, computations are easier when a simplified form of a difference quotient is found before any specific calculations are performed.

■ EXAMPLE 6 For f1x2 = x 3, find a simplified form of the difference quotient. Solution For f1x2 = x 3,

f1x + h2 = 1x + h23 = x 3 + 3x 2h + 3xh2 + h3.

(This is shown in Appendix A at the end of this book.) Then

f1x + h2 - f1x2 = 1x 3 + 3x 2h + 3xh2 + h32 - x 3 = 3x 2h + 3xh2 + h3.

Thus, f1x + h2 - f1x2 h Quick Check 3 Use the result of Example 6 to calculate the slope of the secant line (average rate of change) at x = 2, for h = 0.1, h = 0.01, and h = 0.001.

= =

3x 2h + 3xh2 + h3 h

h13x 2 + 3xh + h22

h 2 = 3x + 3xh + h2, h Z 0.

Again, this is true only for h Z 0.

O Quick Check 3

It is understood that h Z 0. Factoring out h; h>h = 1

O

1.3

• Average Rates of Change

127

The next two examples illustrate the development of the difference quotients for a simple rational function (Example 7) and for the square-root function (Example 8). These two forms are very common in calculus.

■ EXAMPLE 7 For f1x2 = 1>x, find a simplified form of the difference quotient. Solution For f1x2 = 1>x, f1x + h2 =

1 . x + h

Then f1x + h2 - f1x2 =

= =

1 1 x x + h 1 #x 1 x + h - # x x + h x + h x x - 1x + h2

Here we are multiplying by 1 to get a common denominator.

x1x + h2 x - x - h = x1x + h2 =

-h . x1x + h2

Thus, f1x + h2 - f1x2

=

h

-h x1x + h2

h -h # 1 = - 1 , h Z 0. = x1x + h2 h x1x + h2

This is true only for h Z 0.

■ EXAMPLE 8 For f1x2 = 2x, find a simplified form of the difference quotient. Solution For f1x2 = 2x, we have f1x + h2 = 2x + h, so the difference quotient is f1x + h2 - f1x2 h

=

2x + h - 2x . h

Algebraic simplification of this difference quotient leads to f1x + h2 - f1x2 h

=

1 2x + h + 2x

, h Z 0.

Demonstration of this simplification is left as Exercise 54. You should note this simplified difference quotient as it will be seen again in Section 1.5. In all of the above cases where we simplified a difference quotient using algebra, we ended up with two variables: x and h, where h cannot be 0. Although h cannot be exactly 0, we may let h be as close to 0 as we desire. This is a limit! In the next section, we will take that final step: allowing h to approach 0 as a limit.

128

CHAPTER 1

• Differentiation

Section Summary • An average rate of change is the slope of a line between two points. If the two points are 1x 1, y12 and 1x 2, y22, then y2 - y1 the average rate of change is . x2 - x1 • If the two points are solutions to a single function, an f1x + h2 - f1x2 equivalent form of the slope formula is , h where h is the horizontal difference between the two

x-values. This is called the difference quotient. The line connecting these two points is called a secant line. • The difference quotient is the same as the slope formula. Both give the slope of the line between two points. • The difference quotient gives the average rate of change between two points on a graph, represented by the secant line. • It is preferable to simplify a difference quotient algebraically before evaluating it for particular values of x and h.

EXERCISE SET

1.3

0.1

5

0.01

1. f 1x2 = 4x 2

2. f 1x2 = 5x 2

4. f 1x2 = - 5x 2

5. f 1x2 = x 2 + x

6. f 1x2 = x 2 - x

7. f 1x2 =

8. f 1x2 =

9. f 1x2 = - 2x + 5

11. f 1x2 = 1 - x 3

13. f 1x2 = x 2 - 3x 15. f 1x2 = x 2 + 4x - 3

2005

2009

19. Professional services.

3. f 1x2 = - 4x 2

2 x

2000 −2.0

9 x

10. f 1x2 = 2x + 3 12. f 1x2 = 12x 3

14. f 1x2 = x 2 - 4x 16. f 1x2 = x 2 - 3x + 5

APPLICATIONS Business and Economics For Exercises 17–24, use each graph to estimate the average rate of change of the percentage of new employees in that type of

8.0 4.0 2000 −4.0

2005

2009

21. Education.

30.0 20.0 10.0 2005

2009

23. Mining and logging. 30.0 20.0 10.0 2000

2005

10.0 2000 −10.0

2005

2009

20. Health care. 30.0 20.0 10.0 2000

2005

2009

22. Government.

40.0

2000

Percent of New Employees

5

2.0

20.0

Percent of New Employees

1

4.0

2009

Percent of New Employees

5

6.0

18. Construction.

8.0 6.0 4.0 2.0 2000

2005

2009

24. Manufacturing. Percent of New Employees

2

h

Percent of New Employees

5

f 1x + h2 - f 1x2

Percent of New Employees

h

17. Total employment.

Percent of New Employees

x

employment from 2000 to 2005, from 2005 to 2009, and from 2000 to 2009. (Source: Bureau of Labor Statistics.)

Percent of New Employees

For each function in Exercises 1–16, (a) find a simplified form of the difference quotient and then (b) complete the following table.

2000 −10.0 −20.0 −30.0 −40.0

2005

2009

Exercise Set 1.3

25. Use the following graph to find the average rate of change in U.S. energy consumption from 1970 to 1980, from 1980 to 1990, and from 2000 to 2009.

28. Advertising results. The following graph shows a typical response to advertising. After an amount a is spent on advertising, the company sells N1a2 units of a product. N

U.S. ENERGY CONSUMPTION

(4, 700)

.60

Number of units sold

95

.90

98 .67

84

.29

78 .84

67

U.S. energy consumption (in quadrillions of BTUs)

700 105 90 75 60 45 30 15 0

’80

’90

’00

(2, 480)

500 400

(1, 300)

300 200

’09 0

(Source: U.S. Energy Information Administration.)

26. Use the following graph to find the average rate of change of the U.S. trade deficit with Japan from 1990 to 1995, from 1995 to 2000, and from 2000 to 2009. U.S. TRADE DEFICIT WITH JAPAN

Goods and services (in billions of dollars)

(3, 600)

600

100 ’70

Year

41.1

40

40.1

20 1990

1995

2000

2009

Year (Source: U.S. Census Bureau, Statistical Abstract of the United States, 2010.)

27. Utility. Utility is a type of function that occurs in economics. When a consumer receives x units of a certain product, a certain amount of pleasure, or utility U, is derived. The following is a graph of a typical utility function.

Utility (pleasure units)

U 200 (4, 161) (3, 138)

150

30. Compound interest. The amount of money, A1t2, in a savings account that pays 6% interest, compounded quarterly for t years, when an initial investment of $2000 is made, is given by A1t2 = 2000 11.01524t.

a) Find A132. b) Find A152. c) Find A152 - A132. A152 - A132 , and interpret this result. d) Find 5 - 3

A1t2 = 5000 11.142t.

(1, 70)

50

0

where x is the number of years after 1991 and p1x2 is in dollars. (Source: Based on data from www.teammarketing.com.) a) Find p142. b) Find p1172. c) Find p1172 - p142. p1172 - p142 d) Find , and interpret this result. 17 - 4

31. Credit card debt. When a balance of $5000 is owed on a credit card and interest is being charged at a rate of 14% per year, the total amount owed after t years, A1t2, is given by

U

(2, 109) 100

a) Find the average rate of change of N as a changes from 0 to 1; from 1 to 2; from 2 to 3; from 3 to 4. b) Why do you think the average rates of change are decreasing as x increases?

p1x2 = 0.03x 2 + 0.56x + 8.63,

59.1

60

Find 1 2 3 4 Number of units of product

a 1 2 3 4 Money spent (in thousands of dollars)

29. Baseball ticket prices. Based on data from Major League Baseball, the average price of a ticket to a major league game can be approximated by

81.6

80

0

129

x

a) Find the average rate of change of U as x changes from 0 to 1; from 1 to 2; from 2 to 3; from 3 to 4. b) Why do you think the average rates of change are decreasing as x increases?

A132 - A122 3 - 2

, and interpret this result.

32. Credit card debt. When a balance of $3000 is owed on a credit card and interest is charged at a rate of 17% per year, the total amount owed after t years, A1t2, is given by A1t2 = 3000 11.172t.

Find

A142 - A132 4 - 3

, and interpret this result.

CHAPTER 1

• Differentiation

33. Total cost. Suppose that Sport Stylz Inc. determines that the cost, in dollars, of producing x cellphone-sunglasses is given by C1x2 = - 0.05x 2 + 50x. Find

C13012 - C13002

, and interpret the significance of 301 - 300 this result to the company. 34. Total revenue. Suppose that Sports Stylz Inc. determines that the revenue, in dollars, from the sale of x cellphonesunglasses is given by R1x2 = - 0.01x 2 + 1000x. R13012 - R13002

, and interpret the significance of 301 - 300 this result to the company. Find

Life and Physical Sciences

40 36 32 28 24 20 16 12 8 4 0 Birth 6 12 18 24 30 36

Age (in months)

Median weight of baby boys (in pounds)

Median weight of baby girls (in pounds)

35. Growth of a baby. The median weights of babies at age t months are graphed below. 40 36 32 28 24 20 16 12 8 4 0 Birth 6 12 18 24 30 36

Age (in months)

(Source: Developed by the National Center for Health Statistics in collaboration with the National Center for Chronic Disease Prevention and Health Promotion, 2000.)

Use the graph of girls’ median weight to estimate: a) The average growth rate of a girl during her first 12 months. (Your answer should be in pounds per month.) b) The average growth rate of a girl during her second 12 months. c) The average growth rate of a girl during her first 24 months. d) Based on your answers in parts (a)–(c) and the graph, estimate the growth rate of a typical 12-month-old girl. Use a straightedge. e) When does the graph indicate that a baby girl’s growth rate is greatest? 36. Growth of a baby. Use the graph of boys’ median weight in Exercise 35 to estimate: a) The average growth rate of a boy during his first 15 months. (Your answer should be in pounds per month.) b) The average growth rate of a boy during his second 15 months. (Your answer should be in pounds per month.)

c) The average growth rate of a boy during his first 30 months. (Your answer should be in pounds per month.) d) Based on your answers in parts (a)–(c) and the graph, estimate the growth rate of a typical boy at exactly 15 months, and explain how you arrived at this figure. 37. Home range. It has been shown that the home range, in hectares, of a carnivorous mammal weighing w grams can be approximated by H1w2 = 0.11w 1.36. (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. 200, Reading, MA: AddisonWesley, 1973; and Harestad, A. S., and Bunnel, F. L., “Home Range and Body Weight—A Reevaluation,” Ecology, Vol. 60, No. 2, pp. 389–402.) a) Find the average rate at which a carnivorous mammal’s home range increases as the animal’s weight grows from 500 g to 700 g. H13002 - H12002 , and interpret this result. b) Find 300 - 200 38. Radar range. The function given by R1x2 = 11.74x 1/4 can be used to approximate the maximum range R1x2, in miles, of an ARSR-3 surveillance radar with a peak power of x watts (W). (Source: Introduction to RADAR Techniques, Federal Aviation Administration, 1988.) a) Find the rate at which the maximum radar range changes as peak power increases from 40,000 W to 60,000 W. R160,0002 - R150,0002 , and interpret this b) Find 60,000 - 50,000 result. 39. Memory. The total number of words, M1t2, that a person can memorize in t minutes is shown in the following graph. M Number of words memorized

130

40 (32, 25)

30 20

(24, 25) (16, 20)

10

0

(36, 25)

(8, 10)

8

16 24 32 36 40 Time (in minutes)

t

a) Find the average rate of change of M as t changes from 0 to 8; from 8 to 16; from 16 to 24; from 24 to 32; from 32 to 36. b) Why do the average rates of change become 0 after 24 min? 40. Average velocity. Suppose that in t hours, a truck travels s1t2 miles, where s1t2 = 10t 2.

Exercise Set 1.3

41. Average velocity. In t seconds, an object dropped from a certain height will fall s1t2 feet, where s1t2 = 16t 2. a) Find s152 - s132. b) What is the average rate of change of distance with respect to time during the period from 3 to 5 sec? This is also average velocity.

DISAPPEARING BANKS Number of federally insured commercial banks

a) Find s152 - s122. What does this represent? b) Find the average rate of change of distance with respect to time as t changes from t1 = 2 to t2 = 5. This is known as average velocity, or speed.

14,000 12,000

8,000 6,000

6,911 ‘87 ‘89 ‘91 ‘93 ‘95 ‘97 ‘99 ‘01 ‘03 ‘05 ‘07 ‘09

Year (Source: Federal Deposit Insurance Corp.)

2250

Social Sciences

2000

NASDAQ COMPOSITE INDEX (August 2009 to February 2010)

y 2300

2274

2176

2200

2269 2057

2150 2100

2171

2009

2050

2049

1950

43. Population growth. The two curves below describe the numbers of people in two countries at time t, in years.

1900 Aug

1969 0

Sep

1

Oct

2

Nov

3

Dec

4

Jan

5

Feb

6

Mar x

(Source: Nasdaq.)

P

A B

500 Population (in millions)

13,723

10,000

42. Gas mileage. At the beginning of a trip, the odometer on a car reads 30,680, and the car has a full tank of gas. At the end of the trip, the odometer reads 31,077. It takes 13.5 gal of gas to refill the tank. a) What is the average rate at which the car was traveling, in miles per gallon? b) What is the average rate of gas consumption in gallons per mile?

Explain the difference between these graphs in as many ways as you can. Be sure to mention average rates of change.

(4, 500)

400 300

131

(3, 375)

45. Rising cost of college. Like the cost of most things, the cost of a college education has gone up over the past 35 years. The graphs below display the yearly costs of 4-year colleges in 2008 dollars—indicating that the costs prior to 2008 have been adjusted for inflation.

(1, 290) (2, 250)

200 (3, 200)

RISING COLLEGE COSTS

(1, 125)

100

0

1

2 3 Time (in years)

4

t

a) Find the average rate of change of each population with respect to time t as t changes from 0 to 4. This is often called the average growth rate. b) If the calculation in part (a) were the only one made, would we detect the fact that the populations were growing differently? Explain. c) Find the average rates of change of each population as t changes from 0 to 1; from 1 to 2; from 2 to 3; from 3 to 4. d) For which population does the statement “the population grew consistently at a rate of 125 million per year” convey accurate information? Why?

SYNTHESIS 44. Business: comparing rates of change. The following two graphs show the number of federally insured banks and the Nasdaq Composite Stock Index over a 6-month period.

Average charges at four-year institutions (in 2008 dollars)

30,000

Four-year private

20,000

10,000

0 1975–76

Four-year public

1989–90 School year

2007–08

(Source: National Center for Educational Statistics, Annual Digest of Educational Statistics, 2008.)

a) In what school year did the cost of a private 4-year college increase the most? b) In what school year(s) did the cost of a public 4-year college increase the most? c) Assuming an annual inflation rate of 3%, calculate the cost of a year at a public and at a private 4-year college in 1975. Express the costs in 1975 dollars.

132

CHAPTER 1

• Differentiation

Find the simplified difference quotient for each function listed. 47. f 1x2 = ax + bx + c

46. f 1x2 = m x + b

2

48. f 1x2 = ax + bx 3

49. f 1x2 = x

2

50. f 1x2 = x 5 52. f 1x2 =

1 53. f 1x2 = 1 - x

x2

h a) =

b) =

=

d) =

h

e) =

54. Below are shown all of the steps in the simplification of the difference quotient for f 1x2 = 2x (see Example 8). Provide a brief justification for each step. f 1x + h2 - f 1x2

h A 2x + h + 2x B

4

51. f 1x2 = ax 5 + bx 4

1

x + h - x

c) =

h A 2x + h + 2x B 1 2x + h + 2x

For Exercises 55 and 56, find the simplified difference quotient. 55. f 1x2 = 22x + 1

2x + h - 2x h

56. f 1x2 =

1 2x

2x + h - 2x # 2x + h + 2x a b h 2x + h + 2x

Answers to Quick Checks

x + h + 2x2x + h - 2x2x + h - x

1. (a) It rained 1>2 inch per hour. (b) Your car gets 12.5 miles per gallon. (c) The temperature dropped 2 degrees per hour. 2. (a) 21; (b) 7; (c) 3.64 3. 12.61, 12.0601, 12.006001

h A 2x + h + 2x B

1.4 OBJECTIVES • Find derivatives and values of derivatives. • Find equations of tangent lines.

Differentiation Using Limits of Difference Quotients

The slope of the secant line connecting two points 1x, f1x22 and 1x + h, f1x + h22 on the graph of a function y = f1x2 represents the average rate of change of f1x2 over the interval 3x, x + h4. This rate is given by the difference quotient f1x + h2 - f1x2 h

,

h Z 0.

In Section 1.3, we worked out difference quotients for several functions, simplifying them as much as possible. These simplified difference quotients contain the variables x and h, and we can calculate the slope of a secant line by evaluating the difference quotient for a given x-value and a given h-value. Recall that h represents the horizontal distance between x and x + h. Although h cannot equal zero, we can consider the case where h approaches zero as a limit. In this section, we explore this possibility.

Tangent Lines A line that touches a circle at exactly one point is called a tangent line. The word “tangent” derives from the Latin tangentem, meaning “touch,” whereas the word “secant” derives from the Latin secantem, meaning “cut.” In the figure below, the secant line cuts through the circle, while the tangent line touches, but does not cut through, the circle. Secant line

Tangent line

1.4

• Differentiation Using Limits of Difference Quotients

133

These notions can be extended to any smooth curve: a tangent line touches a curve at a single point only, in the same way as the tangent line touches the circle in the figure at the bottom of p. 132. In Fig. 1, the line L touches the curve exactly once in the small interval containing P, the point of tangency. We are not concerned with the behavior of the line far from the point of tangency. We see that L does pass through the curve elsewhere, but it is still considered a tangent line to the curve at point P. y P L

x

FIGURE 1

In Figure 2, line M crosses the curve only at point P, but is not tangent to the curve; it does not touch the curve in the desired manner. y

M M is not a tangent line. P

x

FIGURE 2

The power lines run tangent to the tower brackets at the turn.

In Figure 3, all of the lines except for L 1 and L 2 are tangent lines. y

TECHNOLOGY CONNECTION

L1

Exploratory Graph y1 = 3x 5 - 20x 3 with the viewing window 3 - 3, 3, - 80, 804, with Xscl = 1 and Yscl = 10. Then also graph the lines y2 = - 7x - 10, y3 = - 30x + 13, and y4 = - 45x + 28. Which line appears to be tangent to the graph of y1 at 11, - 172? If necessary, zoom in near 11, - 172 to refine your guess.

T4 L2 T1

T5

T3 T2 x

FIGURE 3

One more observation: if a curve is smooth (has no corners), then each point on the curve will have a unique tangent line; that is, exactly one tangent line is possible at any given point.

134

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Differentiation Using Limits We now define tangent line so that it makes sense for any curve. To do this, we use the notion of limit. To obtain the line tangent to the curve at point P, consider secant lines through P and neighboring points Q1, Q2, and so on. As the Q’s approach P, the secant lines approach line T. Each secant line has a slope. The slopes m 1, m 2, m 3, and so on, of the secant lines approach the slope m of line T. We define line T as the tangent line, the line that contains point P and has slope m, where m is the limit of the slopes of the secant lines as the points Q approach P. Slope m 1 Q1

Slope m 2 Secant lines

Q2

Slope m 3 Q3

Slope m 4 T

Q4

Slope m = instantaneous rate of change at P

Tangent line

P

Think of the sequence of secant lines as an animation: as the points Q move closer to the fixed point P, the resulting secant lines “lie down” on the tangent line. How might we calculate the limit m? Suppose that P has coordinates 1x, f1x22. Then the first coordinate of Q is x plus some number h, or x + h. The coordinates of Q are 1x + h, f1x + h22, as shown in Fig. 4.

Q(x + h, f(x + h))

f(x + h)

Slope =

f(x + h) – f(x) h

Q1 Q2

P(x, f(x)) P

f(x) x

x+h

x

FIGURE 4

Q4

Q3

Slope m = lim f(x + h) – f(x) h h→0 T

x + h1 x + h2 x + h3 x + h4

FIGURE 5

Í ! From Section 1.3, we know that the slope of the secant line PQ is given by f1x + h2 - f1x2 h

.

Now, as we see in Fig. 5, as the Q’s approach P, the values of x + h approach x. That is, h approaches 0. Thus, we have the following.

The slope of the tangent line at 1x, f1x22 = m = lim

h :0

f1x + h2 - f1x2

h This limit is also the instantaneous rate of change of f1x2 at x.

.

1.4

• Differentiation Using Limits of Difference Quotients

135

The formal definition of the derivative of a function f can now be given. We will designate the derivative at x as f ¿1x2, rather than m. The notation f ¿1x2 is read “the derivative of f at x,” “f prime at x,” or “f prime of x.” “Nothing in this world is so powerful as an idea whose time has come.” Victor Hugo

DEFINITION For a function y = f1x2, its derivative at x is the function f ¿ defined by f ¿1x2 = lim

f1x + h2 - f1x2

h :0

h

,

provided that the limit exists. If f ¿1x2 exists, then we say that f is differentiable at x. We sometimes call f ¿ the derived function.

Let’s now calculate some formulas for derivatives. That is, given a formula for a function f, we will attempt to find a formula for f ¿.

f '(x) = slope at (x, f(x)) f(x)

(x, f(x))

x

There are three steps in calculating a derivative.

1. Write the difference quotient, 1f1x + h2 - f1x22>h. 2. Simplify the difference quotient. 3. Find the limit as h approaches 0.

■ EXAMPLE 1 For f1x2 = x 2, find f ¿1x2. Then find f ¿1- 32 and f ¿142. Solution We have 1. 2.

f1x + h2 - f1x2 h f1x + h2 - f1x2 h

=

1x + h22 - x 2 h

x 2 + 2xh + h2 - x 2 h 2 h12x + h2 2xh + h = = h h = 2x + h, h Z 0 =

Evaluating f 1x + h2 and f 1x2

s Simplifying

3. We want to find lim

h :0

f1x + h2 - f1x2 h

= lim 12x + h2. h :0

Recall that in Section 1.3, we calculated the slope of a secant line by evaluating the difference quotient at a particular x-value and a series of h-values that grew closer to zero: h = 0.1, h = 0.01, and so on. If we allow h to approach 0 as close as we desire, we have the derivative. That is, for lim 12x + h2 = 2x,

h:0

136

CHAPTER 1

• Differentiation

we have f ¿1x2 = 2x. Using the fact that f ¿1x2 = 2x, it follows that

f ¿1- 32 = 2 # 1- 32 = - 6, and f ¿142 = 2 # 4 = 8.

This tells us that at x = - 3, the curve has a tangent line whose slope is f ¿1- 32 = - 6, and at x = 4, the tangent line has slope f ¿142 = 8. We can also say: • • • •

The tangent line to the curve at the point 1- 3, 92 has slope - 6. The tangent line to the curve at the point 14, 162 has slope 8. The instantaneous rate of change at x = - 3 is - 6. The instantaneous rate of change at x = 4 is 8. y

TECHNOLOGY CONNECTION

20

Exploratory

18

To see exactly how the definition of derivative works and to check the results of Example 1, let y1 = 11- 3 + x22 - 1- 3222>x and note that we are using x in place of h. Use the TABLE feature and enter smaller and smaller values for x. Then repeat the procedure for y2 = 114 + x22 4 22>x. How do these results confirm those in Example 1?

(4, 16)

16 14 12

(−3, 9)

10 8 6

f (x) = x2

4

m = −6

m=8

2

−5 −4 −3 −2 −1

1

2

3

4

5

x

■ EXAMPLE 2 For f1x2 = x 3, find f ¿1x2. Then find f ¿1- 12 and f ¿11.52. Solution 1. We have f1x + h2 - f1x2 h

=

1x + h23 - x 3 h

.

2. In Example 6 of Section 1.3 (on p. 126), we showed how this difference quotient can be simplified to f1x + h2 - f1x2 h

= 3x 2 + 3xh + h2,

h Z 0.

3. We then have f ¿1x2 = lim

h:0

f1x + h2 - f1x2 h

= lim 13x 2 + 3xh + h22 = 3x 2. h:0

Thus, f ¿1- 12 = 31- 122 = 3 and f ¿11.52 = 311.522 = 6.75.

1.4

• Differentiation Using Limits of Difference Quotients

137

y 5 4

(1.5, 3.375)

3 2

m = 6.75

1 −2

−1

(−1, −1)

Quick Check 1 Use the results from Examples 1 and 2 to find the derivative of f1x2 = x 3 + x 2, and then calculate f ¿1- 22 and f ¿142. Interpret these results.

−1

1

x

2

−2

m=3

−3 −4

f(x) = x3

−5

O Quick Check 1 A common error in simplifying a difference quotient is to write f1x2 + h, which is incorrect! The expression x + h must remain a unit when substituting for x in the function. For example, if f1x2 = x 2, we write f1x + h2 = 1x + h22. Once we have set up the expression for the difference quotient correctly, we can then simplify it using normal algebraic techniques.

■ EXAMPLE 3 For f1x2 = 3x - 4, find f ¿1x2 and f ¿122. Solution We follow the three steps given above.

y 7

1.

6 5

2.

4 3

f1x + h2 - f1x2 h f1x + h2 - f1x2 h

m=3

2 1 −2 −1

−1

1

2

3

4

x

3. lim

h :0

−2 −3

=

31x + h2 - 4 - 13x - 42 h

3x + 3h - 4 - 3x + 4 h 3h = = 3, h Z 0; h =

f1x + h2 - f1x2 h

The parentheses are important.

Using the distributive law Simplifying

= lim 3 = 3, since 3 is a constant. h :0

Thus, if f1x2 = 3x - 4, then f ¿1x2 = 3 and f ¿122 = 3.

f(x) = 3x − 4

The result of Example 3 suggests that, for a straight line, the slope of a tangent line is the slope of the straight line itself. That is, a general formula for the derivative of a linear function is

f1x2 = mx + b f ¿1x2 = m.

The formula can be verified in a manner similar to that used in Example 3. Examples 1–3 and Example 4, which follows, involve a somewhat lengthy process, but in Section 1.5 we will develop some faster techniques. It is very important in this section, however, to fully understand the concept of a derivative. 1

■ EXAMPLE 4 For f1x2 = : x a) Find f ¿1x2. b) Find f ¿122. c) Find an equation of the tangent line to the curve at x = 2.

O

138

CHAPTER 1

• Differentiation

TECHNOLOGY CONNECTION

Solution

Curves and Tangent Lines

a) 1. We have

EXERCISES

1. For f 1x2 = 3>x, find f ¿1x2, f ¿1- 22, and f ¿ A

- 12

f1x + h2 - f1x2

B.

h

=

31>1x + h24 - 11>x2 h

.

2. In Example 7 of Section 1.3 (on p. 127), we showed that this difference quotient simplifies to

2. Find an equation of the tangent line to the graph of

f 1x2 = 3>x at A - 2, - 12 B and an equation of the tangent line to the curve at A - 12, - 6 B . Then graph the curve f 1x2 = 3>x and both tangent lines. Use ZOOM to view the graphs near the points of tangency.

f1x + h2 - f1x2 h

=

-1 , h Z 0. x1x + h2

3. We want to find

3. To check your equations of tangent lines, first graph

f 1x2 = 3>x. Select Tangent from the DRAW menu, and enter first x = - 2 and then x = - 12.

lim

f1x + h2 - f1x2

h :0

= lim

h :0

h

-1 . x 1x + h2

As h : 0, we have x + h : x. Thus, -1 -1 -1 = # = 2. x x h : 0 x1x + h2 x

f ¿1x2 = lim b) Since f ¿1x2 = - 1>x 2, we have f ¿122 =

-1 1 = - . 2 4 2

This is the slope of the tangent line at x = 2.

c) We can find an equation of the tangent line at x = 2 if we know the line’s slope and a point that is on the line. In part (b), we found that the slope at x = 2 is - 14. To find a point on the line, we compute f122: f122 =

1 . 2

CAUTION! Be careful to use f when computing y-values and f ¿ when computing slope.

We have

Point: A 2, 12 B , Slope: - 14.

This is 1x 1, f 1x 122. This is f ¿1x 12.

We substitute into the point–slope equation (see Section R.4): y - y1 = m1x - x 12 y - 12 = - 411x - 22

y = - 14 x + 12 + = - 14 x + 1.

1 2

s

Rewriting in slope–intercept form

The equation of the tangent line to the curve at x = 2 is

y 1 f(x) = − x

y = - 14 x + 1.

2

y = − −14 x + 1 2

1

1

1 1

Quick Check 2 Repeat Example 4a for 2 f1x2 = - . What are the x similarities in your method?

2

O Quick Check 2

(2, −12 ) 2

x

O

139

• Differentiation Using Limits of Difference Quotients

1.4

In Example 4, note that since f102 does not exist for f1x2 = 1>x, we cannot evaluate the difference quotient f10 + h2 - f102 h

.

Thus, f ¿102 does not exist. We say that “f is not differentiable at 0.” When a function is not defined at a point, it is not differentiable at that point. In general, if a function is discontinuous at a point, it is not differentiable at that point.

Sometimes a function f is continuous at a point, but its derivative f ¿ is not defined at this point. The function f1x2 = ƒ x ƒ is an example. It is continuous at x = 0 since it meets all the requirements for continuity there. But what about a tangent line at this point? y

f(x) = |x|

x

(0, 0)

Suppose that we try to draw a tangent line at 10, 02. A function like this with a corner (not smooth) would seem to have many tangent lines at 10, 02, and thus many slopes. The derivative at such a point would not be unique. Let’s try to calculate the derivative at 0. Since f1x2 = ƒ x ƒ = e

x, for x Ú 0, - x, for x 6 0,

it follows from our earlier work with lines that f ¿1x2 = e

1, for x 7 0, - 1, for x 6 0.

Then, since lim f ¿1x2 Z lim- f ¿1x2,

x : 0+

x :0

it follows that f ¿102 does not exist. In general, if a function has a “corner,” it will not have a derivative at that point. The following graphs show examples of “corners.” y

y

a

x

y

a

x

a1

a2

x

140

CHAPTER 1

• Differentiation y

A function will also fail to be differentiable at a point if it has a vertical tangent at that point. For example, the function shown to the right has a vertical tangent at point a. Recall that since the slope of a vertical line is undefined, there is no derivative at such a point.

Vertical tangent x

a

The function given by f1x2 = ƒ x ƒ illustrates the fact that although a function may be continuous at each point in an interval I, it may not be differentiable at each point in I. That is, continuity does not imply differentiability. However, differentiability does guarantee continuity. That is, if f ¿1a2 exists, then f is continuous at a. The function f1x2 = x 2 is an example of a function that is differentiable over the interval 1- q , q 2 and is therefore continuous everywhere. Thus, when a function is differentiable over an interval, it is not just continuous, but is also smooth in the sense that there are no “corners” in its graph.

■ EXAMPLE 5 Below is the graph of a function y = t1x2. List the points in the graph at which the function t is not differentiable. y y = t(x)

a

Quick Check 3

O

Where is f1x2 = ƒ x + 6 ƒ not differentiable? Why?

b

c

d

e

f

g

h

i

j

x

Solution A function is not differentiable at a point if there is (1) a discontinuity, (2) a corner, or (3) a vertical tangent at that point. Therefore, the function y = t1x2 is not differentiable at x = b, x = c, and x = f since the function is discontinuous at these points; it is not differentiable at x = d and x = i since there are corners at these points; and it is not differentiable at x = j as there is a vertical tangent line at this point (the slope is undefined). The function is differentiable at x = a, x = e, x = g, and x = h.

O Quick Check 3

TECHNOLOGY CONNECTION

Calculus Apps for the iPhone or iPod Touch A Technology Connection in Section R.5 introduced two apps for the iPhone and iPod Touch: iPlot and Graphicus (both available from the iTunes Store). Here we begin to cover uses of Graphicus in calculus.

Graphicus This app can graph most of the functions we encounter in this book and draw tangent lines at points on a curve. Let’s consider the function f 1x2 = x 4 - 2x 2. To graph a function and examine its tangent lines, first open Graphicus. Touch the blank rectangle at the top of the screen and enter the function as y(x)=x^4-2x^2. Press + at the upper right. You will see the graph, as shown in Fig. 1.

Notice the seven icons at the bottom of the screen. The first one is for zooming. Touch it to zoom in and zoom out. Touch the screen with two fingers and move vertically or horizontally to adjust the windows and scaling. The second icon is for visualizing tangent lines. Touch it and then touch the graph at a point, and you will see a highlighted point on the graph and a tangent line to the curve at that point (Fig. 2). Move your finger left to right along the graph and you will see highlighted various other points on the curve and tangent lines at those points (Figs. 3 and 4). Think about the slopes of the tangent lines as you move from left to right. When are they positive, negative, or 0? Note where the slope changes from positive to 0 to negative.

(continued)

Exercise Set 1.4

FIGURE 2

FIGURE 1

EXERCISES Use Graphicus to graph each function. You may need to use the zoom function to alter the window. Visualize tangent lines, noting where they are positive, negative, and 0. Then try to determine where the slope changes from positive to 0 to negative. 1. f 1x2 = 2x 3 - x 4

2. f 1x2 = x1200 - x2

3. f 1x2 = x 3 - 6x 2

141

FIGURE 4

FIGURE 3

4. f 1x2 = - 4.32 + 1.44x + 3x 2 - x 3 5. g1x2 = x24 - x 2 6. g1x2 =

4x x2 + 1

7. f 1x2 =

x 2 - 3x x - 1

8. f 1x2 = ƒ x + 2 ƒ - 3. What happens to the tangent line

at the point 1- 2, - 32?

Section Summary • A tangent line is a line that touches a (smooth) curve at a single point, the point of tangency. See Fig. 3 (on p. 133) for examples of tangent lines (and lines that are not considered tangent lines). • The derivative of a function f1x2 is defined by f ¿1x2 = lim

f1x + h2 - f1x2

h:0

h

.

• The slope of the tangent line to the graph of y = f1x2 at x = a is the value of the derivative at x = a; that is, the slope of the tangent line at x = a is f ¿1a2. • Slopes of tangent lines are interpreted as instantaneous rates of change.

• The equation of a tangent line at x = a is found by simplifying y - f1a2 = f ¿1a21x - a2. • If a function is differentiable at a point x = a, then it is continuous at x = a. That is, differentiability implies continuity. • However, continuity at a point x = a does not imply differentiability at x = a. A good example is the absolutevalue function, f1x2 = ƒ x ƒ , or any function whose graph has a corner. Continuity alone is not sufficient to guarantee differentiability. • A function is not differentiable at a point x = a if: (1) there is a discontinuity at x = a, (2) there is a corner at x = a, or (3) there is a vertical tangent at x = a.

EXERCISE SET

1.4 In Exercises 1–16: a) Graph the function. b) Draw tangent lines to the graph at points whose x-coordinates are 2, 0, and 1. f1x  h2  f1x2 c) Find f œ 1x2 by determining lim . h:0 h

d) Find f œ 122, f œ 102, and f œ 112. These slopes should match those of the lines you drew in part (b). 1. f 1x2 = 32 x 2

2. f 1x2 = 12 x 2

5. f 1x2 = x 3

6. f 1x2 = - x 3

3. f 1x2 = - 2x 2

4. f 1x2 = - 3x 2

142

• Differentiation

CHAPTER 1

7. f 1x2 = 2x + 3 9. f 1x2 =

1 2x

8. f 1x2 = - 2x + 5

10. f 1x2 =

- 3

3 4x

- 2

11. f 1x2 = x + x

12. f 1x2 = x - x

15. f 1x2 =

16. f 1x2 =

2

28.

f

2

14. f 1x2 = 5x 2 - 2x + 7

13. f 1x2 = 2x 2 + 3x - 2 1 x

2 x

17. Find an equation of the tangent line to the graph of f 1x2 = x 2 at (a) 13, 92; (b) 1 - 1, 12; (c) 110, 1002. See Example 1. 18. Find an equation of the tangent line to the graph of f 1x2 = x 3 at (a) 1 - 2, - 82; (b) 10, 02; (c) 14, 642. See Example 2. 19. Find an equation of the tangent line to the graph of f 1x2 = 2>x at (a) 11, 22; (b) 1- 1, - 22; (c) 1100, 0.022. See Exercise 16. 20. Find an equation of the tangent line to the graph of f 1x2 = - 1>x at (a) 1- 1, 12; (b) A 2, - 12 B ; (c) A - 5, 15 B . 21. Find an equation of the tangent line to the graph of f 1x2 = 4 - x 2 at (a) 1- 1, 32; (b) 10, 42; (c) 15, - 212. 22. Find an equation of the tangent line to the graph of f 1x2 = x 2 - 2x at (a) 1- 2, 82; (b) 11, - 12; (c) 14, 82.

23. Find f ¿1x2 for f 1x2 = mx + b.

For Exercises 25–28, list the points in the graph at which each function is not differentiable. 25.

x2 x3 x4

x5

x6

x7

x8

x9 x10

29. Draw a graph that is continuous, but not differentiable, at x = 3. 30. Draw a graph that has a horizontal tangent line at x = 5. 31. Draw a graph that is differentiable and has horizontal tangent lines at x = 0, x = 2, and x = 4. 32. Draw a graph that has horizontal tangent lines at x = 2 and x = 5 and is continuous, but not differentiable, at x = 3. 33. Draw a graph that is smooth, but not differentiable, at x = 1. 34. Draw a graph that is smooth for all x, but not differentiable at x = - 1 and x = 2.

APPLICATIONS 35. The postage function. Consider the postage function defined in Exercises 84–88 of Exercise Set 1.1, on p. 108. At what values in the domain is the function not differentiable? 36. The taxicab fare function. Consider the taxicab fare function defined in Exercises 81–83 of Exercise Set 1.1, on p. 108. At what values is the function not differentiable?

f

x4

x1

Business and Economics

24. Find f ¿1x2 for f 1x2 = ax 2 + bx.

x 0 x1 x 2 x 3

x0

x 5 x 6 x 7 x 8 x 9 x10 x11

x12

37. Baseball ticket prices. Consider the model for average Major League Baseball ticket prices in Exercise 29 of Exercise Set 1.3, on p. 129. At what values is the function not differentiable? The values of the Dow Jones Industrial Average for the week of January 4–11, 2010, are graphed below, where x is the day of the month.

26.

x 0 x1

x2

x 3 x4

x5

x6

x7

x8

27.

Dow Jones Industrial Average

f

y 10,760 10,720 10,680 10,640 10,600 10,560

4

5

6

7

8

9

10

11

x

Day of the month

f

(Source: www.moneycentral.msn.com.)

x0

x1

x2

x3

x4

x5

38. On what day did the Dow Jones Industrial Average show the greatest rate of increase? On what day did it show the greatest rate of decrease? Give the rates.

Exercise Set 1.4

39. Is the function differentiable at the given x-values? Why or why not?

143

a) For what x-value(s) is this function not differentiable? b) Describe the simplest way to find f ¿142. 50. Consider the function g given by

SYNTHESIS 40. Which of the lines in the following graph appear to be tangent lines? Try to explain why or why not. y

L4

L2

L1

L5

g1x2 =

x2 + x . 2x

a) For what x-value(s) is this function not differentiable? b) What is g¿132? Describe the simplest way to determine this. 51. Consider the function h given by h1x2 = ƒ x - 3 ƒ + 2.

L3

a) For what x-value(s) is the function not differentiable? b) Evaluate h¿102, h¿112, h¿142, and h¿1102. Is there a shortcut you can use to find these slopes?

L6 x

52. Consider the function k given by 41. On the following graph, use a blue colored pencil to draw each secant line from point P to the points Q. Then use a red colored pencil to draw a tangent line to the curve at P. Describe what happens. y P

k1x2 = 2 ƒ x + 5 ƒ . a) For what x-value(s) is this function not differentiable? b) Evaluate k¿1- 102, k¿1- 72, k¿1- 22, and k¿102. Is there a shortcut you can use to find these slopes? x 2 + 4x + 3 . A student recognizes that this x + 1 function can be simplified as follows: 1x + 121x + 32 x 2 + 4x + 3 = = x + 3. f 1x2 = x + 1 x + 1

53. Let f 1x2 =

Q5 Q4

Q3

Since y = x + 3 is a line with slope 1, the student makes the following conclusions: f ¿1- 22 = 1, f ¿1- 12 = 1, f ¿102 = 1, f ¿112 = 1. Where did the student make an error?

Q2

3

Q1

x

For Exercises 42–48, find f œ 1x2 for the given function. 42. f 1x2 = x 4 (See Exercise 49 in Section 1.3.)

1 43. f 1x2 = (See Exercise 53 in Section 1.3.) 1 - x 44. f 1x2 = x 5 (See Exercise 50 in Section 1.3.) 45. f 1x2 =

1 x2

(See Exercise 52 in Section 1.3.)

46. f 1x2 = 2x (See Example 8 in Section 1.3.) 47. f 1x2 = 22x + 1 (See Exercise 55 in Section 1.3.) 48. f 1x2 =

1 2x

(See Exercise 56 in Section 1.3.)

49. Consider the function f given by f 1x2 =

x2 - 9 . x + 3

54. Let g1x2 = 2x. A student graphed this function, and the graph appeared to be smooth and continuous for all real numbers x. The student concluded that g1x2 is differentiable for all x, which is false. Identify the error, and explain why the conclusion is false. What is the correct conclusion regarding the differentiability of g1x2? 55. Let F be a piecewise-defined function given by F1x2 = e

x 2 + 1, for x … 2, 2x + 1, for x 7 2.

a) Verify that F is continuous at x = 2. b) Is F differentiable at x = 2? Explain why or why not. 56. Let G be a piecewise-defined function given by G1x2 = e

for x … 1, x 3, 3x - 2, for x 7 1.

a) Verify that G is continuous at x = 1. b) Is G differentiable at x = 1? Explain why or why not. 57. Let H be a piecewise-defined function given by H1x2 = e

2x 2 - x, for x … 3, mx + b, for x 7 3.

Determine the values of m and b that make H differentiable at x = 3.

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TECHNOLOGY CONNECTION 58–63. Use a calculator to check your answers to Exercises 17–22. 64. Business: growth of an investment. A company determines that the value of an investment is V, in millions of dollars, after time t, in years, where V is given by V1t2 = 5t 3 - 30t 2 + 45t + 52t. Note: Calculators often use only the variables y and x, so you may need to change the variables when entering this function. a) Graph V over the interval 30, 54. b) Find the equation of the secant line passing through the points 11, V1122 and 15, V1522. Then graph this secant line using the same axes as in part (a). c) Find the average rate of change of the investment between year 1 and year 5.

1.5

d) Repeat parts (b) and (c) for the following pairs of points: 11, V1122 and 14, V1422; 11, V1122 and 13, V1322; 11, V1122 and 11.5, V11.522. e) What appears to be the slope of the tangent line to the graph at the point 11, V1122? f) Approximate the rate at which the value of the investment is changing after 1 yr. 65. Use a calculator to determine where f ¿1x2 does not exist, 3 if f 1x2 = 2x - 5.

Answers to Quick Checks 1. f ¿1x2 = 3x 2 + 2x; f ¿1- 22 = 8, f ¿142 = 56 2 2. f ¿1x2 = 2 3. At x = - 6, the graph has a corner. x

Differentiation Techniques: The Power and Sum–Difference Rules

OBJECTIVES

Leibniz Notation

• Differentiate using the Power Rule or the Sum– Difference Rule.

Let y be a function of x. A common way to express “the derivative of y with respect to x” is the notation

• Differentiate a constant or a constant times a function. • Determine points at which a tangent line has a specified slope.

dy dx

.

This notation was invented by the German mathematician Leibniz. Using this notation, we can write the following sentence: If y = f1x2, then the derivative of y with respect to x is

Historical Note: The German mathematician and philosopher Gottfried Wilhelm von Leibniz (1646–1716) and the English mathematician, philosopher, and physicist Sir Isaac Newton (1642–1727) are both credited with the invention of calculus, though each performed his work independently. Newton used the dot # notation y for dy>dt, where y is a function of time; this notation is still used, though it is not as common as Leibniz notation.

dy dx

= f ¿1x2.

In practice, we often use prime notation, such as y¿ or f ¿1x2, to represent a derivative when there is no confusion as to which variables are involved. The dy>dx notation is a little more formal than the prime notation but has the same meaning. We will use both types of notation often. When we wish to evaluate a derivative at a number, we write dy dx

`

x=2

= f ¿122.

The vertical line is interpreted as “evaluated at,” so the above expression is read as “the derivative of y with respect to x evaluated at x = 2 is the value f ¿122.” We can also write d f1x2. dx This is identical in meaning to dy>dx and is another way to denote the derivative of the function. When placed next to a function, d>dx is treated as a command to find the function’s derivative. Using functions from previous sections, we can write d 2 x = 2x, dx

d 3 x = 3x 2, dx

d 1 1 a b = - 2, dx x x

and so on.

1.5

• Differentiation Techniques: The Power and Sum–Difference Rules

145

We will use all of these derivative forms often, and with practice, their use will become natural. In Chapter 2, we will discuss specific meanings of dy and dx.

The Power Rule In Section 1.4, we calculated the derivative for some simple power functions. Look at the following table and see if you can identify a pattern: Function

Derivative

x2 x3 x4

2x 1 3x 2 4x 3 - 1 # x -2 =

1 = x -1 x 1 = x -2 x2

- 2 # x -3 =

Example 1 in Section 1.4 Example 2 in Section 1.4 Exercise 42 in Section 1.4

-1 x2 -2 x3 1

1 -1>2 x = 2 22x

2x = x 1>2

Example 4 in Section 1.4 Exercise 45 in Section 1.4 Exercise 46 in Section 1.4

The pattern can be described as follows: “To find the derivative of a power function, bring the exponent to the front of the variable as a coefficient and reduce the exponent by 1.” 1. Write the exponent as the coefficient.

d k x = k # xk-1 dx

2. Subtract 1 from the exponent.

This rule is summarized as the following theorem.

THEOREM 1

The Power Rule For any real number k, if y = x k, then d k x = k # x k-1. dx We proved this theorem for the cases where k = 2, 3, and - 1 in Examples 1, 2, and 4 in Section 1.4 and for other cases as exercises at the end of that section. The proof of this theorem for the case where k is any positive integer is very elegant. Proof. Let f1x2 = x k. We need to find the expanded form for f1x + h2 = 1x + h2k so that we can set up the difference quotient. When 1x + h2k is multiplied (expanded), a pattern becomes evident, as the following shows: 1x 1x 1x 1x

+ + + +

h21 h22 h23 h24

= = = =

x x2 x3 x4

+ + + +

h, 2xh + h2, 3x 2h + 3xh2 + h3, 4x 3h + 6x 2h2 + 4xh3 + h4.

The first term is x k, and the second term is kx k - 1h. The terms in the shaded triangle all contain h to the power of 2 or greater. Calling these “shaded terms,” we can summarize the above expansion as follows: 1x + h2k = x k + kx k - 1h + 1shaded terms2.

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CHAPTER 1

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We now substitute for 1x + h2k in the difference quotient: f1x + h2 - f1x2 h

=

1x + h2k - x k h

=

x k + kx k - 1h + 1shaded terms2 - x k h

.

The x k terms in the numerator sum to 0, and h is factored out. The “shaded terms” now contain h to the power of 1 or greater, and we refer to them as the “reduced shaded terms.” The h’s in the numerator and the denominator cancel, and we have h3kx k - 1 + 1reduced shaded terms24 h

= kx k - 1 + 1reduced shaded terms2.

When we take the limit as h : 0, the “reduced shaded terms” become 0: f ¿1x2 = lim kx k - 1 + 1reduced shaded terms2 = kx k - 1. ■ h:0

Although we have proved the Power Rule only for the case where k is a positive integer, it is valid for all real numbers k. However, a complete proof of this fact is outside the scope of this book.

■ EXAMPLE 1 Differentiate each of the following: a) y = x 5;

c) y = x -4.

b) y = x;

Solution Quick Check 1

O

a) Differentiate:

1i2 y = x 15; 1ii2 y = x -7.

d b) Explain why 1p22 = 0, dx not 2p.

d 5 x = 5 # x 5 - 1 = 5x 4 Using the Power Rule dx d b) x = 1 # x1-1 = 1 # x0 = 1 dx d -4 1 c) x = - 4 # x -4 - 1 = - 4x -5, or - 4 # 5 , or dx x a)

-

4 x5

O Quick Check 1 The Power Rule also allows us to differentiate expressions with rational exponents.

■ EXAMPLE 2 Differentiate: 5

a) y = 2x;

b) y = x 0.7.

Solution a)

Quick Check 2 Differentiate: 4 a) y = 2x; b) y = x -1.25.

b)

d 5 d 1 2x = x 1>5 = # x 11>52 - 1 dx dx 5 1 1 1 1 1 1 = x -4>5, or # 4>5 , or # 5 , or 5 5 5 x 5 2x 4 52x 4 d 0.7 x = 0.7x 10.72 - 1 = 0.7x -0.3 dx

O Quick Check 2

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• Differentiation Techniques: The Power and Sum–Difference Rules

1.5

147

TECHNOLOGY CONNECTION

Numerical Differentiation and Tangent Lines

We can also use the Tangent feature from the DRAW menu to draw the tangent line at the point where the derivative was found. Both the line and its equation will appear on the calculator screen.

Consider f 1x2 = x24 - x 2, graphed below. y = x 4 – x2

4 4

–3 –3

3

3 X = 1.0212766 Y = 1.11304782385X+.61944_ –4 –4

EXERCISES

To find the value of dy>dx at a point, we select dy/dx from the CALC menu.

For each of the following functions, use dy/dx to find the derivative, and then draw the tangent line at the given point. When selecting the viewing window, be sure to include the specified x-values.

4 Y1= X (4 – X 2)

–3

1. f 1x2 = x1200 - x2;

3

X=0

x = 24, x = 138, x = 150, x = 190

2. f 1x2 = x 3 - 6x 2;

Y=0

x = - 2, x = 0, x = 2, x = 4, x = 6.3

–4

3. f 1x2 = - 4.32 + 1.44x + 3x 2 - x 3;

Next we key in the desired x-value or use the arrow keys to move the cursor to the desired point. We then press [ to obtain the value of the derivative at the given x-value.

x = - 0.5, x = 0.5, x = 2.1

4

4 Y1= X (4–X 2)

–3

3

X = 1.0212766

Y = 1.7561783 –4

–3

3

dy/dx = 1.1130478 –4

In Section 1.3, we found the simplified difference quotient for two common functions: f1x2 = 1>x and f1x2 = 2x. By taking the limit of each difference quotient as h : 0, we find the derivative of the function as follows: • For f1x2 = 1>x, we have f ¿1x2 = lim

h:0

-1 1 = - 2. x1x + h2 x

(See Example 7 in Section 1.3.)

• For f1x2 = 2x, we have f ¿1x2 = lim a h:0

1 2x + 2x + h

b =

1 22x

.

(See Example 8 in Section 1.3.)

These two functions are very common in calculus, so it may be helpful to memorize their derivative forms. The Power Rule can be used to confirm these results. • For f1x2 = 1>x, we rewrite the function as a power: f1x2 = x -1. The Power Rule then gives f ¿1x2 = 1- 12x 1-12-1 = 1- 12x -2 = -

1 . x2

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CHAPTER 1

• Differentiation

• For f1x2 = 2x, we rewrite the function as a power: f1x2 = x 1>2. The Power Rule then gives f ¿1x2 =

1 11>22 - 1 1 1 1 x = x -1>2 = = . 1>2 2 2 2x 22x

TECHNOLOGY CONNECTION

The Derivative of a Constant Function

Exploratory

Consider the constant function given by F1x2 = c. Note that the slope at each point on its graph is 0.

Graph the constant function y = - 3. Then find the derivative of this function at x = - 6, x = 0, and x = 8. What do you conclude about the derivative of a constant function?

y

F(x) = c

F(x + h) = c

x

x+h

x

This suggests the following theorem.

THEOREM 2 The derivative of a constant function is 0. That is,

d c = 0. dx

Proof. Let F be the function given by F1x2 = c. Then F1x + h2 - F1x2 h

=

c - c h

=

0 = 0. h

The difference quotient for this function is always 0. Thus, as h approaches 0, the limit of the difference quotient is 0, so F¿1x2 = 0. ■

The Derivative of a Constant Times a Function Now let’s consider differentiating functions such as f1x2 = 5x 2 and g1x2 = - 7x 4. Note that we already know how to differentiate x 2 and x 4. Let’s look for a pattern in the results of Section 1.4 and its exercise set. Function

Derivative

5x 2 3x -1 3 2 2x 1 # x3

10x - 3x -2 3x 3x 2

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• Differentiation Techniques: The Power and Sum–Difference Rules

149

Perhaps you have discovered the following theorem.

THEOREM 3 The derivative of a constant times a function is the constant times the derivative of the function. Using derivative notation, we can write this as d # d 3c f1x24 = c # f1x2. dx dx

Proof. Let F1x2 = cf1x2. Then F ¿1x2 = lim

h:0

= lim

F1x + h2 - F1x2

Using the definition of a derivative

h cf1x + h2 - cf1x2

h:0

= c # lim c

h f1x + h2 - f1x2

h:0

h

= c # f ¿1x2.

Substituting

d

Factoring and using the Limit Properties Using the definition of a derivative

■

Combining this rule with the Power Rule allows us to find many derivatives.

■ EXAMPLE 3 Find each of the following derivatives: a)

d 7x 4; dx

b)

d 1- 9x2; dx

c)

d 1 a b. dx 5x 2

Solution d 4 d 7x = 7 x 4 = 7 # 4 # x 4 - 1 = 28x 3 With practice, this may be done in one step. dx dx d d b) 1- 9x2 = - 9 x = - 9 # 1 = - 9 dx dx d 1 d 1 -2 1 d -2 c) a b = a x b = # x dx 5x 2 dx 5 5 dx 1 = 1- 22x -2 - 1 5 2 2 = - x -3, or - 3 5 5x a)

Quick Check 3 Differentiate each of the following: a) y = 10x 9; b) y = px 3; 2 c) y = . 3x 4

O Quick Check 3 A common mistake is to write an expression such as 1>15x 22 as 15x2-2, which is incorrect. The exponent 2 applies only to the x; the 5 is part of the coefficient 15. Carefully examine Example 3c, noting how the final answer is simplified. Recall from Section 1.4 that a function’s derivative at x is also its instantaneous rate of change at x.

■ EXAMPLE 4 Life Science: Volume of a Tumor. The volume V of a spherical tumor can be approximated by V1r2 = 43 pr 3, where r is the radius of the tumor, in centimeters.

O

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• Differentiation

a) Find the rate of change of the volume with respect to the radius. b) Find the rate of change of the volume at r = 1.2 cm.

Solution dV 4 = V¿1r2 = 3 # # pr 2 = 4pr 2 dr 3

a)

(This expression turns out to be equal to the tumor’s surface area.) cm3 b) V¿11.22 = 4p11.222 = 5.76p L 18 = 18 cm2 cm When the radius is 1.2 cm, the volume is changing at the rate of 18 cm3 for every change of 1 cm in the radius.

The Derivative of a Sum or a Difference In Exercise 11 of Exercise Set 1.4, you found that the derivative of f1x2 = x 2 + x f ¿1x2 = 2x + 1.

is

Note that the derivative of x 2 is 2x, the derivative of x is 1, and the sum of these derivatives is f ¿1x2. This illustrates the following. TECHNOLOGY CONNECTION

Exploratory Let y1 = x1100 - x2, and y2 = x2100 - x 2. Using the Y-VARS option from the VARS menu, enter Y3 as Y1 + Y2. Find the derivative of each of the three functions at x = 8 using the numerical differentiation feature. Compare your answers. How do you think you can find the derivative of a sum?

THEOREM 4 The Sum–Difference Rule Sum. The derivative of a sum is the sum of the derivatives: d d d 3f1x2 + g1x24 = f1x2 + g1x2. dx dx dx Difference. The derivative of a difference is the difference of the derivatives: d d d 3f1x2 - g1x24 = f1x2 g1x2. dx dx dx

Proof. The proof of the Sum Rule relies on the fact that the limit of a sum is the sum of the limits. Let F1x2 = f1x2 + g1x2. Then lim

h:0

F1x + h2 - F1x2 h

= lim

3f1x + h2 + g1x + h24 - 3f1x2 + g1x24

h:0

= lim c h:0

h f1x + h2 - f1x2 h

+

g1x + h2 - g1x2 h

d = f ¿1x2 + g¿1x2.

To prove the Difference Rule, we note that d d d d 1f1x2 - g1x22 = 1f1x2 + 1- 12g1x22 = f1x2 + 1- 12g1x2 dx dx dx dx d = f ¿1x2 + 1- 12 g1x2 = f ¿1x2 - g¿1x2. ■ dx Any function that is a sum or difference of several terms can be differentiated term by term.

1.5

• Differentiation Techniques: The Power and Sum–Difference Rules

151

■ EXAMPLE 5 Find each of the following derivatives: a)

d 15x 3 - 72; dx

b)

d 5 a24x - 2x + b. x dx

Solution a)

b)

Quick Check 4 Differentiate: 3

y = 3x 5 + 22x +

1 3x 2

+ 25.

d d d 15x 3 - 72 = 15x 32 172 dx dx dx d = 5 x3 - 0 dx = 5 # 3x 2 = 15x 2 d d 5 d d 5 a24x - 2x + b = 124x2 12x2 + a b x dx dx dx dx x d 1>2 d d x = 24 # x + 5 # x -1 dx dx dx 1 = 24 # 1 - x 11>22 - 1 + 51- 12x -1-1 2 1 = 24 - x -1>2 - 5x -2 2 1 5 = 24 - 2 x 22x

O Quick Check 4 A word of caution! The derivative of f1x2 + c, a function plus a constant, is just the derivative of the function, f ¿1x2. The derivative of c # f1x2,

a function times a constant, is the constant times the derivative c # f ¿1x2.

That is, the constant is retained for a product, but not for a sum.

Slopes of Tangent Lines It is important to be able to determine points at which the tangent line to a curve has a certain slope, that is, points at which the derivative attains a certain value.

O

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• Differentiation

■ EXAMPLE 6 Find the points on the graph of f1x2 = - x 3 + 6x 2 at which the tangent line is horizontal.

Solution The derivative is used to find the slope of a tangent line, and a horizontal tangent line has slope 0. Therefore, we are seeking all x for which f ¿1x2 = 0: f ¿1x2 = 0

Setting the derivative equal to 0

d 1- x 3 + 6x 22 = 0 dx - 3x 2 + 12x = 0.

Differentiating

We factor and solve: - 3x1x - 42 = 0 - 3x = 0 or x - 4 = 0 x = 4. x = 0 or We are to find the points on the graph, so we must determine the second coordinates from the original equation, f1x2 = - x 3 + 6x 2. f102 = - 03 + 6 # 02 = 0. f142 = - 4 3 + 6 # 4 2 = - 64 + 96 = 32.

Thus, the points we are seeking are 10, 02 and 14, 322, as shown on the graph. y 50 (4, 32)

40

m=0 30 f(x) = –x 3 + 6x 2

20 (0, 0) 10 –3

–2

–1 –10

2 1 m=0

3

4

6

5

7

x

– 20

■ EXAMPLE 7 Find the points on the graph of f1x2 = - x 3 + 6x 2 at which the tangent line has slope 9.

Solution We want to find values of x for which f ¿1x2 = 9. That is, we want to find x such that - 3x 2 + 12x = 9.

As in Example 6, note that

d 1- x 3 + 6x 22 = - 3x 2 + 12x. dx

To solve, we add - 9 on both sides and get - 3x 2 + 12x - 9 = 0. We then multiply both sides of the equation by - 13, giving x 2 - 4x + 3 = 0, which is factored as follows:

1x - 321x - 12 = 0.

1.5

TECHNOLOGY CONNECTION

EXERCISE 1

1. Graph y = 3 x 3 - 2x 2 + 4x,

and draw tangent lines at various points. Estimate points at which the tangent line is horizontal. Then use calculus, as in Examples 6 and 7, to find the exact results.

153

• Differentiation Techniques: The Power and Sum–Difference Rules

We have two solutions: x = 1 or x = 3. We need the actual coordinates: when x = 1, we have f112 = - 1123 + 61122 = 5. Therefore, at the point 11, 52 on the graph of f1x2, the tangent line has a slope of 9. In a similar way, we can state that the tangent line at the point 13, 272 has a slope of 9 as well. All of this is illustrated in the following graph. y 40 f(x) = −x 3 + 6x 2

m=9

30

(3, 27)

20 10

Quick Check 5 For the function in Example 7, find the x-values for which f ¿1x2 = - 15.

–3

–2

–1

(1, 5) m=9 1

2

3

4

5

6

x

7

O Quick Check 5 Analyzing a Function by Its Derivative Some functions are always increasing or always decreasing. For example, the function f1x2 = x 3 + 2x is always increasing. That is, at no time does the graph of this function run “downhill” or lie flat. It is steadily increasing: all tangent lines have positive slopes. How can we use the function’s derivative to demonstrate this fact? y 5 4 3 2 1 −2

−1

−1 −2 −3 −4 −5

f(x) = x 3 + 2x 1

2

x

The derivative of this function is dy>dx = f ¿1x2 = 3x 2 + 2. For any x-value, x 2 will be nonnegative; the expression 3x 2 + 2 is thus positive for all x. It is impossible to set this derivative equal to any negative quantity and solve for x (try it). The graph of the function shows the always increasing trend that the derivative proves has to be true.

■ EXAMPLE 8 Let f1x2 = - x 3 - 5x + 1. Is this function always increasing or always decreasing? Use its derivative to support your conjecture.

Solution The graph of f is shown at the right.

y 5 4 3 2 1 −2

−1

−1 −2 −3 −4 −5

f(x) = −x 3 − 5x + 1 1

2

x

Based on the graph alone, the function appears to be always decreasing, but how do we know we aren’t missing something, since we are looking at only a small portion of the graph? The graph alone is not enough to “prove” our observation. We need to use the derivative: f ¿1x2 = - 3x 2 - 5.

O

154

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• Differentiation

Since x 2 is always 0 or positive, - 3x 2 is always negative or 0. Subtracting 5 from - 3x 2 will always give a negative result. Therefore, the derivative is always negative for all real numbers x. This means all tangent lines to this graph have a negative (“downhill”) slope. Thus, the graph is always decreasing. Examples 6, 7, and 8 illustrate ways to use the derivative to analyze the behavior of a function much more accurately than can be done by observation alone. In fact, our eyes can deceive us! For example, the graph of f1x2 = x 3 - x 2 appears to be always increasing if viewed on the standard window of the TI-83. However, it does have a small interval where it is decreasing, which will be shown in Exercise 133.

Section Summary • Common forms of notation for the derivative of a dy d function are y¿, f ¿1x2, , and f1x2. dx dx d • The Power Rule for differentiation is 3x k4 = kx k - 1, dx for all real numbers k. d c = 0. • The derivative of a constant is zero: dx

• The derivative of a constant times a function is the constant times the derivative of the function: d # d 3c f1x24 = c # f1x2. dx dx • The derivative of a sum (or difference) is the sum (or difference) of the derivatives of the terms: d d d 3f1x2 ; g1x24 = f1x2 ; g1x2. dx dx dx

EXERCISE SET

1.5 Find

dy . dx

1. y = x 7

2. y = x 8

3. y = - 3x

4. y = - 0.5x

5. y = 12

6. y = 7

7. y = 2x 15

8. y = 3x 10

9. y = x -6

10. y = x -8

11. y = 4x -2

12. y = 3x -5

13. y = x 3 + 3x 2

14. y = x 4 - 7x

15. y = 82x

16. y = 42x

17. y = x

18. y = x 0.7

0.9

19. y = 12 x 4>5 21. y = 23. y =

7 x3 4x 5

20. y = - 4.8x 1>3 22. y = 24. y =

6 x4 3x 4

29.

d 3 5 x B A - 22 dx

30.

d 4 3 x B A- 2 dx

31.

d 15x 2 - 7x + 32 dx

32.

d 16x 2 - 5x + 92 dx

Find f œ 1x2.

33. f 1x2 = 0.6x 1.5

34. f 1x2 = 0.3x 1.2

35. f 1x2 =

2x 3

36. f 1x2 =

4

38. f 1x2 =

37. f 1x2 = 39. f 1x2 =

7x 3 5 - x 2>3 x

41. f 1x2 = 4x - 7 43. f 1x2 =

x

4>3

4

40. f 1x2 =

44. f 1x2 =

45. f 1x2 = - 0.01x 2 - 0.5x + 70 46. f 1x2 = - 0.01x 2 + 0.4x + 50

25.

3 d 4 a 2x - b x dx

26.

d 2 5 a 2x - b x dx

47. y = 3x -2>3 + x 3>4 + x 6>5 +

27.

d 2 a 2x b dx 2x

28.

d 4 3 a 2x + b dx 2x

48. y = x -3>4 - 3x 2>3 + x 5>4 +

2 5x 6 4 - x 3>5 x

42. f 1x2 = 7x - 14

Find y œ .

Find each derivative.

3x 4

8 x3 2 x4

x 3>2 3

155

Exercise Set 1.5

49. y =

2 x x 2

50. y =

7 x + x 7

51. If f 1x2 = x 2 + 4x - 5, find f ¿1102. 52. If f 1x2 = 2x, find f ¿142. 53. If y =

4

, find

x2

54. If y = x +

2 x

3

dy dx

`

, find

dy dx

`

58. If y =

3x

, find

3

, find

2 5x

83. y = - 0.025x 2 + 4x

84. y = - 0.01x 2 + 2x

APPLICATIONS x=1

3

4

82. y = 6x - x 2

86. y = 13 x 3 - x 2 - 4x + 1

x = -2

56. If y = 2x + 2x, find 1

81. y = 20x - x 2 85. y = 13 x 3 + 2x 2 + 2x

55. If y = x 3 + 2x - 5, find

57. If y =

For each function, find the points on the graph at which the tangent line has slope 1.

dy dx dy dx

` `

dy dx

dy dx

`

`

Life Sciences 87. Healing wound. The circular area A, in square centimeters, of a healing wound is approximated by

x = -2

A1r2 = 3.14r 2, where r is the wound’s radius, in centimeters. a) Find the rate of change of the area with respect to the radius. b) Explain the meaning of your answer to part (a).

x = 64

x = -1

88. Healing wound. The circumference C, in centimeters, of a healing wound is approximated by

x=4

59. Find an equation (in y = mx + b form) of the tangent line to the graph of f 1x2 = x 3 - 2x + 1 a) at 12, 52; b) at 1- 1, 22; c) at 10, 12. 60. Find an equation of the tangent line to the graph of f 1x2 = x 2 - 2x a) at 11, 02; b) at 14, 142; c) at 19, 782. 61. Find an equation of the tangent line to the graph of 1 f 1x2 = 2 x a) at 11, 12; b) at A 3, 19 B ; c) at A - 2, 14 B .

C1r2 = 6.28r, where r is the wound’s radius, in centimeters. a) Find the rate of change of the circumference with respect to the radius. b) Explain the meaning of your answer to part (a). 89. Growth of a baby. The median weight of a boy whose age is between 0 and 36 months can be approximated by the function w1t2 = 8.15 + 1.82t - 0.0596t 2 + 0.000758t 3, where t is measured in months and w is measured in pounds.

62. Find the equation of the tangent line to the graph of g 1x2 = 2x 2 a) at 1- 1, 12;

w

b) at 11, 12;

c) at 18, 42.

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. 63. y = x 2 - 3

64. y = - x 2 + 4

65. y = - x 3 + 1

66. y = x 3 - 2

67. y = 3x 2 - 5x + 4

68. y = 5x 2 - 3x + 8

Median weight (in pounds)

3

30 25 20 15 10 5

69. y = - 0.01x 2 - 0.5x + 70 70. y = - 0.01x 2 + 0.4x + 50

0

71. y = 2x + 4

72. y = - 2x + 5

73. y = 4

74. y = - 3

5

10

15 20 25 Age (in months)

30

35

t

(Source: Centers for Disease Control. Developed by the National Center for Health Statistics in collaboration with the National Center for Chronic Disease Prevention and Health Promotion, 2000.)

75. y = - x 3 + x 2 + 5x - 1 76. y = - 13 x 3 + 6x 2 - 11x - 50 77. y = 13 x 3 - 3x + 2

79. f 1x2 = 13 x 3 + 12 x 2 - 2

80. f 1x2 = 13 x 3 - 3x 2 + 9x - 9

78. y = x 3 - 6x + 1

Use this approximation to find the following for a boy with median weight: a) The rate of change of weight with respect to time. b) The weight of the baby at age 10 months. c) The rate of change of the baby’s weight with respect to time at age 10 months.

156

• Differentiation

CHAPTER 1

90. Temperature during an illness. The temperature T of a person during an illness is given by T1t2 = - 0.1t 2 + 1.2t + 98.6, where T is the temperature, in degrees Fahrenheit, at time t, in days. a) Find the rate of change of the temperature with respect to time. b) Find the temperature at t = 1.5 days. c) Find the rate of change at t = 1.5 days. 91. Heart rate. The equation 6000 R1v2 = v can be used to determine the heart rate, R, of a person whose heart pumps 6000 milliliters (mL) of blood per minute and v milliliters of blood per beat. (Source: Mathematics Teacher, Vol. 99, No. 4, November 2005.) Heart rate (in beats per minute)

R(v)

a) Find the rate of change of resistance with respect to r, the radius of the blood vessel. b) Find the resistance at r = 1.2 mm. c) Find the rate of change of S with respect to r when r = 0.8 mm.

Social Sciences 93. Population growth rate. The population of a city grows from an initial size of 100,000 to a size P given by P1t2 = 100,000 + 2000t 2, where t is in years. a) Find the growth rate, dP>dt. b) Find the population after 10 yr. c) Find the growth rate at t = 10. d) Explain the meaning of your answer to part (c). 94. Median age of women at first marriage. The median age of women at first marriage can be approximated by the linear function A1t2 = 0.08t + 19.7,

120

where A1t2 is the median age of women marrying for the first time at t years after 1950. a) Find the rate of change of the median age A with respect to time t. b) Explain the meaning of your answer to part (a).

100 80

R(v) =

6000 v

60 40

General Interest 60

70

80

90

100

110

120

130

v

Output per beat (in milliliters)

a) Find the rate of change of heart rate with respect to v, the output per beat. b) Find the heart rate at v = 80 mL per beat. c) Find the rate of change at v = 80 mL per beat.

95. View to the horizon. The view V, or distance in miles, that one can see to the horizon from a height h, in feet, is given by V = 1.222h.

92. Blood flow resistance. The equation S1r2 =

1 r4

can be used to determine the resistance to blood flow, S, of a blood vessel that has radius r, in millimeters (mm). (Source: Mathematics Teacher, Vol. 99, No. 4, November 2005.) S(r) Resistance to blood flow

3

2

S(r)  1

1

1 r4

2

3

Radius of blood vessel (in millimeters)

4

r

a) Find the rate of change of V with respect to h. b) How far can one see to the horizon from an airplane window at a height of 40,000 ft? c) Find the rate of change at h = 40,000. d) Explain the meaning of your answers to parts (a) and (c).

Exercise Set 1.5

96. Baseball ticket prices. The average price, in dollars, of a ticket for a Major League baseball game x years after 1990 can be estimated by p1x2 = 9.41 - 0.19x + 0.09x 2. a) Find the rate of change of the average ticket price with respect to the year, dp>dx. b) What is the average ticket price in 2010? c) What is the rate of change of the average ticket price in 2010?

157

119. y = 1x + 123 120. Use Theorem 1 to prove that the derivative of 1 is 0. 121. When might Leibniz notation be more convenient than function notation? 122. Write a short biographical paper on Leibniz and/or Newton. Emphasize the contributions each man made to many areas of science and society.

SYNTHESIS

TECHNOLOGY CONNECTION

For each function, find the interval(s) for which f œ 1x2 is positive.

Graph each of the following. Then estimate the x-values at which tangent lines are horizontal.

97. f 1x2 = x 2 - 4x + 1

123. f 1x2 = x 4 - 3x 2 + 1

98. f 1x2 = x 2 + 7x + 2 99. f 1x2 =

1 3 3x

124. f 1x2 = 1.6x 3 - 2.3x - 3.7

- x - 3x + 5 2

125. f 1x2 = 10.2x 4 - 6.9x 3

100. Find the points on the graph of

126. f 1x2 =

y = x 4 - 43 x 2 - 4 at which the tangent line is horizontal. y = 2x 6 - x 4 - 2 Use the derivative to help show whether each function is always increasing, always decreasing, or neither. 102. f 1x2 = x 5 + x 3

103. f 1x2 = x 3 + 2x

1 104. f 1x2 = , x Z 0 x

105. f 1x2 = 2x, x Ú 0

106. The function f 1x2 = x 3 + ax is always increasing if a 7 0, but not if a 6 0. Use the derivative of f to explain why this observation is true. Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand.

111. y = 112. y =

x5 - x3 x

108. y = 1x - 121x + 12 110. y =

2

5x 2 - 8x + 3 8

x5 + x

128. f 1x2 = x 4 - 3x 2 + 1

131. f 1x2 =

132. f 1x2 =

4x x + 1 2

130. f 1x2 = x 4 - x 3

5x 2 + 8x - 3 3x 2 + 2

133. The function f 1x2 = x 3 - x 2 (mentioned after Example 8) appears to be always increasing, or possibly flat, on the default viewing window of the TI-83. a) Graph the function in the default window; then zoom in until you see a small interval in which f is decreasing. b) Use the derivative to determine the point(s) at which the graph has horizontal tangent lines. c) Use your result from part (b) to infer the interval for which f is decreasing. Does this agree with your calculator’s image of the graph? d) Is it possible there are other intervals for which f is decreasing? Explain why or why not.

Answers to Quick Checks

x2 x 5 - 3x 4 + 2x + 4 x2

113. y = 1- 4x23

114. y = 27x

116. y = 1x - 322

3

115. y = 28x 117. y = a 2x -

127. f 1x2 = 20x 3 - 3x 5

129. f 1x2 = x 3 - 2x - 2

at which the tangent line is horizontal.

109. y =

3x 2 + 2

For each of the following, graph f and f œ and then determine f œ 112. For Exercises 131 and 132, use nDeriv on the TI-83.

101. Find the points on the graph of

107. y = 1x + 321x - 22

5x 2 + 8x - 3

1 2x

b

2

118. y = 12x + 2x22 3

1. (a) (i) y¿ = 15x 14, (ii) y¿ = - 7x -8; (b) because p2 is a constant 1 2. (a) y¿ = 14 x -3>4 = 4 ; (b) y¿ = - 1.25x -2.25 42x 3 8 3. (a) y¿ = 90x 8; (b) y¿ = 3px 2; (c) y¿ = - 5 3x 2 2 4 4. y¿ = 15x + 3 3x 3 32x 2 5. x = - 1 and x = 5

158

CHAPTER 1

• Differentiation

1.6

Differentiation Techniques: The Product and Quotient Rules

OBJECTIVES

The Product Rule

• Differentiate using the Product and Quotient Rules.

A function can be written as the product of two other functions. For example, the function F1x2 = x 3 # x 4 can be viewed as the product of the two functions f1x2 = x 3 and g1x2 = x 4, yielding F1x2 = f1x2 # g1x2. Is the derivative of F1x2 the product of the derivatives of its factors, f1x2 and g1x2? The answer is no. To see this, note that the product of x 3 and x 4 is x 7, and the derivative of this product is 7x 6. However, the derivatives of the two functions are 3x 2 and 4x 3, and the product of these derivatives is 12x 5. This example shows that, in general, the derivative of a product is not the product of the derivatives. The following is a rule for finding the derivative of a product.

• Use the Quotient Rule to differentiate the average cost, revenue, and profit functions.

THEOREM 5

The Product Rule

Let F1x2 = f1x2 # g1x2. Then F¿1x2 =

d d d 3f1x2 # g1x24 = f1x2 # c g1x2 d + g1x2 # c f1x2 d. dx dx dx

The derivative of a product is the first factor times the derivative of the second factor, plus the second factor times the derivative of the first factor.

The proof of the Product Rule is outlined in Exercise 123 at the end of this section. Let’s check the Product Rule for x 2 # x 5. There are five steps: d 2 5 (x  x ) dx 1 2 3 4 = x 2 · 5x 4 + x 5 · 2x = 5x 6 + 2x 6 5 = 7x 6

1. Write down the first factor. 2. Multiply it by the derivative of the second factor. 3. Write down the second factor. 4. Multiply it by the derivative of the first factor. 5. Add the result of steps (1) and (2) to the result of steps (3) and (4). Usually we try to write the results in simplified form. In Examples 1 and 2, we do not simplify in order to better emphasize the steps being performed.

■ EXAMPLE 1 Find

d 31x 4 - 2x 3 - 7213x 2 - 5x24. Do not simplify. dx

Solution We let f1x2 = x 4 - 2x 3 - 7 and g1x2 = 3x 2 - 5x. We differentiate each

of these, obtaining f ¿1x2 = 4x 3 - 6x 2 and g¿1x2 = 6x - 5. By the Product Rule, the derivative of the given function is then f1x2

#

g1x2

f1x2

#

g¿1x2

+

g1x2

#

f ¿1x2

d 31x 4 - 2x 3 - 7213x 2 - 5x24 = 1x 4 - 2x 3 - 7216x - 52 + 13x 2 - 5x214x 3 - 6x 22 dx In this example, we could have first multiplied the polynomials and then differentiated. Both methods give the same solution after simplification. It makes no difference which factor of the given function is called f1x2 and which is called g1x2. We usually let the first function listed be f1x2 and the second function be g1x2, but if we switch the names, the process still gives the same answer. Try it by repeating Example 1 with g1x2 = x 4 - 2x 3 - 7 and f1x2 = 3x 2 - 5x.

1.6

Quick Check 1 Use the Product Rule to differentiate each of the following functions. Do not simplify. a) y = 12x 5 + x - 1213x - 22 5 b) y = A 2x + 1 B A 2x - x B

• Differentiation Techniques: The Product and Quotient Rules

159

■ EXAMPLE 2 For F1x2 = 1x 2 + 4x - 112 A 7x 3 - 2x B , find F¿1x2. Do not simplify. Solution We rewrite this as

F1x2 = 1x 2 + 4x - 11217x 3 - x 1>22.

Then, using the Product Rule, we have

F¿1x2 = 1x 2 + 4x - 112 A 21x 2 - 12 x -1>2 B + 17x 3 - x 1>2212x + 42.

O Quick Check 1 The Quotient Rule The derivative of a quotient is not the quotient of the derivatives. To see why, consider x 5 and x 2. The quotient x 5>x 2 is x 3, and the derivative of this quotient is 3x 2. The individual derivatives are 5x 4 and 2x, and the quotient of these derivatives, 5x 4>12x2, is 15>22x 3, which is not 3x 2. The rule for differentiating quotients is as follows.

THEOREM 6 If Q1x2 =

The Quotient Rule

N1x2 D1x2

, then

Q¿1x2 =

D1x2 # N¿1x2 - N1x2 # D¿1x2 3D1x242

.

The derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. (If we think of the function in the numerator as the first function and the function in the denominator as the second function, then we can reword the Quotient Rule as “the derivative of a quotient is the second function times the derivative of the first function minus the first function times the derivative of the second function, all divided by the square of the second function.”)

A proof of this result is outlined in Exercise 101 of Section 1.7 (on p. 176). The Quotient Rule is illustrated below. 1

d dx

 D(x)  N(x)

2 3 4 5 D(x) · N'(x) – N(x) · D'(x) = ————————————— [D(x)]2 6

There are six steps: 1. 2. 3. 4. 5. 6.

Write the denominator. Multiply the denominator by the derivative of the numerator. Write a minus sign. Write the numerator. Multiply it by the derivative of the denominator. Divide by the square of the denominator.

O

160

CHAPTER 1

• Differentiation

■ EXAMPLE 3 For Q1x2 = x 5>x 2, find Q¿1x2. Solution We have already seen that x 5>x 2 = x 3 and

d 3 x = 3x 2, but we wish to dx practice using the Quotient Rule. We have D1x2 = x 2 and N1x2 = x 5: D1x2 # N¿1x2 - N1x2 # D¿1x2

Q¿1x2 =

x2

#

5x 4

- x5 1x 222

#

2x

3D1x242

=

3x 6 5x 6 - 2x 6 = 4 = 3x 2. 4 x x

■ EXAMPLE 4 Differentiate: f1x2 = Solution f ¿1x2 = = = =

This checks with the result above.

1 + x2 . x3

x 3 # 2x - 11 + x 22 # 3x 2 1x 322

2x 4 - 3x 2 - 3x 4 - x 4 - 3x 2 = x6 x6 x 21- x 2 - 32 x2 # x4

- x2 - 3 x4

Using the Quotient Rule

Factoring Removing a factor equal to 1:

■ EXAMPLE 5 Differentiate: f1x2 =

x 2 - 3x . x - 1

Solution We have f ¿1x2 = Quick Check 2 1 - 3x a) Differentiate: f1x2 = 2 . x + 2 Simplify your result. b) Show that d ax + 1 a - b c d = . dx bx + 1 1bx + 122

1x - 1212x - 32 - 1x 2 - 3x2 # 1 1x - 122

=

2x 2 - 5x + 3 - x 2 + 3x 1x - 122

Using the distributive law

=

x 2 - 2x + 3 . 1x - 122

Simplifying

It is not necessary to multiply out 1x - 122.

O Quick Check 2

Using the Quotient Rule

x2 = 1 x2

O

1.6

• Differentiation Techniques: The Product and Quotient Rules

161

TECHNOLOGY CONNECTION

Checking Derivatives Graphically To check Example 5, we first enter the function: y1 =

You should verify that had we miscalculated the derivative as, say, y2 = 1x 2 - 2x - 82>1x - 122, neither the tables nor the graphs of y2 and y3 would agree.

x 2 - 3x . x - 1

2 y2 = x – 2x –2 8 (x – 1)

10

Then we enter the possible derivative: y2 =

x 2 - 2x + 3 1x - 122

y3

.

–10

For the third function, we enter

y2 –10

y3 = nDeriv1y1, x, x2. Next, we deselect y1 and graph y2 and y3. We use different graph styles and the Sequential mode to see each graph as it appears on the screen. y2 =

x2 –

2x + 3 , y = nDeriv(y , x, x) 3 1 (x – 1)2 10

–10

10

–10

Since the graphs appear to coincide, it appears that y2 = y3 and we have a check. This is considered a partial check, however, because the graphs might not coincide at a point not in the viewing window. We can also use a table to check that y2 = y3. X 5.97 5.98 5.99 6 6.01 6.02 6.03 X = 5.97

10

Y2

Y3

1.081 1.0806 1.0803 1.08 1.0797 1.0794 1.079

1.081 1.0806 1.0803 1.08 1.0797 1.0794 1.079

EXERCISES 1. For the function

f 1x2 =

x 2 - 4x , x + 2

use graphs and tables to determine which of the following seems to be the correct derivative. - x 2 - 4x - 8 a) f ¿1x2 = 1x + 222 2 x - 4x + 8 b) f ¿1x2 = 1x + 222 2 x + 4x - 8 c) f ¿1x2 = 1x + 222 2–5. Check the results of Examples 1–4 in this section.

Application of the Quotient Rule The total cost, total revenue, and total profit functions, discussed in Section R.4, pertain to the accumulated cost, revenue, and profit when x items are produced. Because of economies of scale and other factors, it is common for the cost, revenue (price), and profit for, say, the 10th item to differ from those for the 1000th item. For this reason, a business is often interested in the average cost, revenue, and profit associated with the production and sale of x items.

162

CHAPTER 1

• Differentiation

DEFINITION If C1x2 is the cost of producing x items, then the average cost of producing C1x2 ; x items is x if R1x2 is the revenue from the sale of x items, then the average revenue R1x2 from selling x items is ; x if P1x2 is the profit from the sale of x items, then the average profit from selling P1x2 x items is . x

■ EXAMPLE 6 Business. Paulsen’s Greenhouse finds that the cost, in dollars, of growing x hundred geraniums is modeled by 4

C1x2 = 200 + 1002x. If the revenue from the sale of x hundred geraniums is modeled by R1x2 = 120 + 902x, find each of the following. a) The average cost, the average revenue, and the average profit when x hundred geraniums are grown and sold. b) The rate at which average profit is changing when 300 geraniums are being grown and sold.

Solution a) We let AC, AR, and AP represent average cost, average revenue, and average profit, respectively. Then AC1x2 =

4 C1x2 200 + 1002x = ; x x

AR1x2 =

R1x2 120 + 902x = ; x x

AP1x2 =

4 R1x2 - C1x2 P1x2 - 80 + 902x - 1002x = = . x x x

b) To find the rate at which average profit is changing when 300 geraniums are being grown, we calculate AP ¿132 (remember that x is in hundreds): AP ¿1x2 = =

d - 80 + 90x 1>2 - 100x 1>4 B R x dx x A 12 # 90x 1>2 - 1 -

1 4

# 100x 1>4 - 1 B

- 1- 80 + 90x 1>2 - 100x 1>42 # 1

x2

45x 1>2 - 25x 1>4 + 80 - 90x 1>2 + 100x 1>4 75x 1>4 - 45x 1>2 + 80 = ; x2 x2 4 7523 - 4523 + 80 AP ¿132 = L 11.20. 32 =

When 300 geraniums are being grown, the average profit is increasing by $11.20 per hundred plants, or about 11.2 cents per plant.

Exercise Set 1.6

163

TECHNOLOGY CONNECTION

Using Y-VARS

EXERCISES

One way to save keystrokes on most calculators is to use the Y-VARS option on the VARS menu. To check Example 6, we let y1 = 200 + 100x 0.25 and y2 = 120 + 90x 0.5. To express the profit function as y3, we press E and move the cursor to enter y3. Next we press O and select Y-VARS and then FUNCTION. From the FUNCTION menu we select Y2, which then appears on the E screen. After pressing c, we repeat the procedure to get Y1 on the E screen.

1. Use the Y-VARS option to enter y4 = y1>x, y5 = y2>x,

and y6 = y3>x, and explain what each of the functions represents.

2. Use nDeriv from the MATH menu or dy/dx from the CALC menu to check part (b)of Example 6.

Plot 1 Plot 2 Plot 3 Y 1 = 200 + 100 X 0.25 Y 2 = 120 + 90 X 0.5 Y3 = Y2 − Y1 Y4 = Y5 = Y6 =

Section Summary • The Product Rule is d d d 3f1x2 # g1x24 = f1x2 # 3g1x24 + g1x2 # 3f1x24. dx dx dx

• Be careful to note the order in which you write out the factors when using the Quotient Rule. Because the Quotient Rule involves subtraction and division, the order in which you perform the operations is important.

• The Quotient Rule is d f1x2 c d = dx g1x2

g1x2 #

d d 3f1x24 - f1x2 # 3g1x24 dx dx . 3g1x242

EXERCISE SET

1.6 Differentiate two ways: first, by using the Product Rule; then, by multiplying the expressions before differentiating. Compare your results as a check. 1. y = x 5 # x 6 2. y = x

9

#x

3. f 1x2 = 12x + 5213x - 42

4. g1x2 = 13x - 2214x + 12 5. G1x2 = 4x 21x 3 + 5x2 6. F1x2 = 3x 41x 2 - 4x2 8. y = A 42x + 3 B x

10. f1x2 = 12x + 5213x 2 - 4x + 12 11. F1t2 =

4

7. y = A 32x + 2 B x

9. g1x2 = 14x - 3212x 2 + 3x + 52

2 3

A 2t + 2 B A 3t - 42t + 7 B

12. G1t2 = A 2t + 32t + 5 B A 2t + 4 B Differentiate two ways: first, by using the Quotient Rule; then, by dividing the expressions before differentiating. Compare your results as a check. 13. y =

x7

14. y =

x3

15. f 1x2 =

2x 5 + x 2 x

x6 x4

16. g1x2 =

3x 7 - x 3 x

164

CHAPTER 1

8x 3 - 1 2x - 1

17. G1x2 = 19. y =

• Differentiation

18. F1x2 =

t 2 - 16 t + 4

20. y =

x 3 + 27 x + 3

t 2 - 25 t - 5

100. Find an equation of the tangent line to the graph of y = 4x>11 + x 22 at (a) 10, 02; (b) 1- 1, - 22.

APPLICATIONS

Differentiate each function.

21. f 1x2 = 13x 2 - 2x + 5214x 2 + 3x - 12

Business and Economics

22. g1x2 = 15x 2 + 4x - 3212x 2 - 3x + 12 23. y =

5x 2 - 1

24. y =

2x + 3 3

25. G1x2 = A 8x + 2x B 15x 2 + 32

3x 4 + 2x x3 - 1

26. F1x2 = 1- 3x 2 + 4x2 A 72x + 1 B 27. g1t2 =

28. f 1t2 =

t + 5t 3 3 - t

29. F1x2 = 1x + 322

t - 2t 4 5 + 2t

30. G1x2 = 15x - 422 31. y = 1x 3 - 4x22

33. g1x2 = 5x - 31x 4 - 5x 3 + 10x - 22

34. f 1x2 = 6x - 416x 3 + 10x 2 - 8x + 32

37. y =

x + 1 2

x3 - 1

105. Average profit. Use the information in Exercises 101 and 103 to determine the rate at which Summertime Fabrics’ average profit per jacket is changing when 400 jackets have been produced and sold.

2 b 1t 2 - 32 t

- 5x 2

5 b t 38. y =

3

39. y =

2x - 7 2x + 3

41. f 1x2 =

x x -1 + 1

40. y =

x - 1 3

x2 + 1

+ 4x 3

2x + 4 3

2x - 5

42. f 1x2 =

x -1 x + x -1

43. F1t2 =

1 t - 4

44. G1t2 =

1 t + 2

45. f 1x2 =

3x 2 + 2x

46. f 1x2 =

3x 2 - 5x

47. g1t2 =

x + 1 2

- t + 3t + 5 2

t 2 - 2t + 4

102. Average cost. Tongue-Tied Sauces, Inc., finds that the cost, in dollars, of producing x bottles of barbecue sauce is given by C1x2 = 375 + 0.75x 3>4. Find the rate at which the average cost is changing when 81 bottles of barbecue sauce have been produced.

104. Average revenue. Tongue-Tied Sauces, Inc., finds that the revenue, in dollars, from the sale of x bottles of barbecue sauce is given by R1x2 = 7.5x 0.7. Find the rate at which average revenue is changing when 81 bottles of barbecue sauce have been produced.

32. y = 13x 2 - 4x + 522

36. G1t2 = 13t 5 - t 22at -

101. Average cost. Summertime Fabrics finds that the cost, in dollars, of producing x jackets is given by C1x2 = 950 + 152x. Find the rate at which the average cost is changing when 400 jackets have been produced.

103. Average revenue. Summertime Fabrics finds that the revenue, in dollars, from the sale of x jackets is given by R1x2 = 852x. Find the rate at which average revenue is changing when 400 jackets have been produced.

[Hint: 1x + 322 = 1x + 321x + 32.]

35. F1t2 = at +

99. Find an equation of the tangent line to the graph of y = x 2 + 3>1x - 12 at (a) x = 2; (b) x = 3.

48. f 1t2 =

x - 1 2

3t + 2t - 1 2

- t 2 + 4t + 1

49–96. Use a graphing calculator to check the results of Exercises 1–48. 97. Find an equation of the tangent line to the graph of y = 8>1x 2 + 42 at (a) 10, 22; (b) 1- 2, 12. 98. Find an equation of the tangent line to the graph of y = 2x>1x + 12 at (a) x = 1; (b) x = 14.

106. Average profit. Use the information in Exercises 102 and 104 to determine the rate at which Tongue-Tied Sauces’ average profit per bottle of barbecue sauce is changing when 81 bottles have been produced and sold. 107. Average profit. Sparkle Pottery has determined that the cost, in dollars, of producing x vases is given by C1x2 = 4300 + 2.1x 0.6. If the revenue from the sale of x vases is given by R1x2 = 65x 0.9, find the rate at which the average profit per vase is changing when 50 vases have been made and sold. 108. Average profit. Cruzin’ Boards has found that the cost, in dollars, of producing x skateboards is given by C1x2 = 900 + 18x 0.7. If the revenue from the sale of x skateboards is given by R1x2 = 75x 0.8, find the rate at which the average profit per skateboard is changing when 20 skateboards have been built and sold. 109. Gross domestic product. The U.S. gross domestic product (in billions of dollars) can be approximated using the function P1t2 = 567 + t136t 0.6 - 1042, where t is the number of the years since 1960.

Exercise Set 1.6

a) Find the rate of change of the temperature with respect to time. b) Find the temperature at t = 2 hr. c) Find the rate of change of the temperature at t = 2 hr.

P(t)

Gross domestic product (in billions of dollars)

12,000

P(t) = 567 + t (36t 0.6 – 104)

10,000

165

8000

SYNTHESIS

6000

Differentiate each function. 4000

112. f 1x2 =

2000

10

0

20

30

40

50

Number of years since 1960 (Source: U.S. Bureau of Economic Analysis.)

a) Find P¿1t2. b) Find P¿1452. c) In words, explain what P¿1452 means.

Social Sciences 110. Population growth. The population P, in thousands, of a small city is given by P1t2 =

500t

,

2t 2 + 9

115. g1x2 = 1x 3 - 82 # 116. f 1t2 = 1t 5 + 32 #

x2 + 1 x2 - 1

t3 - 1 t3 + 1

1x - 121x 2 + x + 12 x 4 - 3x 3 - 5

-1 x . and g1x2 = x + 1 x + 1 a) Compute f ¿1x2. b) Compute g¿1x2. c) What can you conclude about f and g on the basis of your results from parts (a) and (b)?

500t P(t) = — —— 2t2 + 9

50

114. f 1x2 = x13x 3 + 6x - 2213x 4 + 72

118. Let f 1x2 =

P 60

(Hint: Simplify before differentiating.)

+ 5

113. y1t2 = 5t1t - 1212t + 32

117. f 1x2 =

where t is the time, in years.

4 x2

t

3 2x

7 -

40 30 20 10 2

4

6

8 10 12

t

a) Find the growth rate. b) Find the population after 12 yr. c) Find the growth rate at t = 12 yr.

Life and Physical Sciences 111. Temperature during an illness. The temperature T of a person during an illness is given by T1t2 =

4t t2 + 1

+ 98.6,

where T is the temperature, in degrees Fahrenheit, at time t, in hours. T 100.5

4t— + 98.6 T(t) = — t 2 +1

100 99.5 99 98.5 5

10 15 20 25

t

119. Let f 1x2 =

x2

and g1x2 =

1

. x - 1 x - 1 a) Compute f ¿1x2. b) Compute g¿1x2. c) What can you conclude about the graphs of f and g on the basis of your results from parts (a) and (b)? 2

2

120. Write a rule for finding the derivative of f 1x2 # g1x2 # h1x2. Describe the rule in words. 121. Is the derivative of the reciprocal of f 1x2 the reciprocal of the derivative of f ¿1x2? Why or why not? 122. Sensitivity. The reaction R of the body to a dose Q of medication is often represented by the general function R1Q2 = Q2 a

Q k - b, 2 3

where k is a constant and R is in millimeters of mercury (mmHg) if the reaction is a change in blood pressure or in degrees Fahrenheit (°F) if the reaction is a change in temperature. The rate of change dR>dQ is defined to be the body’s sensitivity to the medication. a) Find a formula for the sensitivity. b) Explain the meaning of your answer to part (a).

166

CHAPTER 1

• Differentiation

123. A proof of the Product Rule appears below. Provide a justification for each step. f 1x + h2g1x + h2 - f 1x2g 1x2 d 3 f 1x2 # g1x24 = lim a) h:0 dx h f 1x + h2g1x + h2 - f 1x + h2g 1x2 + f 1x + h2g 1x2 - f 1x2g 1x2 = lim b) h:0 h f 1x + h2g1x + h2 - f 1x + h2g 1x2 f 1x + h2g 1x2 - f 1x2g 1x2 = lim + lim c) h:0 h:0 h h g1x + h2 - g 1x2 f 1x + h2 - f 1x2 d + lim cg 1x2 # d = lim c f 1x + h2 # d) h:0 h:0 h h g1x + h2 - g 1x2 f 1x + h2 - f 1x2 e) + g 1x2 # lim = f 1x2 # lim h:0 h:0 h h f) = f 1x2 # g¿1x2 + g1x2 # f ¿1x2 = f 1x2 # c

g)

d d g1x2 d + g1x2 # c f 1x2 d dx dx

TECHNOLOGY CONNECTION

132. Use a graph to decide which of the following seems to be the correct derivative of the function in Exercise 131.

124. Business. Refer to Exercises 102, 104, and 106. At what rate is Tongue-Tied Sauces’ profit changing at the break-even point? At what rate is the average profit per bottle of barbecue sauce changing at that point?

y1 = y2 =

125. Business. Refer to Exercises 101, 103, and 105. At what rate is Summertime Fabrics’ profit changing at the break-even point? At what rate is the average profit per jacket changing at that point?

y3 = y4 =

For the function in each of Exercises 126–131, graph f and f ¿. Then estimate points at which the tangent line to f is horizontal. If no such point exists, state that fact. 126. f 1x2 = x 21x - 221x + 22

130. f 1x2 =

x3 - 1

129. f 1x2 =

x + 1 2

0.01x 2 x + 0.0256 4

131. f 1x2 =

1.7 OBJECTIVES • Find the composition of two functions. • Differentiate using the Extended Power Rule or the Chain Rule.

x2 + 1 4 - 4x 2

1x 2 + 122 4x 2 - 4 1x 2 + 122

Answers to Quick Checks

1. (a) y¿ = 12x 5 + x - 12132 + 13x - 22110x 4 + 12 1 1 5 (b) y¿ = A 2x + 1 B a 5 - 1 b + A 2x - x B a b 52x 4 22x 2 3x - 2x - 6 2. (a) y¿ = 1x 2 + 222 1bx + 121a2 - 1ax + 121b2 abx + a - abx - b (b) = 2 1bx + 12 1bx + 122 a - b = 1bx + 122

2 127. f 1x2 = ax + b 1x 2 - 32 x 128. f 1x2 =

2 x 4 - 4x

0.3x 0.04 + x 2 4x x + 1 2

The Chain Rule The Extended Power Rule Some functions are considered simple. Simple is a subjective concept, and it may be best to illustrate what it means with examples. The following functions are considered simple and are similar to those we have seen in Sections 1.5 and 1.6: f1x2 = 2x,

g1x2 = 3x 2 - 5x,

h1x2 = 22x,

j1x2 =

2x - 1 . x2 - 3

On the other hand, these functions are not considered simple: f1x2 = 1x 3 + 2x25,

g1x2 = 22x + 5,

h1x2 = a

3x - 7 3 b . 4x 2 + 1

1.7

167

• The Chain Rule

In these cases, we see that the variable x is part of one or more expressions that are raised to some power. How can we use the concepts of differentiation from Sections 1.5 and 1.6 to differentiate functions of this form? In this section, we will introduce and discuss the Chain Rule, but we begin our discussion with a special case of the Chain Rule called the Extended Power Rule.

The Extended Power Rule The function y = 1 + x 2 is considered simple, and its derivative can be found directly from the Power Rule. However, if we nest this function in some manner, for example, y = 11 + x 223, we now have a more complicated form. How do we determine the derivative? We might guess the following: d ? 311 + x 2234 = 311 + x 222. dx

Remember, this is a guess.

To check this, we expand the function y = 11 + x 223: y = = = =

11 + x 223 11 + x 22 # 11 + x 22 # 11 + x 22 11 + 2x 2 + x 42 # 11 + x 22 1 + 3x 2 + 3x 4 + x 6.

Multiplying the first two factors Multiplying by the third factor

Taking the derivative of this function, we have y¿ = 6x + 12x 3 + 6x 5. Now we can factor out 6x:

y¿ = 6x11 + 2x 2 + x 42.

We rewrite 6x as 3 # 2x and factor the expression within the parentheses: y¿ = 311 + x 222 # 2x

Thus, it seems our original guess was close: it lacked only the extra factor, 2x, which is the derivative of the expression inside the parentheses. The correct derivative of y = 11 + x 223 is y¿ = 311 + x 222 # 2x, which suggests a general pattern for differentiating functions of this form, in which an expression is raised to a power k.

THEOREM 7 The Extended Power Rule Suppose that g1x2 is a differentiable function of x. Then, for any real number k, d d 3g1x24k = k3g1x24k-1 # g1x2. dx dx

The Extended Power Rule allows us to differentiate functions such as y = 11 + x 2289 without having to expand the expression 1 + x 2 to the 89th power (very time-consuming) and functions such as y = 11 + x 221>3, for which “expanding” to the 13 power is impractical. Let’s differentiate 11 + x 325. There are three steps to carry out. 1. Mentally block out the “inside” function, 1 + x 3. 2. Differentiate the “outside” function, 11 + x 325. 3. Multiply by the derivative of the “inside” function.

11 + x 325 511 + x 324 511 + x 324 # 3x 2 = 15x 211 + x 324

Step (3) is quite commonly overlooked. Do not forget it!

Simplified

168

CHAPTER 1

• Differentiation

TECHNOLOGY CONNECTION

Exploratory One way to check your differentiation of y1 is to enter your derivative as y2 and see if the graph of y2 coincides with the graph of y3 = nDeriv1y1, x, x2. Use this approach to check Example 1. Be sure to use the Sequential mode and different graph styles for the two curves.

The Extended Power Rule is best illustrated by examples. Carefully examine each of the following examples, noting the three-step process for applying the rule.

■ EXAMPLE 1 Differentiate: f1x2 = 11 + x 321>2. Solution d 1 11 + x 321>2 = 11 + x 321>2 - 1 # 3x 2 dx 2 = =

3x 2 11 + x 32-1>2 2 3x 2 221 + x 3

■ EXAMPLE 2 Differentiate: y = 11 - x 223 + 15 + 4x22. Solution Here we combine the Sum–Difference Rule and the Extended Power Rule: dy dx

= 311 - x 2221- 2x2 + 215 + 4x21 # 4.

We differentiate each term using the Extended Power Rule.

Since dy>dx = 311 - x 2221- 2x2 + 215 + 4x2 # 4, it follows that

Quick Check 1

O

a) Use the Extended Power Rule to differentiate y = 1x 4 + 2x 2 + 123. b) Explain why d 31x 2 + 4x + 1244 = dx 41x 2 + 4x + 123 # 2x + 4 is incorrect.

dy dx

= - 6x11 - x 222 + 815 + 4x2 = - 6x11 - 2x 2 + x 42 + 40 + 32x = - 6x + 12x 3 - 6x 5 + 40 + 32x = 40 + 26x + 12x 3 - 6x 5.

O Quick Check 1 ■ EXAMPLE 3 Differentiate: f1x2 = 13x - 52417 - x210. Solution Here we combine the Product Rule and the Extended Power Rule:

f ¿1x2 = 13x - 524 # 1017 - x291- 12 + 17 - x210 413x - 523132 = - 1013x - 52417 - x29 + 17 - x210 1213x - 523 = 213x - 52317 - x293- 513x - 52 + 617 - x24 We factor out = 213x - 52 17 - x2 1- 15x + 25 + 42 - 6x2 = 213x - 52317 - x29167 - 21x2. 3

9

■ EXAMPLE 4 Differentiate: f1x2 =

213x - 52317 - x29.

4 x + 3 . Ax - 2

Solution Here we use the Quotient Rule to differentiate the inside function:

d 4 x + 3 d x + 3 1>4 1 x + 3 1>4 - 1 1x - 221 - 11x + 32 = a b = a b c d dx A x - 2 dx x - 2 4 x - 2 1x - 222 1 x + 3 -3>4 x - 2 - x - 3 = a b c d 4 x - 2 1x - 222

Quick Check 2 Differentiate: f1x2 =

12x 2 - 12

13x 4 + 222

= .

O Quick Check 2

1 x + 3 -3/4 -5 a b c d, or 4 x - 2 1x - 222

-5 . 41x + 32 1x - 225/4 3/4

O

1.7

• The Chain Rule

169

Composition of Functions and the Chain Rule Before discussing the Chain Rule, let’s consider composition of functions. One author of this text exercises three times a week at a local YMCA. When he recently bought a pair of running shoes, he found a label on which the numbers at the bottom indicate equivalent shoe sizes in five countries.

UPC

0615060001

072177

11 (T1)

1CA RUNNING SHOES 6

39781 02305

11 D Author Marv Bittinger and his size-1112 running shoes

14

RWH I /BLACK/SILVER

11 GB

46 F

11

295J

This label suggests that there are functions that convert one country’s shoe sizes to those used in another country. There is, indeed, a function g that gives a correspondence between shoe sizes in the United States and those in France: 4x + 92 , 3

g1x2 =

where x is the U.S. size and g1x2 is the French size. Thus, a U.S. size 1112 corresponds to a French size g111122

=

4 # 1112 + 92 3

, or 46.

There is also a function f that gives a correspondence between shoe sizes in France and those in Japan. The function is given by f1x2 =

15x - 100 , 2

where x is the French size and f1x2 is the corresponding Japanese size. Thus, a French size 46 corresponds to a Japanese size f1462 =

15 # 46 - 100 , or 295. 2

It seems reasonable to conclude that a shoe size of 1112 in the United States corresponds to a size of 295 in Japan and that some function h describes this correspondence. Can we find a formula for h? g g 1x2 =

f

4x + 92 3

f 1x2 =

15x - 100 2

U.S.

France

Japan

8 9 10 11 1112

41 13 42 23 44 45 13 46

260 270 280 290 295

h h1x2 = ?

170

CHAPTER 1

• Differentiation

A shoe size x in the United States corresponds to a shoe size g1x2 in France, where g1x2 =

4x + 92 . 3

Thus, 14x + 922>3 represents a shoe size in France. If we replace x in f1x2 with 14x + 922>3, we can find the corresponding shoe size in Japan: 15a f1g1x22 = =

4x + 92 b - 100 3 2

514x + 922 - 100

=

20x + 460 - 100 2

2 20x + 360 = = 10x + 180. 2

This gives a formula for h: h1x2 = 10x + 180. As a check, a shoe size of 1112 in the United States corresponds to a shoe size of h A 1112 B = 10 A 1112 B + 180 = 295 in Japan. The function h is the composition of f and g, symbolized by f  g and read as “f composed with g,” or simply “f circle g.”

DEFINITION The composed function f  g, the composition of f and g, is defined as 1f  g21x2 = f1g1x22.

We can visualize the composition of functions as shown below. Inputs, x x

f° g g g(x) f

f( g(x))

Outputs, f(g(x)) or (f ° g)(x)

A composition machine for functions f and g

To find 1f  g21x2, we substitute g1x2 for x in f1x2. The function g1x2 is nested within f1x2.

■ EXAMPLE 5 For f1x2 = x 3 and g1x2 = 1 + x 2, find 1f  g21x2 and 1g  f21x2. Solution Consider each function separately: f1x2 = x 3

This function cubes each input.

g1x2 = 1 + x 2.

This function adds 1 to the square of each input.

and

1.7

• The Chain Rule

171

a) The function f  g first does what g does (adds 1 to the square) and then does what f does (cubes). We find f1g1x22 by substituting g1x2 for x: 1f  g21x2 = f1g1x22 = f11 + x 22 = 11 + x 223 = 1 + 3x 2 + 3x 4 + x 6.

Using g 1x2 as an input

b) The function g  f first does what f does (cubes) and then does what g does (adds 1 to the square). We find g1f1x22 by substituting f1x2 for x: 1g  f21x2 = g1f1x22 = g1x 32 = 1 + 1x 322 = 1 + x 6.

Using f 1x2 as an input

■ EXAMPLE 6 For f1x2 = 2x and g1x2 = x - 1, find 1f  g21x2 and 1g  f21x2. Quick Check 3 For the functions in Example 6, find: a) 1f  f21x2; b) 1g  g21x2.

Solution

1f  g21x2 = f1g1x22 = f1x - 12 = 2x - 1 1g  f21x2 = g1f1x22 = g A 2x B = 2x - 1

O Quick Check 3 Keep in mind that, in general, 1f  g21x2 Z 1g  f21x2. We see this fact demonstrated in Examples 5 and 6. How do we differentiate a composition of functions? The following theorem tells us.

THEOREM 8

The Chain Rule The derivative of the composition f  g is given by d d 31f  g21x24 = 3f1g1x224 = f ¿1g1x22 # g¿1x2. dx dx

As we noted earlier, the Extended Power Rule is a special case of the Chain Rule. Consider f1x2 = x k. For any other function g1x2, we have 1f  g21x2 = 3g1x24k, and the derivative of the composition is d 3g1x24k = k3g1x24k-1 # g¿1x2. dx The Chain Rule often appears in another form. Suppose that y = f1u2 and u = g1x2. Then dy dx

=

dy du # . du dx

To better understand the Chain Rule, suppose that a video game manufacturer wished to determine its rate of profit, in dollars per minute. One way to find this rate would be to multiply the rate of profit, in dollars per item, by the production rate, in items per minute. That is,

B with respect to time R = B to number of games produced R # B produced with respect to time. R Change in profits

Change in profits with respect

Change in number of games

O

172

CHAPTER 1

• Differentiation

■ EXAMPLE 7 For y = 2 + 2u and u = x 3 + 1, find dy>du, du>dx, and dy>dx. Solution First we find dy>du and du>dx: dy du

=

1 -1/2 u and 2

du = 3x 2. dx

Then dy dx

= =

Quick Check 4

=

If y = u2 + u and dy u = x 2 + x, find . dx

dy du # du dx 1 #

3x 2 22u 3x 2 22x 3 + 1

Substituting x 3 + 1 for u

.

O Quick Check 4 ■ EXAMPLE 8 Business. A new product is placed on the market and becomes very popular. Its quantity sold N is given as a function of time t, where t is measured in weeks: N1t2 =

250,000t 2 12t + 122

t 7 0.

,

Differentiate this function. Then use the derivative to evaluate N¿1522 and N¿12082, and interpret these results.

Solution To determine N¿1t2, we use the Quotient Rule along with the Extended Power Rule:

N¿1t2 =

250,000t 2

d B R = dt 12t + 122 = = =

Therefore,

N¿1t2 =

12t + 122 #

d d 3250,000t 24 - 250,000t 2 # 312t + 1224 dt dt 312t + 12242

12t + 122 # 1500,000t2 - 250,000t 2 # 212t + 121 # 2 12t + 12

4

12t + 1221500,000t2 - 1,000,000t 212t + 12

The Extended Power Rule is used here.

12t + 124

500,000t12t + 12312t + 12 - 2t4 12t + 124

500,000t

12t + 12

3

.

.

The expression 500,000t12t + 12 is factored out in the numerator; the 2t terms inside the square brackets sum to 0.

The factor 12t + 12 in the numerator cancels 12t + 12 in the denominator.

We evaluate N¿1t2 at t = 52: N¿1522 =

500,0001522

121522 + 123

L 22.5.

Thus, after 52 weeks (1 yr), the quantity sold is increasing by about 22.5 units per week. For t = 208 weeks (4 yr), we get N¿12082 =

500,00012082

1212082 + 123

L 1.4.

O

Exercise Set 1.7

173

After 4 yr, the quantity sold is increasing at about 1.4 units per week. What is happening here? Consider the graph of N1t2: N(t)

Quantity sold

62,500

N(52) = 61,315 units sold N‘(52) = 22.5 units sold per week

N

50

100

150

N(208) = 62,201 units sold N‘(208) = 1.4 units sold per week

200

t

Time (in weeks)

We see that the slopes of the tangent lines, representing the change in numbers of units sold per week, are leveling off as t increases. Perhaps the market is becoming saturated with this product: while sales continue to increase, the rate of the sales increase per week is leveling off.

Section Summary

• The Extended Power Rule tells us that if y = 3f1x24k, then y¿ =

d 3f1x24k = k3f1x24k - 1 # f ¿1x2. dx

• The Chain Rule is used to differentiate a composition of functions. If F1x2 = 1f  g21x2 = f1g1x22,

• The composition of f1x2 with g1x2 is written 1f  g21x2 and is defined as 1f  g21x2 = f1g1x22. • In general, 1f  g21x2 Z 1g  f21x2.

then F¿1x2 =

d 31f  g21x24 = f ¿1g1x22 # g¿1x2. dx

EXERCISE SET

1.7 Differentiate each function. 1. y = 12x + 122

s

2. y = 13 - 2x22

15. y =

Check by expanding and then differentiating.

3. y = 17 - x255

4. y = 18 - x2100

5. y = 21 + 8x

6. y = 21 - x

7. y = 23x - 4

8. y = 24x + 1

2

9. y = 18x - 62 2

-40

11. y = 1x - 42812x + 326

2

10. y = 14x 2 + 12-50

13. y =

13x + 822

17 - 5x23

16. y =

17. f 1x2 = 11 + x 323 - 12 + x 824 18. f 1x2 = 13 + x 325 - 11 + x 724 19. f 1x2 = x 2 + 1200 - x22 20. f 1x2 = x 2 + 1100 - x22 21. g1x2 = 2x + 1x - 323 22. G1x2 = 22x - 1 + 14 - x22 3

12. y = 1x + 52714x - 1210 1

4x 2

14. y =

1

14x + 522

23. f 1x2 = - 5x12x - 324 24. f 1x2 = - 3x15x + 426

7x 3

14 - 9x25

174

• Differentiation

CHAPTER 1

25. g1x2 = 13x - 12712x + 125

57. Find an equation for the tangent line to the graph of y = 2x 2 + 3x at the point 11, 22.

26. F1x2 = 15x + 22412x - 328 27. f 1x2 = x 2 24x - 1

58. Find an equation for the tangent line to the graph of y = 1x 3 - 4x210 at the point 12, 02.

28. f 1x2 = x 3 25x + 2

59. Find an equation for the tangent line to the graph of y = x22x + 3 at the point 13, 92.

3

29. G1x2 = 2x 5 + 6x 4

30. F1x2 = 2x 2 - 5x + 2 31. f 1x2 = a

3x - 1 b 5x + 2

4

32. f 1x2 = a

4 - x A3 + x

33. g1x2 =

35. f 1x2 = 12x - 3x + 4x + 12 2

x + 1

b

3 + 2x A5 - x

34. g1x2 =

3

2x 2

3

37. g1x2 = a

2x + 3 -4 b 5x - 1

38. h1x2 = a

1 - 3x -5 b 2 - 7x

39. f 1x2 =

x2 + x

40. f 1x2 =

4 - x3

12x + 324

41. f 1x2 =

42. f 1x2 =

13x - 225

3

A x - x2

15x - 427

16x + 123

43. f 1x2 = 1212x + 122/313x - 425/4 3

44. y = 62x + x1x - 6x2 Find

2

4

3

dy du dy , , and . du dx dx

45. y = 2u and u = x 2 - 1 46. y =

15 u3

u + 1 and u = 1 + 2x u - 1

49. y = u1u + 12 and u = x - 2x 3

50. y = 1u + 121u - 12 and u = x 3 + 1 dy Find for each pair of functions. dx 51. y = 5u 2 + 3u and u = x 3 + 1 52. y = u3 - 7u 2 and u = x 2 + 3 3

53. y = 22u + 5 and u = x 2 - x 54. y = 27 - 3u and u = x 2 - 9 55. Find

56. Find

dy dt dy dt

f 1x2 =

x2

11 + x25

.

a) Find f ¿1x2 using the Quotient Rule and the Extended Power Rule. b) Note that f 1x2 = x 211 + x2-5. Find f ¿1x2 using the Product Rule and the Extended Power Rule. c) Compare your answers to parts (a) and (b). 62. Consider g1x2 = a

6x + 1 2 b . 2x - 5

a) Find g¿1x2 using the Extended Power Rule. b) Note that g1x2 =

36x 2 + 12x + 1 4x 2 - 20x + 25

.

Find g¿1x2 using the Quotient Rule. c) Compare your answers to parts (a) and (b). Which approach was easier, and why? In Exercises 63–66, find f1x2 and g1x2 such that h1x2  1 f  g21x2. Answers may vary.

and u = 2x + 1

47. y = u50 and u = 4x 3 - 2x 2 48. y =

61. Consider

100

36. f 1x2 = 17x 4 + 6x 3 - x2204

Bx 2 - x

60. Find an equation for the tangent line to the graph of 2x + 3 3 y = a b at the point 12, 3432. x - 1

63. h1x2 = 13x 2 - 725 65. h1x2 =

x + 1 3

x3 - 1

64. h1x2 = 66. h1x2 =

if y = if y =

u2 + u

and u = 5 + 3t.

1 3u 5 - 7

and u = 7t 2 + 1.

27x + 2

A 2x + 5 B 4

Do Exercises 67–70 in two ways. First, use the Chain Rule to find the answer. Next, check your answer by finding f 1g1x22, taking the derivative, and substituting. 67. f 1u2 = u 3, g1x2 = u = 2x 4 + 1 Find 1 f  g2¿1- 12. u + 1 , g1x2 = u = 2x u - 1 Find 1 f  g2¿142.

68. f 1u2 =

69. f 1u2 = 2u, g1x2 = u = 1 + 3x 2 Find 1 f  g2¿122. 3

1

1

70. f 1u2 = 2u5, g1x2 = u = Find 1 f  g2¿1- 102.

3 - x 4 + x

Exercise Set 1.7

For Exercises 71–74, use the Chain Rule to differentiate each function. You may need to apply the rule more than once. 71. f 1x2 = 12x 3 + 14x - 52226 72. f 1x2 =

A - x 5 + 4x + 22x + 1 B 3

74. f 1x2 = 22x + 1x 2 + x24 3

APPLICATIONS Business and Economics 75. Total revenue. A total-revenue function is given by R1x2 = 10002x 2 - 0.1x, where R1x2 is the total revenue, in thousands of dollars, from the sale of x items. Find the rate at which total revenue is changing when 20 items have been sold. C1x2 = 20001x + 22

+ 700,

where C1x2 is the total cost, in thousands of dollars, of producing x items. Find the rate at which total cost is changing when 20 items have been produced. 77. Total profit. Use the total-cost and total-revenue functions in Exercises 75 and 76 to find the rate at which total profit is changing when x items have been produced and sold. 78. Total cost. A company determines that its total cost, in thousands of dollars, for producing x items is C1x2 = 25x 2 + 60,

81. Compound interest. If $1000 is invested at interest rate i, compounded annually, in 3 yr it will grow to an amount A given by (see Section R.1) A = $100011 + i23.

82. Compound interest. If $1000 is invested at interest rate i, compounded quarterly, in 5 yr it will grow to an amount, A, given by A = $1000 a1 +

x1t2 = 20t + 40. How fast will costs be rising 4 months from now? 79. Consumer credit. The total outstanding consumer credit of the United States (in billions of dollars) can be modeled by the function C1x2 = 0.21x 4 - 5.92x 3 + 50.53x 2 - 18.92x + 1114.93, where x is the number of years since 1995.

C(x) = 0.21x4 − 5.92x3 + 50.53x2 − 18.92x + 1114.93 C(x) 3000 2500

i 20 b . 4

a) Find the rate of change, dA>di. b) Interpret the meaning of dA>di. 83. Consumer demand. Suppose that the demand function for a product is given by D1p2 =

and it plans to boost production t months from now according to the function

Consumer credit outstanding (in billions of dollars)

Find the rate at which the utility changes with respect to the number of tickets bought.

a) Find the rate of change, dA>di. b) Interpret the meaning of dA>di.

76. Total cost. A total-cost function is given by 1/3

a) Find dC>dx. b) Interpret the meaning of dC>dx. c) Using this model, estimate how quickly outstanding consumer credit was rising in 2010. 80. Utility. Utility is a type of function that occurs in economics. When a consumer receives x units of a product, a certain amount of pleasure, or utility, U, is derived. Suppose that the utility related to the number of tickets x for a ride at a county fair is 2x + 1 . U1x2 = 80 A 3x + 4

73. f 1x2 = 3x 2 + 21 - 3x

2

175

80,000 , p

and that price p is a function of time given by p = 1.6t + 9, where t is in days. a) Find the demand as a function of time t. b) Find the rate of change of the quantity demanded when t = 100 days. 84. Business profit. A company is selling laptop computers. It determines that its total profit, in dollars, is given by P1x2 = 0.08x 2 + 80x, where x is the number of units produced and sold. Suppose that x is a function of time, in months, where x = 5t + 1. a) Find the total profit as a function of time t. b) Find the rate of change of total profit when t = 48 months.

Life and Physical Sciences

2000 1500 1000 0

4

8

12

Number of years since 1995 ( Source: Federal Reserve Board.)

16

x

85. Chemotherapy. The dosage for Carboplatin chemotherapy drugs depends on several parameters of the particular drug as well as the age, weight, and sex of the patient. For female patients, the formulas giving the dosage for such drugs are D = 0.85A1c + 252 and

c = 1140 - y2

w , 72x

176

• Differentiation

CHAPTER 1

where A and x depend on which drug is used, D is the dosage in milligrams (mg), c is called the creatine clearance, y is the patient’s age in years, and w is the patient’s weight in kilograms (kg). (Source: U.S. Oncology.) a) Suppose that a patient is a 45-year-old woman and the drug has parameters A = 5 and x = 0.6. Use this information to write formulas for D and c that give D as a function of c and c as a function of w. b) Use your formulas from part (a) to compute dD>dc. c) Use your formulas from part (a) to compute dc>dw. d) Compute dD>dw. e) Interpret the meaning of the derivative dD>dw.

SYNTHESIS If f1x2 is a function, then 1 f  f 21x2  f 1 f 1x22 is the composition of f with itself. This is called an iterated function, and the composition can be repeated many times. For example, 1 f  f  f 21x2  f 1 f 1f 1x222. Iterated functions are very useful in many areas, including finance (compound interest is a simple case) and the sciences (in weather forecasting, for example). For the each function, use the Chain Rule to find the derivative. 86. If f 1x2 = x 2 + 1, find

d 31 f  f 21x24. dx

87. If f 1x2 = x + 2x, find 88. If f 1x2 = x 2 + 1, find

91. y = 2x 3 + 6x + 1 # x 5 92. s = 2t 4 + 3t 2 + 8 # 3t 4

97. y = a

x2 - x - 1 x + 1 2

94. y = A x21 + x 2 B 3 96. w = b

3

Therefore, . . . . 102. The Extended Power Rule (for positive integer powers) can be verified using the Product Rule. For example, if y = 3 f 1x242, then the Product Rule is applied by recognizing that 3 f 1x242 = 3 f 1x24 # 3 f 1x24. Therefore, d 13 f 1x24 # 3 f 1x242 = f 1x2 # f ¿1x2 + f ¿1x2 # f 1x2 dx = 2f 1x2 # f ¿1x2. d a) Use the Product Rule to show that 3 f 1x243 = dx 33 f 1x242 # f ¿1x2. [Hint: 3 f 1x243 = 3 f 1x242 # f 1x2.4 d b) Use the Product Rule to show that 3 f 1x244 = dx 43 f 1x243 # f ¿1x2.

106. g1x2 =

3

21 - x 95. y = 1 - x

Q1x2 = N1x2 # 3D1x24-1.

105. f 1x2 = x24 - x 2

90. y = 212x - 322 + 1

2

.

Find the derivative of each of the following functions analytically. Then use a calculator to check the results.

Differentiate.

b

D1x2

104. f 1x2 = 26x 3 - 3x 2 - 48x + 45; 3- 5, 54

d 31 f  f  f 21x24. dx Do you see a shortcut?

2x - 1

Then

N1x2

103. f 1x2 = 1.68x29.2 - x 2; 3- 3, 34

3

93. y = a

Q1x2 =

For the function in each of Exercises 103 and 104, graph f and f ¿ over the given interval. Then estimate points at which the tangent line is horizontal.

d 31 f  f  f 21x24 dx

3

Proof. Let

TECHNOLOGY CONNECTION

d 31 f  f 21x24. dx

89. If f 1x2 = 2x, find

x

the Power Rule. Complete the proof, giving reasons for each step.

u 21 + u2

98. g1x2 =

x 2 - 4x B 2x + 1

99. f 1t2 = 33t + 2t 100. F1x2 = 36x13 - x25 + 244 101. The following is the beginning of an alternative proof of the Quotient Rule that uses the Product Rule and

107. f 1x2 =

4x 2x - 10

A 22x - 1 + x 3 B 5

Answers to Quick Checks

1. (a) y¿ = 31x 4 + 2x 2 + 12214x 3 + 4x2 (b) The result lacks parentheses around 2x + 4. It should be written: y¿ = 41x 2 + 4x + 12312x + 42. - 36x 5 + 24x 3 + 8x 2. y¿ = 13x 4 + 223 3. (a) f 1 f 1x22 = 32x = 2x; (b) g1g1x22 = 1x - 12 - 1 = x - 2 dy du dy # = 12u + 1212x + 12 = 4. = dx du dx 121x 2 + x2 + 1212x + 12 = 12x 2 + 2x + 1212x + 12 4

1.8

1.8 OBJECTIVES

177

Higher-Order Derivatives Consider the function given by y = f1x2 = x 5 - 3x 4 + x. Its derivative f ¿ is given by

• Find derivatives of higher order. • Given a formula for distance, find velocity and acceleration.

• Higher-Order Derivatives

y¿ = f ¿1x2 = 5x 4 - 12x 3 + 1. The derivative function f ¿ can also be differentiated. We can think of the derivative of f ¿ as the rate of change of the slope of the tangent lines of f. It can also be regarded as the rate at which f ¿1x2 is changing. We use the notation f – for the derivative 1 f ¿2¿. That is, f –1x2 =

d f ¿1x2. dx

We call f – the second derivative of f. For f1x2 = x 5 - 3x 4 + x, the second derivative is given by y – = f –1x2 = 20x 3 - 36x 2. Continuing in this manner, we have f ‡1x2 = 60x 2 - 72x, f 1x2 = 120x - 72, f ¿1x2 = 120.

The third derivative of f The fourth derivative of f The fifth derivative of f

When notation like f ‡1x2 gets lengthy, we abbreviate it using a number or n in parentheses. Thus, f 1n21x2 is the nth derivative. For the function above, f 1421x2 f 1521x2 f 1621x2 f 1n21x2

= = = =

120x - 72, 120, 0, and 0, for any integer n Ú 6.

Leibniz notation for the second derivative of a function given by y = f1x2 is d 2y dx

2

, or

d dy a b, dx dx

read “the second derivative of y with respect to x.” The 2’s in this notation are not exponents. If y = x 5 - 3x 4 + x, then d 2y dx 2

= 20x 3 - 36x 2.

Leibniz notation for the third derivative is d 3y>dx 3; for the fourth derivative, d y>dx 4; and so on: 4

d 3y dx 3 d 4y dx 4 d 5y dx 5

= 60x 2 - 72x, = 120x - 72, = 120.

178

CHAPTER 1

• Differentiation

■ EXAMPLE 1 For y = 1>x, find d 2y>dx 2. Solution We have y = x -1, so = - 1 # x -1 - 1 = - x -2, or

dy dx

-

1 . x2

Then d 2y dx 2

= 1- 22 1- 12x -2 - 1 = 2x -3, or

2 . x3

■ EXAMPLE 2 For y = 1x 2 + 10x220, find y¿ and y–. Solution To find y¿, we use the Extended Power Rule: y¿ = 201x 2 + 10x21912x + 102 = 201x 2 + 10x219 # 21x + 52 = 401x + 10x2 1x + 52. 2

19

Factor out a 2. 20 * 2 = 40

To find y–, we use the Product Rule and the Extended Power Rule: Quick Check 1 a) Find y–: (i) y = - 6x 4 + 3x 2; 2 (ii) y = 3 ; x (iii) y = 13x 2 + 122. b) Find

d4 1 c d. dx 4 x

y– = = = = = =

401x 2 401x 2 401x 2 401x 2 401x 2 401x 2

+ + + + + +

10x219112 + 1x + 5219 # 401x 2 + 10x21812x + 102 10x219 + 7601x + 52 1x 2 + 10x218 21x + 52 19 * 40 = 760 10x219 + 15201x + 5221x 2 + 10x218 760 * 2 = 1520 10x218 31x 2 + 10x2 + 381x + 5224 Factoring 18 2 2 10x2 3x + 10x + 381x + 10x + 2524 10x218139x 2 + 390x + 9502.

O Quick Check 1 Velocity and Acceleration We have already seen that a function’s derivative represents its instantaneous rate of change. When the function relates a change in distance to a change in time, the instantaneous rate of change is called speed, or velocity.* The letter v is generally used to stand for velocity.

DEFINITION The velocity of an object that is s1t2 units from a starting point at time t is given by Velocity = v1t2 = s¿1t2 = lim

h:0

s1t + h2 - s1t2 h

.

*In this text, the words “speed” and “velocity” are used interchangeably. In physics and engineering, this is not done, since velocity requires direction and speed does not.

O

1.8

• Higher-Order Derivatives

179

■ EXAMPLE 3 Physical Science: Velocity. Suppose that an object travels so that its distance s, in miles, from its starting point is a function of time t, in hours, as follows: s1t2 = 10t 2. a) Find the average velocity between the times t = 2 hr and t = 5 hr. b) Find the (instantaneous) velocity when t = 4 hr.

Solution a) From t = 2 hr to t = 5 hr, we have s152 - s122 Difference in miles = Difference in hours 3 2 # 10 5 mi - 10 # 2 2 mi mi = = 70 . 3 hr hr b) The instantaneous velocity is given by lim

s1t + h2 - s1t2

h :0

h

= s¿1t2.

We know how to find this limit quickly from the special techniques learned in Section 1.5. Thus, s¿1t2 = 20t, and s¿142 = 20 # 4 = 80

mi . hr

Often velocity itself is a function of time. When a jet takes off or a vehicle comes to a sudden stop, the change in velocity is easily felt by passengers. The rate at which velocity changes is called acceleration. Suppose that Car A reaches a speed of 65 mi/hr in 8.4 sec and Car B reaches a speed of 65 mi/hr in 8 sec; then B has a faster acceleration than A. We generally use the letter a for acceleration. It is useful to think of acceleration as the rate at which velocity is changing.

DEFINITION Acceleration = a1t2 = v¿1t2 = s–1t2.

■ EXAMPLE 4 Physical Science: Distance, Velocity, and Acceleration. For s1t2 =

10t 2, find v1t2 and a1t2, where s is the distance from the starting point, in miles, and t is in hours. Then find the distance, velocity, and acceleration when t = 4 hr.

Solution We have s1t2 = 10t 2. Thus, v1t2 = s¿1t2 = 20t and a1t2 = v¿1t2 = s–1t2 = 20. It follows that

and

s142 = 101422 = 160 mi, v142 = 20142 = 80 mi>hr, a142 = 20 mi>hr2.

If this distance function applies to motion of a vehicle, then at time t = 4 hr, the vehicle has traveled 160 mi, the velocity is 80 mi>hr, and the acceleration is 20 miles per hour per hour, which we abbreviate as 20 mi>hr2.

180

CHAPTER 1

• Differentiation

Note from Example 4 that since acceleration represents the rate at which velocity is changing, the units in which it is measured involve a unit of time squared: mi>hr Change in velocity mi # 1 = = = mi>hr2. Change in time hr hr hr

■ EXAMPLE 5 Free Fall. When an object is dropped, the distance it falls in t seconds, assuming that air resistance is negligible, is given by s1t2 = 4.905t 2, where s1t2 is in meters (m). If a stone is dropped from a cliff, find each of the following, assuming that air resistance is negligible: (a) how far it has traveled 5 sec after being dropped, (b) how fast it is traveling 5 sec after being dropped, and (c) the stone’s acceleration after it has been falling for 5 sec.

Solution a) After 5 sec, the stone has traveled s152 = 4.9051522 = 4.9051252 = 122.625 m. b) The speed at which the stone falls is given by v1t2 = s¿1t2 = 9.81t. Quick Check 2 A pebble is dropped from a hot-air balloon. Find how far it has fallen, how fast it is falling, and its acceleration after 3.5 sec. Let s1t2 = 16t 2, where t is in seconds and s is in feet.

Thus,

v152 = 9.81 # 5 = 49.05 m>sec.

c) The stone’s acceleration after t sec is constant: a1t2 = v¿1t2 = s–1t2 = 9.81 m>sec2. Thus, s–152 = 9.81 m>sec2.

O Quick Check 2 In Example 8 in Section 1.7, N1t2 represented the quantity sold N of a product after t weeks on the market. Its first derivative was always positive (always increasing) indicating that sales were always increasing. But sales were leveling off toward zero. In the following example, we see how the second derivative can help us understand this observation.

■ EXAMPLE 6 Business. In Example 8 in Section 1.7, the function N1t2 =

250,000t 2 ,

12t + 122 t 7 0, represented the quantity sold N of a product after t weeks on the market. Its derivative is N¿1t2 =

500,000t

12t + 123

.

Recall that N¿1t2 represented the rate of change in number of units sold per week. Find N–1t2; then use it to calculate N–1522 and N–12082 and interpret these results.

Solution We use Quotient Rule along with the Extended Power Rule: d 3N¿1t24 = dt = =

12t + 123 #

d d 3500,000t4 - 1500,000t2 # 312t + 1234 dt dt 312t + 12342 Extended

12t + 1231500,0002 - 1500,000t213212t + 122 # 2

Power Rule

312t + 12342

12t + 122312t + 121500,0002 - 61500,000t24 12t + 126

.

Factoring out 12t + 122

O

1.8

• Higher-Order Derivatives

181

After simplification, we have N–1t2 =

- 2,000,000t + 500,000 12t + 124

.

At t = 52, we have N–1522 =

- 2,000,0001522 + 500,000 321522 + 144

L - 0.852.

Thus, after 52 weeks (1 yr), the rate of the rate of sales is decreasing at - 0.852 units per week per week. In other words, although sales are increasing during the 52nd week (remember, N¿1522 7 02, the rate at which sales are increasing is decreasing. To put it in most basic terms: sales are increasing but not as fast as before. This is more evident when t = 208: N–12082 =

- 2,000,00012082 + 500,000 3212082 + 144

L - 0.014.

After 208 weeks (4 yr), sales have nearly leveled off. Remember, sales are increasing at the 208th week (recall that N¿12082 7 0), but since the market is nearly saturated with this product, the rate at which sales are increasing has slowed to near 0. The change in the slopes of these tangent lines is indicated by the second derivative N”(t).

N(t)

Quantity sold

62,500

N(t) =

50

250,000 t 2 ,t>0 (2t + 1) 2

100

150

200

t

Time (in weeks)

There are many important real-world applications that make use of the second derivative. In Chapter 2, we will examine a number of these, including applications in the fields of economics, health care, and the natural and physical sciences.

Section Summary • The second derivative is the derivative of the first derivative of a function. In symbols, f –1x2 = 3f ¿1x24¿. • The second derivative describes the rate of change of the rate of change. In other words, it describes the rate of change of the first derivative.

• A real-life example of a second derivative is acceleration. If s1t2 represents distance as a function of time of a moving object, then v1t2 = s¿1t2 describes the speed (velocity) of the object. Any change in the speed of the object is the acceleration: a1t2 = v¿1t2 = s–1t2. • The common notation for the nth derivative of a function dn is f 1n21x2 or n f1x2. dx

182

• Differentiation

CHAPTER 1

TECHNOLOGY CONNECTION

Exploratory

2. The x-values in the table are increasing. What is the

Many calculators have a tangent-drawing feature. This feature can be used to explore the behavior of the second derivative. Graph the function f 1x2 = x 2 in the standard window. Select DRAW and then select Tangent. Choose an x-value by typing it in or using the arrow keys to trace along the graph. Press [, and a tangent line will be drawn at the selected x-value. In the lower-left corner of the screen, the equation of the tangent line is given. The slope of the line is the coefficient of x. You can create a table of slope values at various x-values, and from this table, infer how the second derivative helps describe the shape of a graph. (This activity can be easily adapted for Graphicus and iPlot.)

EXERCISES

1. Let f 1x2 = x 2. Use the tangent-drawing feature to com-

plete the following table: x

corresponding behavior of the slopes (increasing or decreasing)? 3. How is the graph “turning”? What conclusion can you

make about the behavior of the slopes as x increases in value? 4. Make a table of slopes for f 1x2 = - x 2. Analyze their

behavior relative to the “turning” of this function’s graph. 5. What general conclusion can you make about the

second derivative and the “turning” of a graph? This “turning” behavior of a graph is known as concavity, and it is a very useful concept in the analysis of functions. We will develop more about the second derivative and concavity in Chapter 2.

Slope at x

-4 -2 0 2 4

EXERCISE SET

1.8 19. f 1x2 = 1x 2 + 3x27

Find d 2y/d x 2.

21. f 1x2 = 12x 2 - 3x + 1210

1. y = x + 9

2. y = x - 7

3. y = 2x - 5x

4. y = 5x + 4x

5. y = 4x + 3x - 1

6. y = 4x - 5x + 7

7. y = 7x + 2

8. y = 6x - 3

5

4

4 2

9. y =

1

3 2

10. y =

x2

Find f fl 1x2.

15. f 1x2 = x

1>5

17. f 1x2 = 4x - 3

3 x

14. f 1x2 = x 3 16. f 1x2 = x

4

3

26. y = x 3>2 - 5x

29. y = 2x 5>4 + x 1>2

30. y = 3x 4>3 - x 1>2

5 x

31. y =

2 x

3

+

1 x

2

33. y = 1x + 3214x - 12 2

1>3

18. f 1x2 = 2x - 2

24. f 1x2 = 21x 2 - 122

25. y = x 2>3 + 4x

27. y = 1x 3 - x23>4

x3

12. y = 2x

13. f 1x2 = x 4 +

23. f 1x2 = 21x 2 + 123 Find y–.

1 4

11. y = 2x

22. f 1x2 = 13x 2 + 2x + 125

20. f 1x2 = 1x 3 + 2x26

35. y =

3x + 1 2x - 3

28. y = 1x 4 + x22>3

32. y =

3 x

4

-

1 x

34. y = 1x 3 - 2215x + 12 36. y =

2x + 3 5x - 1

183

Exercise Set 1.8

37. For y = x 4, find d 4y>dx 4. 38. For y = x 5, find d 4y>dx 4. 39. For y = x 6 - x 3 + 2x, find d 5y>dx 5. 40. For y = x 7 - 8x 2 + 2, find d 6y>dx 6. 41. For f1x2 = x - 2 - x 1>2, find f 1421x2. 42. For f 1x2 = x

-3

+ 2x

1>3

, find f

1x2.

152

50. Free fall. (See Exercise 49.) Suppose a worker drops a bolt from a scaffold high above a work site. Assuming that air resistance is negligible, find (a) how far the bolt falls in 2 sec, (b) how fast the bolt is traveling 2 sec after being dropped, and (c) the bolt’s acceleration after it has been falling for 2 sec. 51. Free fall. Find the velocity and acceleration of the stone in Example 5 after it has been falling for 2 sec.

43. For g1x2 = x - 3x - 7x - 6x + 9, find g 1621x2.

52. Free fall. Find the velocity and acceleration of the stone in Example 5 after it has been falling for 3 sec.

44. For g1x2 = 6x 5 + 2x 4 - 4x 3 + 7x 2 - 8x + 3, find g 1721x2.

53. The following graph describes an airplane’s distance from its last point of rest.

4

3

2

APPLICATIONS s(t) Distance (in meters)

Life and Physical Sciences 45. Given s1t2 = t 3 + t, where s is in feet and t is in seconds, find each of the following. a) v1t2 b) a1t2 c) The velocity and acceleration when t = 4 sec

(20, 10,000) 10,000 8000 6000 4000 2000 0

(6, 1000) 2

4

6

8

10

12

14

16

18

20

t

Time (in seconds)

46. Given s1t2 = - 10t 2 + 2t + 5,

a) Is the plane’s velocity greater at t = 6 sec or t = 20 sec? How can you tell? b) Is the plane’s acceleration positive or negative? How can you tell?

where s is in meters and t is in seconds, find each of the following. a) v1t2 b) a1t2 c) The velocity and acceleration when t = 1 sec s1t2 = 3t + 10, where s is in miles and t is in hours, find each of the following. a) v1t2 b) a1t2 c) The velocity and acceleration when t = 2 hr d) When the distance function is given by a linear function, we have uniform motion. What does uniform motion mean in terms of velocity and acceleration? 48. Given s1t2 = t 2 -

1 t + 3, 2

where s is in meters and t is in seconds, find each of the following. a) v1t2 b) a1t2 c) The velocity and acceleration when t = 1 sec 49. Free fall. When an object is dropped, the distance it falls in t seconds, assuming that air resistance is negligible, is given by s1t2 = 16t 2, where s1t2 is in feet. Suppose that a medic’s reflex hammer is dropped from a hovering helicopter. Find (a) how far the hammer falls in 3 sec, (b) how fast the hammer is traveling 3 sec after being dropped, and (c) the hammer’s acceleration after it has been falling for 3 sec.

54. The following graph describes a bicycle racer’s distance from a roadside television camera. s(t) Distance (in meters)

47. Given

1000

(100, 1000)

800 600 400 200 0

(10, 280) 10

20

30

40

50

60

70

80

90 100 110 120

t

Time (in seconds)

a) When is the bicyclist’s velocity the greatest? How can you tell? b) Is the bicyclist’s acceleration positive or negative? How can you tell? 55. Sales. A company determines that monthly sales S, in thousands of dollars, after t months of marketing a product is given by S1t2 = 2t 3 - 40t 2 + 220t + 160. a) Find S¿112, S¿122, and S¿142. b) Find S–112, S–122, and S–142. c) Interpret the meaning of your answers to parts (a) and (b).

184

CHAPTER 1

• Differentiation

56. Sales. A business discovers that the number of items sold t days after launching a new sales promotion is given by N1t2 = 2t 3 - 3t 2 + 2t. a) Find N¿112, N¿122, and N¿142. b) Find N–112, N–122, and N–142. c) Interpret the meaning of your answers to parts (a) and (b). 2000t models the 4t + 75 population p of deer in an area after t months. a) Find p¿1102, p¿1502, and p¿11002. b) Find p–1102, p–1502, and p–11002. c) Interpret the meaning of your answers to parts (a) and (b). What is happening to this population of deer in the long term?

57. Population. The function p1t2 =

58. Medicine. A medication is injected into the bloodstream, where it is quickly metabolized. The percent concentration p of the medication after t minutes in the blood2.5t stream is modeled by the function p1t2 = 2 . t + 1 a) Find p¿10.52, p¿112, p¿152, and p¿1302. b) Find p–10.52, p–112, p–152, and p–1302. c) Interpret the meaning of your answers to parts (a) and (b). What is happening to the concentration of medication in the bloodstream in the long term?

SYNTHESIS Find

yÔ

for each function.

59. y =

1 1 - x

61. y =

1 22x + 1

60. y = x21 + x 2 62. y =

3x - 1 2x + 3

Find y fl for each function. 63. y =

2x + 1 2x - 1

71. Free fall. On the moon, all free-fall distance functions are of the form s1t2 = 0.81t 2, where t is in seconds and s is in meters. An object is dropped from a height of 200 meters above the moon. After t = 2 sec, a) How far has the object fallen? b) How fast is it traveling? c) What is its acceleration? d) Explain the meaning of the second derivative of this free-fall function. 72. Hang time. On Earth, an object will have traveled 4.905 m after 1 sec of free fall. Thus, by symmetry, a jumper requires 1 sec to leap 4.905 m high, then another second to come back to the ground. Assume a jumper starts on level ground. Explain why it is impossible for a human being, even Michael Jordan, to stay in the air for (have a “hang time” of) 2 sec. Can a human have a hang time of 1.5 sec? 1 sec? What do you think is the longest possible hang time achievable by humans jumping from level ground? 73. Free fall. Skateboarder Danny Way free-fell 28 ft from the Fender Stratocaster Guitar atop the Hard Rock Hotel & Casino in Las Vegas onto a ramp below. The distance s1t2, in feet, traveled by a body falling freely from rest in t seconds is approximated by s1t2 = 16t 2. Estimate Way’s velocity at the moment he touched down onto the ramp. (Note: You will need the result from Exercise 26 in Section R.1.)

TECHNOLOGY CONNECTION For the distance function in each of Exercises 74–77, graph s, v, and a over the given interval. Then use the graphs to determine the point(s) at which the velocity will switch from increasing to decreasing or from decreasing to increasing. 74. s1t2 = 0.1t 4 - t 2 + 0.4;

75. s1t2 = - t + 3t; 3- 3, 34

76. s1t2 = t 4 + t 3 - 4t 2 - 2t + 4; 77. s1t2 = t - 3t + 2; 3

64. y =

3- 5, 54

3

2

3- 2, 44

x 2x - 1

Answers to Quick Checks

65. For y = x k, find d 5y>dx 5.

1. (a) (i) y– = - 72x 2 + 6, (ii) y– =

66. For y = ax 3 + bx 2 + cx + d, find d 3y>dx 3.

(iii) y– = 108x 2 + 12; (b) y 142 =

Find the first through the fourth derivatives. Be sure to simplify at each stage before continuing. 67. f 1x2 =

x - 1 x + 2

3- 3, 34

68. f 1x2 =

x + 3 x - 2

69. Baseball. A baseball is dropping from a height of 180 ft. For how many seconds must it fall to reach a speed of 50 mi> hr? (Hint: See Exercise 49.) 70. Free fall. All free-fall distance functions follow this form on Earth: s1t2 = 4.905t 2, where t is in seconds and s is in meters. The second derivative always has the same value. What does that value represent?

24 x5

2. Distance = s13.52 = 196 ft; velocity = s¿13.52 = 112 ft>sec; acceleration = s–13.52 = 32 ft>sec2 .

24 , x5

CHAPTER 1 SUMMARY KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 1.1 As x approaches (but is not equal to) a, the limit of f 1x2 is L, written as

lim 12x + 32 = 11

x2 - 1 = 2 x:1 x - 1

Limit Numerically

Limit Graphically

lim

x:4

lim f 1x2 = L.

x:a

3.9 x 6 4e

y

2x + 3

x

4

10.8

3 2

3.99

10.98

3.999

10.998

4.1

11.2

4.01

11.02

4.001

11.002

1

x 7 4e

If x approaches a from the left 1x 6 a2, we have a left-hand limit, written as

G1x2 = e

x:a

lim+ f 1x2.

x:a

If the left-hand and right-hand limits are equal, then the limit as x approaches a exists. If the left-hand and right-hand limits are not equal, then the limit as x approaches a does not exist.

1

3

x

4

Consider the function G given by

lim- f 1x2.

If x approaches a from the right 1x 7 a2, we have a right-hand limit, written as

2 1 1

4 - x, for x 6 3, 2x - 2 + 1, for x Ú 3.

Graph the function and find each limit, if it exists. a) lim G1x2

b) lim G1x2

x:1

x:3

We check the limits from the left and from the right, both numerically and graphically. a)

Limit Numerically x : 1- 1x 6 12

Limit Graphically y

G1x2 6

0.9

3.1

0.99

3.01

0.999

3.001

5

G

3

1

x : 1+ 1x 7 12

G1x2

1.1

2.9

1.01

2.99

1.001

2.999

1 1

1

3

4

5

6

x

These choices can vary.

Both the tables and the graph show that as x gets closer to 1, the outputs G1x2 get closer to 3. Thus, lim G1x2 = 3. x:1

(continued) Chapter 1 Summary

185

186

CHAPTER 1

•

Differentiation

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 1.1 (continued ) b)

Limit Numerically

Limit Graphically y

x : 3 - 1x 6 32

G1x2 6

2.5

1.5

2.9

1.1

2.99

1.01

2.999

1.001

5 4

G

3 2 1

x : 3 + 1x 7 32

1 1

1

3

5

x

6

G1x2

3.5

2.2247

3.1

2.0488

3.01

2.0050

3.001

2.0005

lim G1x2 does not exist.

x:3

Both the tables and the graph indicate that lim- G1x2 Z lim+ G1x2. Since x:3

x:3

the left-hand and right-hand limits differ, lim G1x2 does not exist. x:3

SECTION 1.2 For any rational function F (see Section R.5) with a in its domain, we have lim F1x2 = F1a2.

x:a

Let f 1x2 = 2x 2 + 3x - 1, and let a = 2. We have

lim f 1x2 = f 122 = 21222 + 3122 - 1 = 13.

x:2

Let g1x2 =

x 2 - 16 , and let a = 6. We have x + 4

lim g1x2 =

x:6

A function f is continuous at x = a if the following three conditions are met: 1. f 1a2 exists. (The output at a exists.) 2. lim f 1x2 exists. (The limit as x : a x:a

exists.) 3. lim f 1x2 = f 1a2. (The limit is the x:a

same as the output.) If any one of these conditions is not fulfilled, the function is discontinuous at x = a.

1622 - 16 162 + 4

=

20 = 2. 10

x 2 - 3x - 4 continuous over 3- 3, 34? x + 1 For g to be continuous over 3- 3, 34, it must be continuous at each point in 3- 3, 34. Note that

Is the function g given by g 1x2 =

g 1x2 = =

x 2 - 3x - 4 x + 1 1x + 121x - 42

x + 1 = x - 4, provided x Z - 1.

Since - 1 is not in the domain of g, it follows that g1- 12 does not exist. Thus, g is not continuous over 3- 3, 34.

y 3 2 1 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

1 2 3 4 5 6 7

g(x) =

x2 − 3x − 4 x+1

x

187

Chapter 1 Summary

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 1.3 The average rate of change of y with respect to x between two points 1x 1, y12 and 1x 2, y22 is the slope of the line connecting the points: y2 - y1 x2 - x1

Business. At 1 P.M., a bookstore had revenue of $570 for the day, and at 4 P.M., it had revenue of $900 for the day. Therefore, the average rate of = 330 revenue with respect to time is 9004 -- 570 1 3 = 110, or $110 dollars per hour for the period of time between 1 P.M. and 4 P.M.

.

Let f be a function. Any line connecting two points on the graph of f is called a secant line. Its slope is the average rate of change of f, which is given by the difference quotient: f 1x + h2 - f 1x2 h

SECTION 1.4 The derivative of a function f is defined by h:0

f 1x + h2 = 31x + h22 = 3x 2 + 6xh + 3h2.

The difference quotient for this function simplifies to f 1x + h2 - f 1x2 h

,

where h is the difference between the two input x-values.

f ¿1x2 = lim

Let f 1x2 = 3x 2. Then

f 1x + h2 - f 1x2 h

3x 2 + 6xh + 3h2 - 3x 2 = 6x + 3h. h

Therefore, if x = 2 and h = 0.05, the slope of the secant line is 6122 + 310.052 = 12.15.

Let f 1x2 = 3x 2. Its simplified difference quotient is 6x + 3h. Therefore, the derivative is f ¿1x2 = lim 16x + 3h2 = 6x.

.

The derivative gives the slope of the tangent line to f at x = a, and that slope is interpreted as the instantaneous rate of change. The process of finding a derivative is called differentiation.

=

h:0

The slope of the tangent line at x = 2 is f ¿122 = 6122 = 12. For f 1x2 = - x 2 + 5, find f ¿1x2 and f ¿122. We have

f 1x + h2 - f 1x2 h

= =

- 1x + h22 + 5 - 1- x 2 + 52 h - 1x 2 + 2xh + h22 + 5 + x 2 - 5

h - 2xh - h2 = h = - 2x - h, h Z 0. Since f ¿1x2 = lim

f 1x + h2 - f 1x2

h:0

h

= lim 1- 2x - h2, h:0

we have f ¿1x2 = - 2x.

It follows that f ¿122 = - 2 # 2 = - 4.

y 6

f(x) = −

x2

+5

5 4 3

f '(2) = slope at (2, 1) = −4

2 1 −4 −3 −2 −1 −1

1

2

3

4

x

−2

(continued)

188

CHAPTER 1

• Differentiation

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 1.4 (continued ) Continuity: 1. If a function f is differentiable at x = a, then it is continuous at x = a. (Differentiability implies continuity.) 2. Continuity of a function f at x = a does not necessarily mean that f is differentiable at x = a. Any function whose graph has a corner is continuous but not differentiable at the corner. 3. If a function f is discontinuous at x = a, then it is not differentiable at x = a.

1. Let f 1x2 = 3x 2. Since we know the derivative is f ¿1x2 = 6x and the derivative at x = 2 exists, we can conclude that f 1x2 is continuous at x = 2. y f

24 18

f '(2) exists, so f is continuous at x = 2.

12 6 −1

1

2

3

4

x

2. The absolute-value function is continuous at x = 0 but not differentiable at x = 0, since there is a corner at x = 0. y f (x) = | x |

f is continuous at x = 0, but f '(0) does not exist. x

x 2 - 16 is discontinuous at x = - 4; therefore, x + 4 the derivative g¿ 1x2 is not defined at x = - 4. (Note that the derivative is defined at other values of x.)

3. The function g 1x2 =

y −4

x

g is not continuous at x = −4, so g'(−4) is not defined. −8

SECTION 1.5

If y = f 1x2, the derivative in Leibniz dy d notation is written or f 1x2. Each has dx dx the same meaning as f ¿1x2. The Power Rule: For any real number k, d k x = k # x k - 1. dx

The derivative of a constant is d c = 0. dx

Let y = x 3. Then, in Leibniz notation,

dy dx

= 3x 2.

d 7 x = 7x 6 dx d d 1>2 1 1 2x = x = x -1>2 = dx dx 2 22x d 1 1 d -1 a b = x = - 1 # x -2 = - 2 dx x dx x d 34 = 0 dx d 22 = 0 dx

Chapter 1 Summary

KEY TERMS AND CONCEPTS

189

EXAMPLES d d 3x 8 = 3 # x 8 = 3 # 8x 7 = 24x 7 dx dx 2 d 1 2 4 2 d a 2b = # a 2 b = # 1- 2x -32 = - 3 dx 3x 3 dx x 3 3x

The derivative of a constant times a function is d # d 3c f 1x24 = c # f 1x2. dx dx

d 7 1x + 3x2 = 7x 6 + 3 dx d 15x - x 42 = 5 - 4x 3 dx

The Sum–Difference Rule: d d d 3 f 1x2 ; g 1x24 = f 1x2 ; g 1x2 dx dx dx

SECTION 1.6 The Product Rule: d 3 f 1x2 # g 1x24 dx = f 1x2 # g¿1x2 + g 1x2 # f ¿1x2. The Quotient Rule: d f 1x2 d = c dx g 1x2

g 1x2 # f ¿1x2 - f 1x2 # g¿1x2 3g 1x242

.

SECTION 1.7 The Extended Power Rule: d d 3g 1x24k = k3g 1x24k - 1 # g 1x2. dx dx

The Chain Rule: d d 31 f  g21x24 = 3 f 1g 1x224 dx dx = f ¿1g 1x22 # g¿1x2.

d 1 C 12x + 322x D = 12x + 32 x -1>2 + 2x # 2 dx 2 2x + 3 = + 22x 22x 12x + 523 - 13x - 122 d 3x - 1 17 = a b = 2 dx 2x + 5 12x + 52 12x + 522

If f 1x2 = 12x 5 + 4x27, then

f ¿1x2 = 712x 5 + 4x26110x 4 + 42.

The parentheses are required if the derivative of the “inside” function consists of more than one term. Let y = u 3, where u = 2x 4 + 7. Find We have dy dx

dy dx

.

dy dy dy du du # , = 8x 3 and = 3u 2. Since = dx du dx du dx

= 3u 2 # 8x 3 = 24x 312x 4 + 722.

Note that the Extended Power Rule is a special case of the Chain Rule.

SECTION 1.8 The second derivative is the derivative of the first derivative. d 3 f ¿1x24 = 3 f ¿1x24¿ = f –1x2. dx The second derivative describes the rate of change of the derivative.

Let f 1x2 = 5x 7 + 20x. Then d f 1x2 = f ¿1x2 = 35x 6 + 20, dx and therefore, d2 dx 2

f 1x2 = f –1x2 = 210x 5. (continued)

190

CHAPTER 1

• Differentiation

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 1.8 (continued ) Higher-order derivatives include second, third, fourth, and so on, derivatives of a function. The nth derivative of a function is written as d ny dx

n

For y = 5x 7 + 20x, we have the following: d 3y dx 3 d 4y

= f 1n21x2.

dx 4 d 5y dx 5

A real-life application of the second derivative is acceleration. If s1t2 represents distance as a function of time, then velocity is v1t2 = s¿1t2 and acceleration is the change in velocity: a1t2 = v¿1t2 = s–1t2.

= f 1321x2 = 1050x 4, = f 1421x2 = 4200x 3, = f 1521x2 = 12,600x 2, and so on.

Physical Sciences. A particle moves according to the distance function s1t2 = 5t 3, where t is in seconds and s in feet. Therefore, its velocity function is v1t2 = s¿1t2 = 15t 2 and its acceleration function is a1t2 = v¿1t2 = s–1t2 = 30t. At t = 2 sec, the particle is s122 = 40 ft from the starting point, traveling at v122 = 60 ft>sec and accelerating at a122 = 60 ft>sec2 (it’s speeding up).

CHAPTER 1 REVIEW EXERCISES These review exercises are for test preparation. They can also be used as a practice test. Answers are at the back of the book. The blue bracketed section references tell you what part(s) of the chapter to restudy if your answer is incorrect.

CONCEPT REINFORCEMENT Classify each statement as either true or false.

1. If lim f 1x2 exists, then f 152 must exist. [1.1] x:5

2. If lim f 1x2 = L, then L = f 122. [1.1] x:2

3. If f is continuous at x = 3, then lim f 1x2 = f 132. [1.2] x:3

4. A function’s average rate of change over the interval 32, 84 is the same as its instantaneous rate of change at x = 5. [1.3, 1.4]

7. If f is continuous at 5, then f ¿152 must exist. [1.4] 8. The acceleration function is the derivative of the velocity function. [1.8] Match each function in column A with the rule in column B that would be the most appropriate to use for differentiating the function. [1.5, 1.6] Column A

9. f 1x2 = x

7

10. g1x2 = x + 9

11. F1x2 = 15x - 324 12. G1x2 =

2x + 1 3x - 4

5. A function’s derivative at a point, if it exists, can be found as the limit of a difference quotient. [1.4]

13. H1x2 = f 1x2 # g1x2

6. For f ¿152 to exist, f must be continuous at 5. [1.4]

14. f 1x2 = 2x - 7

Column B a) Extended Power Rule b) Product Rule c) Sum Rule d) Difference Rule e) Power Rule f) Quotient Rule

Chapter 1 Review Exercises

29. Is g continuous at - 2? Why or why not? [1.2]

For Exercises 15–17, consider lim f 1x2, where f 1x2 =

x : -7

30. For f 1x2 = x 3 + x 2 - 2x, find the average rate of change as x changes from - 1 to 2. [1.3]

x + 4x - 21 . x + 7 2

15. Limit numerically. [1.1] a) Complete the following input–output tables. f 1x2

x : - 7-

191

31. Find a simplified difference quotient for g 1x2 = - 3x + 2. [1.3] 32. Find a simplified difference quotient for f 1x2 = 2x 2 - 3. [1.3]

f 1x2

x : - 7+

-8

-6

33. Find an equation of the tangent line to the graph of y = x 2 + 3x at the point 1- 1, - 22. [1.4]

- 7.5

- 6.5

- 7.1

- 6.9

34. Find the points on the graph of y = - x 2 + 8x - 11 at which the tangent line is horizontal. [1.5]

- 7.01

- 6.99

35. Find the points on the graph of y = 5x 2 - 49x + 12 at which the tangent line has slope 1. [1.5]

- 7.001

- 6.999

Find dy/dx.

- 7.0001

- 6.9999

b) Find lim - f 1x2, lim + f 1x2, and lim f 1x2, if each x : -7 x : -7 x : -7 exists. 16. Limit graphically. Graph the function, and use the graph to find the limit. [1.1] 17. Limit algebraically. Find the limit algebraically. Show your work. [1.2] Find each limit, if it exists. If a limit does not exist, state that fact. [1.1, 1.2] 8 x : -2 x

19. lim 14x 3 - x 2 + 7x2

x 2 + 2x - 35 x : -7 x + 7

21. lim

18. lim

x:1

20. lim

x: q

23.

y

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

y

1 2 3 4 5

x

y = g(x)

–5 – 4 –3 –2 –1 –1 –2 –3 –4 –5

1 2 3 4 5

x

y = f(x)

For the function graphed in Exercise 22, answer the following. 24. Find lim g 1x2. [1.2] x:1

25. Find g112. [1.2]

26. Is g continuous at 1? Why or why not? [1.2] 27. Find lim g 1x2. [1.2] x : -2

28. Find g1 - 22. [1.2]

-3 x8

39. y = 15x 2>5 [1.5]

[1.5]

40. y = 0.1x 7 - 3x 4 - x 3 + 6 [1.5] Differentiate. 41. f 1x2 = 42. y =

5 6 x + 8x 4 - 2x [1.5] 12

x3 + x [1.5, 1.6] x

43. y =

44. g 1x2 = 15 - x2212x - 125 [1.6]

x2 + 8 [1.6] 8 - x

45. f 1x2 = 1x 5 - 327 [1.7]

47. For y = x 3 48. For y =

5 4 3 2 1

5 4 3 2 1

38. y =

37. y = 82x [1.5]

46. f 1x2 = x 214x + 223>4 [1.7]

1 + 3 [1.1] x

From the graphs in Exercises 22 and 23, determine whether each function is continuous and explain why or why not. [1.2] 22.

3

36. y = 9x 5 [1.5]

d 4y 2 , find 4 . [1.8] x dx

3 7 x - 10x 3 + 13x 2 + 28x - 2, find y–. [1.8] 42

49. For s1t2 = t + t 4, with t in seconds and s1t2 in feet, find each of the following. [1.8] a) v1t2 b) a1t2 c) The velocity and the acceleration when t = 2 sec 50. Business: average revenue, cost, and profit. Given revenue and cost functions R1x2 = 40x and C1x2 = 52x + 100, find each of the following. Assume R1x2 and C1x2 are in dollars and x is the number of items produced. [1.6] a) The average cost, the average revenue, and the average profit when x items are produced and sold b) The rate at which average cost is changing when 9 items are produced

192

• Differentiation

CHAPTER 1

51. Social science: growth rate. The population of a city grows from an initial size of 10,000 to a size P, given by P = 10,000 + 50t 2, where t is in years. [1.5] a) Find the growth rate. b) Find the number of people in the city after 20 yr 1at t = 202. c) Find the growth rate at t = 20.

52. Find 1 f  g21x2 and 1g  f 21x2, given that f 1x2 = x 2 + 5 and g 1x2 = 1 - 2x. [1.7]

TECHNOLOGY CONNECTION Create an input–output table that includes each of the following limits. Start with ≤Tbl  0.1 and then go to 0.01, 0.001, and 0.0001. When you think you know the limit, graph the functions, and use the TRACE feature to verify your assertion. 54. lim

x:1

2 - 2x + 3 [1.1, 1.5] x - 1

55. lim

x : 11

SYNTHESIS 53. Differentiate y =

x21 + 3x 1 + x3

2x - 2 - 3 [1.1, 1.5] x - 11

56. Graph f and f ¿ over the given interval. Then estimate points at which the tangent line to f is horizontal. [1.5] f 1x2 = 3.8x 5 - 18.6x 3; 3- 3, 34

. [1.7]

CHAPTER 1 TEST For Exercises 1–3, consider lim f 1x2, where f 1x2 =

x:6

Graphical limits. Consider the following graph of function f for Exercises 4–11.

x - 36 . x - 6 2

y

1. Numerical limits. a) Complete the following input–output tables.

5 4

x:6

-

f 1x2

x:6

+

f 1x2

3 2

5

7

5.7

6.5

5.9

6.1

5.99

6.01

5.999

6.001

5.9999

6.0001

b) Find lim- f 1x2, lim+ f 1x2, and lim f 1x2, if each exists. x:6

x:6

f

1

x:6

2. Graphical limits. Graph the function, and use the graph to find the limit. 3. Algebraic limits. Find the limit algebraically. Show your work.

–6 –5 –4 –3 – 2 –1

1

2

3

4

5

6

7

x

–1 –2

Find each limit, if it exists. If a limit does not exist, state that fact. 4. lim f 1x2

5. lim f 1x2

6. lim f 1x2

7. lim f 1x2

8. lim f 1x2

9. lim f 1x2

x : -5

x : -3

x : -1

10. lim f 1x2 x:2

x : -4

x : -2

x:1

11. lim f 1x2 x:3

Chapter 1 Test

Determine whether each function is continuous. If a function is not continuous, state why. 12. 13. y y 5 4 3 2 1

5 4 3 2 1 –5 – 4 –3 –2 –1 –1 –2 –3 –4 –5

1 2 3 4 5

y = g(x)

1 2 3 4 5

x

y = f(x)

Consider the function shown in Exercise 13. 14. Find lim f 1x2.

15. Find f 132.

x:3

17. Find lim f 1x2.

16. Is f continuous at 3?

x:4

18. Find f 142.

19. Is f continuous at 4?

Find each limit, if it exists. If a limit does not exist, state why. 20. lim 13x 4 - 2x 2 + 52

21. lim+

x:1

22. lim

x:0

x:2

36. f 1x2 = x2x 2 + 5 37. For y = x 4 - 3x 2, find

d 3y dx 3

.

38. Business: average revenue, cost, and profit. Given revenue and cost functions

–5 – 4 –3 –2 –1 –1 –2 –3 –4 –5

x

193

x - 2 x1x 2 - 42

7 x

R1x2 = 50x and C1x2 = x 2>3 + 750, where x is the number of items produced and R1x2 and C1x2 are in dollars, find the following: a) The average revenue, the average cost, and the average profit when x items are produced b) The rate at which average cost is changing when 8 items are produced 39. Social sciences: memory. In a certain memory experiment, a person is able to memorize M words after t minutes, where M = - 0.001t 3 + 0.1t 2. a) Find the rate of change of the number of words memorized with respect to time. b) How many words are memorized during the first 10 min (at t = 10)? c) At what rate are words being memorized after 10 min (at t = 10)?

40. Find 1f  g21x2 and 1g  f21x2 for f 1x2 = x 2 - x and g 1x2 = 2x 3.

23. Find a simplified difference quotient for f 1x2 = 2x 2 + 3x - 9.

24. Find an equation of the tangent line to the graph of y = x + 14>x2 at the point 14, 52. 25. Find the point(s) on the graph of y = x 3 - 3x 2 at which the tangent line is horizontal.

SYNTHESIS

41. Differentiate y = 211 - 3x22>311 + 3x21>3. x 3 - 27 . x:3 x - 3

42. Find lim

Find dy/dx. 3

26. y = x 23 28. y =

27. y = 42x + 52x

- 10 x

29. y = x 5>4

30. y = - 0.5x 2 + 0.61x + 90

32. y =

43. Graph f and f ¿ over the interval 30, 54. Then estimate points at which the tangent line to f is horizontal. f 1x2 = 5x 3 - 30x 2 + 45x + 52x; 30, 54

44. Find the following limit by creating a table of values:

Differentiate. 31. y =

TECHNOLOGY CONNECTION

lim

1 3 x - x 2 + 2x + 4 3 3x - 4

x:0

x 33. f 1x2 = 5 - x

x3

34. f 1x2 = 1x + 32 17 - x2 4

35. y = 1x 5 - 4x 3 + x2-5

5

25x + 25 - 5 . x

Start with ¢Tbl = 0.1 and then go to 0.01 and 0.001. When you think you know the limit, graph y =

25x + 25 - 5 , x

and use the TRACE feature to verify your assertion.

Extended Technology Application Path of a Baseball: The Tale of the Tape Have you ever watched a baseball game and seen a home run ball hit an obstruction after it has cleared the fence? Suppose the ball hits a sign at a location that is 60 ft above the ground at a distance of 400 ft from home plate. An announcer or a message on the scoreboard might proclaim, “According to the tale of the tape, the ball would have traveled 442 ft.” How is such a calculation made? The answer is related to the curve formed by the path of a baseball. Whatever the path of a well-hit baseball is, it is not the graph of a parabola, f1x2 = ax 2 + bx + c. A well-hit baseball follows the path of a “skewed” parabola, as shown at the lower right. One reason that the ball’s flight is not parabolic is that a well-hit ball has backspin. This fact, combined with the frictional effect of the ball’s stitches with the air, skews the path of the ball in the direction of its landing.

250 200 150

Not the path of a well-hit baseball! Parabola

100 50 50 100 150 200 250 300 350 400 450 500

250 200 150 100 50 –100 –50

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• Differentiation

50 100 150 200 250 300 350 400 450 500

Extended Technology Application

Let’s see if we can model the path of a baseball. Consider the following data. HORIZONTAL DISTANCE, x (IN FEET)

0

VERTICAL DISTANCE, y (IN FEET)

4.5

50

43

100

82

200

130

285

142

300

134

360

100

400

60

Assume for the given data that 10, 4.52 is the point at home plate at which the ball is hit, roughly 4.5 ft off the ground. Also, assume that the ball has hit a billboard 60 ft above the ground and 400 ft from home plate.

EXERCISES 1. Plot the points and connect them with line segments. This can be done on many calculators by pressing STAT PLOT, turning on PLOT, and selecting the appropriate type. 2. a) Use REGRESSION to find a cubic function y = ax 3 + bx 2 + cx + d that fits the data. b) Graph the function over the interval 30, 5004. c) Does the function closely model the given data? d) Predict the horizontal distance from home plate at which the ball would have hit the ground had it not hit the billboard.

e) Find the rate of change of the ball’s height with respect to its horizontal distance from home plate. f) Find the point(s) at which the graph has a horizontal tangent line. Explain the significance of the point(s). 3. a) Use REGRESSION to find a quartic function y = ax 4 + bx 3 + cx 2 + dx + e that fits the data. b) Graph the function over the interval 30, 5004. c) Does the function closely model the given data? d) Predict the horizontal distance from home plate at which the ball would have hit the ground had it not hit the billboard. e) Find the rate of change of the ball’s height with respect to its horizontal distance from home plate. f) Find the point(s) at which the graph has a horizontal tangent line. Explain the significance of the point(s). 4. a) Although most calculators cannot fit such a function to the data, assume that the equation y = 0.0015x 2202,500 - x 2 has been found using some type of curve-fitting technique. Graph the function over the interval 30, 5004. b) Predict the horizontal distance from home plate at which the ball would have hit the ground had it not hit the billboard. c) Find the rate of change of the ball’s height with respect to its horizontal distance from home plate. d) Find the point(s) at which the graph has a horizontal tangent line. Explain the significance of the point(s). 5. Compare the answers in Exercises 2(d), 3(d), and 4(b). Discuss the relative merits of using the quartic model in Exercise 3 with the model in Exercise 4 to make the prediction.

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Tale of the tape. Actually, scoreboard operators in the major leagues use different models to predict the distance that a home run ball would have traveled. The models are linear and are related to the trajectory of the ball, that is, how high the ball is hit. See the following graph. 28° above horizontal 35° above horizontal 42° above horizontal

Home Run Trajectories 250

Height (in feet)

196

200 150

Towering Medium

100

Low

50 100

200

Obstruction 300

400

500

Distance (in feet)

Suppose that a ball hits an obstruction d feet horizontally from home plate at a height of H feet. Then the estimated horizontal distance D that the ball would have traveled, depending on its trajectory type, is Low trajectory: D = 1.1H + d, Medium trajectory: D = 0.7H + d, Towering trajectory: D = 0.5H + d. 6. For a ball striking an obstacle at d = 400 ft and H = 60 ft, estimate how far the ball would have traveled if it were following a low trajectory, or a medium trajectory, or a towering trajectory. 7. In 1953, Hall-of-Famer Mickey Mantle hit a towering home run in old Griffith Stadium in Washington, D.C., that hit an obstruction 60 ft high and 460 ft from home plate. Reporters asserted at the time that the ball would have traveled 565 ft. Is this estimate valid?

Many thanks to Robert K. Adair, professor of physics at Yale University, for many of the ideas presented in this application.

8. Use the appropriate formula to estimate the distance D for each the following famous long home runs. a) Ted Williams (Boston Red Sox, June 9, 1946): Purportedly the longest home run ball ever hit to right field at Boston’s Fenway Park, Williams’s ball landed in the stands 502 feet from home plate, 30 feet above the ground. Assume a medium trajectory. b) Reggie Jackson (Oakland Athletics, July 13, 1971): Jackson’s mighty blast hit an electrical transformer on top of the right-field roof at old Tiger Stadium in the 1971 All-Star Game. The transformer was 380 feet from home plate, 100 feet up. Assume a towering trajectory. Jackson’s home run was reported to still be on the upward arc when it hit the transformer. c) Richie Sexson (Arizona Diamondbacks, April 26, 2004): Sexson hit a drive that caromed off the center-field scoreboard at Bank One Ballpark in Phoenix. The scoreboard is 414 feet from home plate and 75 feet high. Assume a medium trajectory. The reported distances these balls would have traveled are 527 feet for Williams’s home run, 530 feet for Jackson’s, and 469 feet for Sexson’s (Source: www.hittrackeronline.com). How close are your estimates?

Applications of Differentiation

2

Chapter Snapshot What You’ll Learn

Why It’s Important

2.1

In this chapter, we explore many applications of differentiation.We learn to find maximum and minimum values of functions, and that skill allows us to solve many kinds of problems in which we need to find the largest and/or smallest value in a real-world situation.We also apply our differentiation skills to graphing and to approximating function values.

2.5 2.6 2.7

Where It’s Used

REL ATI NG TH E R ADIUS OF A CAN TO ITS SU RFA CE AREA A s)

2.4

MINIMIZING MATERIAL USED Minimizing the amount of material used is a common goal in manufacturing, as it reduces overall costs as well as increases efficiency. For example, cylindrical food cans come in a variety of sizes. Suppose a can is to have a volume of 500 milliliters. Are there optimal dimensions for the can’s height and radius that will minimize the material needed to produce each can? Can you see how minimizing the material used per can translates into minimized costs and conservation of resources? This problem appears as Example 3 in Section 2.5.

uare centimeter

2.3

Surface area (in sq

2.2

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs Graph Sketching: Asymptotes and Rational Functions Using Derivatives to Find Absolute Maximum and Minimum Values Maximum–Minimum Problems; Business and Economics Applications Marginals and Differentials Implicit Differentiation and Related Rates

500 400

A(r) = 2p r 2 + 1000 r

300

(4.3, 348.73)

200 100 0 2

4

6

Radius (in cent

imeters)

8

r

197

•

Applications of Differentiation

2.1 OBJECTIVES • Find relative extrema of a continuous function using the First-Derivative Test. • Sketch graphs of continuous functions.

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs The graph below shows a typical life cycle of a retail product and is similar to graphs we will consider in this chapter. Note that the number of items sold varies with respect to time. Sales begin at a small level and increase to a point of maximum sales, after which they taper off to a low level, where the decline is probably due to the effect of new competitive products. The company then rejuvenates the product by making improvements. Think about versions of certain products: televisions can be traditional, flat-screen, or high-definition; music recordings have been produced as phonograph (vinyl) records, audiotapes, compact discs, and MP3 files. Where might each of these products be in a typical product life cycle? Does the curve seem appropriate for each product? S PRODUCT LIFE CYCLE

t

⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎨ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

CHAPTER 2

Sales of product

198

Introduction Growth

Maturity

Decline

Rejuvenation

Time

Finding the largest and smallest values of a function—that is, the maximum and minimum values—has extensive applications. The first and second derivatives of a function are calculus tools that provide information we can use in graphing functions and finding minimum and maximum values. Throughout this section we will assume, unless otherwise noted, that all functions are continuous. However, continuity of a function does not guarantee that its first and second derivatives are continuous.

Increasing and Decreasing Functions If the graph of a function rises from left to right over an interval I, the function is said to be increasing on, or over, I. y f f(b)

ing

f(a)

s rea

Inc

a

b

f is an increasing function over I: for all a, b in I, if a < b, then f(a) < f(b).

x

2.1

•

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs

TECHNOLOGY CONNECTION

199

If the graph drops from left to right, the function is said to be decreasing on, or over, I.

Exploratory

y

Graph the function y = - 13x 3 + 6x 2 - 11x - 50

De

cre

g(a)

and its derivative

asi

ng

g(b)

y¿ = - x 2 + 12x - 11

using the window 3- 10, 25, - 100, 1504, with Xscl = 5 and Yscl = 25. Then TRACE from left to right along each graph. As you move the cursor from left to right, note that the x-coordinate always increases. If a function is increasing over an interval, the y-coordinate will increase as well. If a function is decreasing over an interval, the y-coordinate will decrease.

g

a

b

x

g is a decreasing function over I: for all a, b in I, if a < b, then g(a) > g(b).

Over what intervals is the function increasing? Over what intervals is the function decreasing? Over what intervals is the derivative positive? Over what intervals is the derivative negative?

We can describe these phenomena mathematically as follows.

DEFINITIONS

What rules can you propose relating the sign of y¿ to the behavior of y?

A function f is increasing over I if, for every a and b in I, if a 6 b, then f1a2 6 f1b2. (If the input a is less than the input b, then the output for a is less than the output for b.) A function f is decreasing over I if, for every a and b in I, if a 6 b, then f1a2 7 f1b2. (If the input a is less than the input b, then the output for a is greater than the output for b.)

The above definitions can be restated in terms of secant lines. If a graph is increasing over an interval I, then, for all a and b in I such that a 6 b, the slope of the secant line between x = a and x = b is positive. Similarly, if a graph is decreasing over an interval I, then, for all a and b in I such that a 6 b, the slope of the secant line between x = a and x = b is negative: Increasing:

f1b2 - f1a2

7 0.

b - a

y

Decreasing:

f1b2 - f1a2 b - a

6 0.

y f

f(b)

f(a)

Slope of secant line is positive.

Slope of secant line is negative.

f(a)

f

f(b)

a

b

x

a

b

x

The following theorem shows how we can use the derivative (the slope of a tangent line) to determine whether a function is increasing or decreasing.

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•

Applications of Differentiation

THEOREM 1 If f ¿1x2 7 0 for all x in an open interval I, then f is increasing over I. If f ¿1x2 6 0 for all x in an open interval I, then f is decreasing over I.

Theorem 1 is illustrated in the following graph. –1,

4 – 3

y

f(x) =

1

–2

–1

Increasing f '(x) > 0

1 – 3

x 3 – x + 23–

x

(1, 0) Decreasing f '(x) < 0

Increasing f '(x) > 0

f is increasing over the intervals (–∞, –1) and (1, ∞); slopes of tangent lines are positive. f is decreasing over the interval (–1, 1); slopes of tangent lines are negative.

For determining increasing or decreasing behavior using a derivative, the interval I is an open interval; that is, it does not include its endpoints. Note how the intervals on which f is increasing and decreasing are written in the preceding graph: x = - 1 and x = 1 are not included in any interval over which the function is increasing or decreasing. These values are examples of critical values.

Critical Values Consider the graph of a continuous function f in Fig. 1. y f

c1

c2

c3 c4 c5

c6

c7

c8

x

FIGURE 1

Note the following: 1. f ¿1c2 = 0 at x = c1, c2, c4, c7, and c8. That is, the tangent line to the graph is horizontal for these values. 2. f ¿1c2 does not exist at x = c3, c5, and c6. The tangent line is vertical at c3, and there are corner points at both c5 and c6. (See also the discussion at the end of Section 1.4.)

2.1

•

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs

201

DEFINITION A critical value of a function f is any number c in the domain of f for which the tangent line at 1c, f1c22 is horizontal or for which the derivative does not exist. That is, c is a critical value if f1c2 exists and f ¿1c2 = 0 or f ¿1c2 does not exist.

Thus, in the graph of f in Fig. 1: 1. c1, c2, c4, c7, and c8 are critical values because f ¿1c2 = 0 for each value. 2. c3, c5, and c6 are critical values because f ¿1c2 does not exist for each value. Also note that a continuous function can change from increasing to decreasing or from decreasing to increasing only at a critical value. In the graph in Fig. 1, c1, c2, c4, c5, c6, and c7 separate the intervals over which the function changes from increasing to decreasing or from decreasing to increasing. Although c3 and c8 are critical values, they do not separate intervals over which the function changes from increasing to decreasing or from decreasing to increasing.

Finding Relative Maximum and Minimum Values Now consider the graph in Fig. 2. Note the “peaks” and “valleys” at the interior points c1, c2, and c3. y f(b)

Absolute maximum

Relative maximum

Maxima

f

f(c2)

f(c3)

Minima

Absolute minimum

Relative minimum

f(a) f(c1) a

c1

c2

c3

b

x

FIGURE 2

Here f1c22 is an example of a relative maximum (plural: maxima). Each of f1c12 and f1c32 is called a relative minimum (plural: minima). The terms local maximum and local minimum are also used.

DEFINITIONS Let I be the domain of f. f1c2 is a relative minimum if there exists within I an open interval I1 containing c such that f1c2 … f1x2, for all x in I1; and f1c2 is a relative maximum if there exists within I an open interval I2 containing c such that f1c2 Ú f1x2, for all x in I2.

A relative maximum can be thought of loosely as the second coordinate of a “peak” that may or may not be the highest point over all of I. Similarly, a relative minimum can

202

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•

Applications of Differentiation

be thought of as the second coordinate of a “valley” that may or may not be the lowest point on I. The second coordinates of the points that are the highest and the lowest on the interval are, respectively, the absolute maximum and the absolute minimum. For now, we focus on finding relative maximum or minimum values, collectively referred to as relative extrema (singular: extremum). Look again at the graph in Fig. 2. The x-values at which a continuous function has relative extrema are those values for which the derivative is 0 or for which the derivative does not exist—the critical values.

THEOREM 2 If a function f has a relative extreme value f1c2 on an open interval, then c is a critical value, so f ¿1c2 = 0

or

f ¿1c2 does not exist.

A relative extreme point, 1c, f1c22, is higher or lower than all other points over some open interval containing c. A relative minimum point will have a y-value that is lower than that of points both to the left and to the right of it, and, similarly, a relative maximum point will have a y-value that is higher than that of points to the left and right of it. Thus, relative extrema cannot be located at the endpoints of a closed interval, since an endpoint lacks “both sides” with which to make the necessary comparisons. However, as we will see in Section 2.4, endpoints can be absolute extrema. Note that the right endpoint of the curve in Fig. 2 is the absolute maximum point. Theorem 2 is very useful, but it is important to understand it precisely. What it says is that to find relative extrema, we need only consider those inputs for which the derivative is 0 or for which it does not exist. We can think of a critical value as a candidate for a value where a relative extremum might occur. y That is, Theorem 2 does not say that every criti10 cal value will yield a relative maximum or mini9 mum. Consider, for example, the graph of 8 f1x2 = 1x - 123 + 2,

7 6 5 4 3 2 1

shown at the right. Note that f ¿1x2 = 31x - 122, and –1

f ¿112 = 311 - 122 = 0.

–1 –2 –3 –4

f(x) = (x – 1)3 + 2

(1, 2)

1

f '(1) = 0 2

3

4

x

The function has c = 1 as a critical value, but has no relative maximum or minimum at that value. Theorem 2 does say that if a relative maximum or minimum occurs, then the first coordinate of that extremum will be a critical value. How can we tell when the existence of a critical value leads us to a relative extremum? The following graph leads us to a test. y

Relative maximum Relative maximum

Relative minimum

Relative minimum x

Decreasing

Increasing

Decreasing

Increasing

Decreasing

•

2.1

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs

203

Note that at a critical value where there is a relative minimum, the function is decreasing to the left of the critical value and increasing to the right. At a critical value where there is a relative maximum, the function is increasing to the left of the critical value and decreasing to the right. In both cases, the derivative changes signs on either side of the critical value. Graph over the interval 1a, b2

–

Sign of f ¿1x2 for x in 1a, c2

Sign of f ¿1x2 for x in 1c, b2

Increasing or decreasing

Relative minimum

-

+

Decreasing on 1a, c2; increasing on 1c, b2

Relative maximum

+

-

Increasing on 1a, c2; decreasing on 1c, b2

No relative maxima or minima

-

-

Decreasing on 1a, b2

No relative maxima or minima

+

+

Increasing on 1a, b2

f1c2

+

a

c

b

+

–

a

c

b

– – a

c

b

+

+ a

c

b

Derivatives tell us when a function is increasing or decreasing. This leads us to the First-Derivative Test.

THEOREM 3 The First-Derivative Test for Relative Extrema For any continuous function f that has exactly one critical value c in an open interval 1a, b2:

F1. f has a relative minimum at c if f ¿1x2 6 0 on 1a, c2 and f ¿1x2 7 0 on 1c, b2. That is, f is decreasing to the left of c and increasing to the right of c. F2. f has a relative maximum at c if f ¿1x2 7 0 on 1a, c2 and f ¿1x2 6 0 on 1c, b2. That is, f is increasing to the left of c and decreasing to the right of c. F3. f has neither a relative maximum nor a relative minimum at c if f ¿1x2 has the same sign on 1a, c2 as on 1c, b2.

Now let’s see how we can use the First-Derivative Test to find relative extrema and create accurate graphs.

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•

Applications of Differentiation

■ EXAMPLE 1 Graph the function f given by f1x2 = 2x 3 - 3x 2 - 12x + 12, and find the relative extrema.

Solution Suppose that we are trying to graph this function but don’t know any calculus. What can we do? We could plot several points to determine in which direction the graph seems to be turning. Let’s pick some x-values and see what happens. y

x

f1x2

-3

- 33

-2

8

-1

19

30 (–1, 19) 20 m = 0? 10

(–2, 8) –5

0

–4

–3

–2

12

1

-1

2

-8

3

3

4

44

(4, 44)

40

–1

– 10

(0, 12) (1, –1)

(3, 3) 2

(2, –8)

3 4 m = 0?

5

6x

– 20 – 30

(–3, –33)

We plot the points and use them to sketch a “best guess” of the graph, shown as the dashed line in the figure above. According to this rough sketch, it appears that the graph has a tangent line with slope 0 somewhere around x = - 1 and x = 2. But how do we know for sure? We use calculus to support our observations. We begin by finding a general expression for the derivative: f ¿1x2 = 6x 2 - 6x - 12. We next determine where f ¿1x2 does not exist or where f ¿1x2 = 0. Since we can evaluate f ¿1x2 = 6x 2 - 6x - 12 for any real number, there is no value for which f ¿1x2 does not exist. So the only possibilities for critical values are those where f ¿1x2 = 0, locations at which there are horizontal tangents. To find such values, we solve f ¿1x2 = 0: 6x 2 - 6x - 12 x2 - x - 2 1x + 121x - 22 x + 1 = 0 x = -1

= 0 = 0 = 0 or x - 2 = 0 or x = 2.

Dividing both sides by 6 Factoring Using the Principle of Zero Products

The critical values are - 1 and 2. Since it is at these values that a relative maximum or minimum might exist, we examine the intervals on each side of the critical values: A is 1- q , - 12, B is 1- 1, 22, and C is 12, q 2, as shown below. A

B –1

C 2

x

Next, we analyze the sign of the derivative on each interval. If f ¿1x2 is positive for one value in the interval, then it will be positive for all values in the interval. Similarly, if it is negative for one value, it will be negative for all values in the interval. Thus, we choose a test value in each interval and make a substitution. The test values we choose are - 2, 0, and 4.

2.1

•

205

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs

A:

Test - 2, f ¿1- 22 = 61- 222 - 61- 22 - 12 = 24 + 12 - 12 = 24 7 0;

B:

Test 0, f ¿102 = 61022 - 6102 - 12 = - 12 6 0;

C:

Test 4, f ¿142 = 61422 - 6142 - 12 = 96 - 24 - 12 = 60 7 0. A

B

Interval

C

-1

x

⎪

⎪

⎪ ⎬

⎪

⎪ ⎭

⎫

⎪ ⎪ ⎪

⎬ ⎪⎪ ⎪

⎭

⎫ ⎪ ⎪

Result

⎬

Sign of f ¿1x2

⎪

⎪

⎪ Test Value

2

x = -2

x = 0

x = 4

f ¿1- 22 7 0

f ¿102 6 0

f ¿142 7 0

f is increasing on 1- q , - 12

f is decreasing on 1- 1, 22

f is increasing on 12, q 2

Change indicates a relative minimum.

Change indicates a relative maximum.

Therefore, by the First-Derivative Test, f has a relative maximum at x = - 1 given by f1- 12 = 21- 123 - 31- 122 - 121- 12 + 12 = 19

Substituting into the original function

This is a relative maximum.

and f has a relative minimum at x = 2 given by f122 = 21223 - 31222 - 12122 + 12 = - 8.

This is a relative minimum.

Thus, there is a relative maximum at 1- 1, 192 and a relative minimum at 12, - 82, as we suspected from the sketch of the graph. The information we have obtained from the first derivative can be very useful in sketching a graph of the function. We know that this polynomial is continuous, and we know where the function is increasing, where it is decreasing, and where it has relative extrema. We complete the graph by using a calculator to generate some additional function values. The graph of the function, shown below in red, has been scaled to clearly show its curving nature. TECHNOLOGY CONNECTION

f'(x) = 6x 2 – 6x – 12

Exploratory

f(x) = 2x 3 – 3x 2 – 12x + 12

Relative maximum 30

Consider the function f given by f 1x2 = x 3 - 3x + 2. Graph both f and f ¿ using the same set of axes. Examine the graphs using the TABLE and TRACE features. Where do you think the relative extrema of f 1x2 occur? Where is the derivative equal to 0? Where does f 1x2 have critical values?

y 40

(–1, 19) 20 10 –4

–3

–2

–1

–10 –20

1

2

3

4

(2, –8) Relative minimum

–30

Increasing f'(x) > 0

Decreasing f'(x) < 0

Increasing f'(x) > 0

x

206

CHAPTER 2

•

Applications of Differentiation

For reference, the graph of the derivative is shown in blue. Note that f ¿1x2 = 0 where f1x2 has relative extrema. We summarize the behavior of this function as follows, by noting where it is increasing or decreasing, and by characterizing its critical points:

Quick Check 1 Graph the function g given by g1x2 = x 3 - 27x - 6, and find the relative extrema.

• • • • •

The function f is increasing over the interval 1- q , - 12. The function f has a relative maximum at the point 1- 1, 192. The function f is decreasing over the interval 1- 1, 22. The function f has a relative minimum at the point 12, - 82. The function f is increasing over the interval 12, q 2.

O Quick Check 1 Interval notation and point notation look alike. Be clear when stating your answers whether you are identifying an interval or a point. To use the first derivative for graphing a function f: 1. Find all critical values by determining where f ¿1x2 is 0 and where f ¿1x2 is undefined (but f1x2 is defined). Find f1x2 for each critical value. 2. Use the critical values to divide the x-axis into intervals and choose a test value in each interval. 3. Find the sign of f ¿1x2 for each test value chosen in step 2, and use this information to determine where f1x2 is increasing or decreasing and to classify any extrema as relative maxima or minima. 4. Plot some additional points and sketch the graph. The derivative f ¿ is used to find the critical values of f. The test values are substituted into the derivative f ¿, and the function values are found using the original function f.

■ EXAMPLE 2 Find the relative extrema of the function f given by f1x2 = 2x 3 - x4. Then sketch the graph.

Solution First, we must determine the critical values. To do so, we find f ¿1x2: f ¿1x2 = 6x 2 - 4x 3. Next, we find where f ¿1x2 does not exist or where f ¿1x2 = 0. Since f ¿1x2 = 6x 2 - 4x 3 is a polynomial, it exists for all real numbers x. Therefore, the only candidates for critical values are where f ¿1x2 = 0, that is, where the tangent line is horizontal: 6x 2 - 4x 3 2x 213 - 2x2 2x 2 = 0 x2 = 0 x = 0

= 0 Setting f ¿1x2 equal to 0 = 0 Factoring or 3 - 2x = 0 or 3 = 2x or x = 32.

The critical values are 0 and 32. We use these values to divide the x-axis into three intervals as shown below: A is 1- q , 02; B is A 0, 32 B ; and C is A 32, q B . A

B 0

Note that f A 32 B = 2 A 32 B 3 -

A 32 B 4 =

27 16

C –3– 2

x

and f102 = 2 # 03 - 04 = 0 are possible extrema.

O

2.1

•

207

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs

We now determine the sign of the derivative on each interval by choosing a test value in each interval and substituting. We generally choose test values for which it is easy to compute f ¿1x2. A:

Test - 1, f ¿1- 12 = 61- 122 - 41- 123 = 6 + 4 = 10 7 0;

B:

Test 1, f ¿112 = 61122 - 41123 = 6 - 4 = 2 7 0;

C:

Test 2, f ¿122 = 61222 - 41223 = 24 - 32 = - 8 6 0. A

B

C

Interval

3 2

0

x

⎪

⎪

⎪ ⎬

⎪

⎪ ⎭

⎫

⎪ ⎪ ⎪

⎬ ⎪⎪ ⎪

⎭

⎫ ⎪ ⎪

⎬

⎪

⎪

⎪

x = -1

x = 1

x = 2

Sign of f ¿1x2

f ¿1- 12 7 0

f ¿112 7 0

f ¿122 6 0

Result

f is increasing on 1- q , 02

f is increasing on A 0, 32 B

f is decreasing on A 32, q B

Test Value

No change

Change indicates a relative maximum.

Therefore, by the First-Derivative Test, f has no extremum at x = 0 (since f1x2 is increasing on both sides of 0) and has a relative maximum at x = 32. Thus, f A 32 B , or 27 16 , is a relative maximum. We use the information obtained to sketch the graph below. Other function values are listed in the table.

x

f1x2, approximately

y

3 27 –– , — 2 16

2

-1

-3

- 0.5

- 0.31

0

0

0.5

0.19

–2

1

1

–3

1.25

1.46

2

0

We summarize the behavior of f:

Relative maximum

(0, 0)

1

–1

1 –1

Increasing f '(x) > 0

2

Not a relative extremum

x f(x) = 2x 3 – x 4

Increasing Decreasing f '(x) > 0 f '(x) < 0

• The function f is increasing over the interval 1- q , 02. • The function f has a critical point at 10, 02, which is neither a minimum nor a maximum.

208

CHAPTER 2

•

Applications of Differentiation

• The function f is increasing over the interval A 0, 32 B .

A 32, 27 16 B . 3 • The function f is decreasing over the interval A 2, q B . Since f is increasing over the intervals 1- q , 02 and A 0, 32 B , we can say that f is increasing over A - q , 32 B despite the fact that f ¿102 = 0 within this interval. In this case, we • The function f has relative maximum at the point

Quick Check 2

O

Find the relative extrema of the function h given by h1x2 = x 4 - 83 x 3. Then sketch the graph.

can observe that any secant line connecting two points within this interval will have a positive slope.

O Quick Check 2 ■ EXAMPLE 3 Find the relative extrema of the function f given by f1x2 = 1x - 222>3 + 1.

TECHNOLOGY CONNECTION

Then sketch the graph.

EXERCISES

Solution First, we determine the critical values. To do so, we find f ¿1x2:

In Exercises 1 and 2, consider the function f given by f 1x2 = 2 - 1x - 122>3.

2 1x - 22-1>3 3 2 = 3 . 32x - 2

f ¿1x2 =

1. Graph the function using the

viewing window 3 - 4, 6, - 2, 44.

2. Graph the first derivative.

What happens to the graph of the derivative at the critical values?

Next, we find where f ¿1x2 does not exist or where f ¿1x2 = 0. Note that f ¿1x2 does not exist at 2, although f1x2 does. Thus, 2 is a critical value. Since the only way for a fraction to be 0 is if its numerator is 0, we see that f ¿1x2 = 0 has no solution. Thus, 2 is the only critical value. We use 2 to divide the x-axis into the intervals A, which is 1- q , 22, and B, which is 12, q 2. Note that f122 = 12 - 222>3 + 1 = 1. A

B x

2

To determine the sign of the derivative, we choose a test value in each interval and substitute each value into the derivative. We choose test values 0 and 3. It is not necessary to find an exact value of the derivative; we need only determine the sign. Sometimes we can do this by just examining the formula for the derivative: A:

2

Test 0, f ¿102 =

B: Test 3, f ¿132 =

3

320 - 2 2 3

323 - 2

6 0; 7 0.

Interval

x = 0

x = 3

f ¿102 6 0

f ¿132 7 0

f is decreasing on 1- q , 22

f is increasing on 12, q 2

Change indicates a relative minimum.

x

⎪ ⎪ ⎪

⎬

⎪ ⎪

⎭

⎫ ⎪

Result

⎪

Sign of f ¿1x2

⎬ ⎪

⎪

⎪ Test Value

2

•

2.1

209

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs

Since we have a change from decreasing to increasing, we conclude from the FirstDerivative Test that a relative minimum occurs at 12, f1222, or 12, 12. The graph has no tangent line at 12, 12 since f ¿122 does not exist. We use the information obtained to sketch the graph. Other function values are listed in the table. y

-1

3.08

- 0.5

2.84

0

2.59

0.5

2.31

1

2

1.5

1.63

2

1

2.5

1.63

3

2

3.5

2.31

4

2.59

3

f(x) = (x – 2)2/3 + 1

2

Relative minimum

1

–1

(2, 1) 1

2

3

4

x

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪

x

f1x2, approximately

Decreasing f '(x) < 0

Increasing f '(x) > 0

We summarize the behavior of f: Quick Check 3

O

Find the relative extrema of the function g given by g1x2 = 3 - x 1>3. Then sketch the graph.

• The function f is decreasing over the interval 1- q , 22. • The function f has a relative minimum at the point 12, 12. • The function f is increasing over the interval 12, q 2.

O Quick Check 3

TECHNOLOGY CONNECTION

Finding Relative Extrema

Method 1: TRACE

To explore some methods for approximating relative extrema, let’s find the relative extrema of

Beginning with the window shown at left, we press TRACE and move the cursor along the curve, noting where relative extrema might occur.

f 1x2 = - 0.4x 3 + 6.2x 2 - 11.3x - 54.8. We first graph the function, using a window that reveals the curvature.

Y1= –.4X^3 + 6.2X^2 – 11.3X – 54.8

y = – 0.4x 3 + 6.2x 2 – 11.3x – 54.8 150

X = 9.4680851

–10

20

Y = 54.502163

A relative maximum seems to be about y = 54.5 at x = 9.47. We can refine the approximation by zooming in to obtain the following window. We press TRACE and move

–100

(continued)

210

CHAPTER 2

•

Applications of Differentiation

Finding Relative Extrema (continued)

Method 4: fMax or fMin

the cursor along the curve, again noting where the y-value is largest. The approximation is about y = 54.61 at x = 9.31.

This feature calculates a relative maximum or minimum value over any specified closed interval. We see from the initial graph that a relative maximum occurs in the interval 3- 10, 204. Using the fMax option from the MATH menu, we see that a relative maximum occurs on 3- 10, 204 when x L 9.32.

Y1 = –.4X^3 + 6.2X^2 – 11.3X – 54.8

Y1 = –.4X^3 + 6.2X^2 – 11.3X – 54.8

fMax (Y1, X, –10, 20) X = 9.4680851

Y = 54.502163

X = 9.3085106

9.323324165

Y = 54.606815

We can continue in this manner until the desired accuracy is achieved.

Method 2: TABLE We can also use the TABLE feature, adjusting starting points and step values to improve accuracy: TblStart = 9.3

To obtain the maximum value, we evaluate the function at the given x-value, obtaining the following. fMax (Y1, X, –10, 20)

¢ Tbl = .01

9.323324165 Y1(Ans)

X 9.3 9.31 9.32 9.33 9.34 9.35 9.36 X = 9.32

54.60790781

Y1 54.605 54.607 54.608 54.608 54.607 54.604 54.601

The approximation seems to be nearly y = 54.61 at an x-value between 9.32 and 9.33. We could next set up a new table showing function values between f 19.322 and f 19.332 to refine the approximation.

The approximation is about y = 54.61 at x = 9.32. Using any of these methods, we find the relative minimum to be about y = - 60.30 at x = 1.01.

Method 3: MAXIMUM, MINIMUM Using the MAXIMUM option from the CALC menu, we find that a relative maximum of about 54.61 occurs at x L 9.32.

Minimum X = 1.0100107

Y = – 60.3005

EXERCISE 1. Using one of the methods just described, approximate

the relative extrema of the function in Example 1.

Maximum X = 9.3233222

Y = 54.607908

TECHNOLOGY CONNECTION

Finding Relative Extrema with iPlot We can use iPlot to graph a function and its derivative and then find relative extrema. iPlot has the capability of graphing a function and its derivative on the same set of axes, though it does not give a formula for the derivative but merely draws the graph. As an example, let’s consider the function given by f 1x2 = x 3 - 3x + 4.

To graph a function and its derivative, first open the iPlot app on your iPhone or iPad. You will get a screen like the one in Fig. 1. Notice the four icons at the bottom. The Functions icon is highlighted. Press + in the upper right; then enter f 1x2 = x 3 - 3x + 4 using the notation x^3–3*x+4. Press Done at the upper right and then Plot at the lower right (Fig. 2). (continued)

2.1

•

Using First Derivatives to Find Maximum and Minimum Values and Sketch Graphs

FIGURE 2

FIGURE 1

FIGURE 4

FIGURE 3

FIGURE 5

The graph of f 1x2 = x 3 - 3x + 4 is shown in red in Fig. 3. To graph the derivative of f, first click on the Functions icon again, and then press + . You will get the screen shown in Fig. 4. Next, slide the Derivate button to the left. (“Derivate” means “Differentiate.”) Then enter the same function as before, x^3–3*x+4, and press Done. D(x^3–3*x+4) will appear in the second line. Press Plot, and you will see both functions plotted, as shown in Fig. 5. Look over the two graphs, and use Trace to find various function values. Press Prev to jump between the function and its derivative. Look for x-values where the derivative is 0. What happens at these

FIGURE 6

211

values of the original function? Examining the graphs in this way reveals that the graph of f 1x2 = x 3 - 3x + 4 has a relative maximum point at 1- 1, 62 and a relative minimum point at 11, 22. iPlot has an additional feature that allows us to be more certain about these relative extrema. Go back to the original plot of f 1x2 = x 3 - 3x + 4 (Fig. 3), and press Settings. Change the window to 3- 3, 3, 12, - 104 to better see the graph. Press the MinMax button at the bottom. Touch the screen as closely as possible to what might be a relative extremum. See Figs. 6 and 7 for the relative maximum. The relative minimum can be found similarly.

FIGURE 7 (continued)

212

CHAPTER 2

•

Applications of Differentiation

4. f 1x2 = - 4.32 + 1.44x + 3x 2 - x 3

EXERCISES For each function, use iPlot to create the graph and find the derivative. Then explore each graph to look for possible relative extrema. Use MinMax to determine the relative extrema.

5. g1x2 = x24 - x 2

2. f 1x2 = x1200 - x2

7. f 1x2 =

1. f 1x2 = 2x 3 - x4

3. f 1x2 = x 3 - 6x 2

6. g1x2 =

4x x + 1 2

x 2 - 3x x - 1

8. f 1x2 = ƒ x + 2 ƒ - 3

Section Summary • A function f is increasing over an interval I if, for all a and b in I such that a 6 b, f1a2 6 f1b2. Equivalently, the slope of the secant line connecting a and b is positive: f1b2 - f1a2 7 0. b - a • A function f is decreasing over an interval I if, for all a and b in I such that a 6 b, f1a2 7 f1b2. Equivalently, the slope of the secant line connecting a and b is negative: f1b2 - f1a2 6 0. b - a • Using the first derivative, a function is increasing over an open interval I if, for all x in I, the slope of the tangent line at x is positive; that is, f ¿1x2 7 0. Similarly, a function is decreasing over an open interval I if, for all x in I, the slope of the tangent line is negative; that is, f ¿1x2 6 0.

• A critical value is a number c in the domain of f such that f ¿1c2 = 0 or f ¿1c2 does not exist. The point 1c, f1c22 is called a critical point. • A relative maximum point is higher than all other points in some interval containing it. Similarly, a relative minimum point is lower than all other points in some interval containing it. The y-value of such a point is called a relative maximum (or minimum) value of the function. • Minimum and maximum points are collectively called extrema. • Critical values are candidates for possible relative extrema. The First-Derivative Test is used to classify a critical value as a relative minimum, a relative maximum, or neither.

EXERCISE SET

2.1 Find the relative extrema of each function, if they exist. List each extremum along with the x-value at which it occurs. Then sketch a graph of the function. 1. f 1x2 = x 2 + 4x + 5 3. f 1x2 = 5 - x - x 2

2. f 1x2 = x 2 + 6x - 3

4. f 1x2 = 2 - 3x - 2x 2

16. f 1x2 = 12 + 9x - 3x 2 - x 3 17. g1x2 = x 3 - x 4

18. f 1x2 = x 4 - 2x 3

19. f 1x2 = 13 x 3 - 2x 2 + 4x - 1 20. F1x2 = - 13 x3 + 3x 2 - 9x + 2

5. g1x2 = 1 + 6x + 3x 2

21. g1x2 = 2x4 - 20x 2 + 18

6. F1x2 = 0.5x 2 + 2x - 11

22. f 1x2 = 3x 4 - 15x 2 + 12

7. G1x2 = x - x - x + 2

23. F1x2 = 2x - 1

24. G1x2 = 2x + 2

25. f 1x2 = 1 - x 2>3

26. f 1x2 = 1x + 322>3 - 5

3

2

8. g1x2 = x 3 + 12 x 2 - 2x + 5 9. f 1x2 = x - 3x + 6 3

3

10. f 1x2 = x - 3x 3

11. f 1x2 = 3x 2 + 2x 3

12. f 1x2 = x 3 + 3x

13. g1x2 = 2x 3 - 16

14. F1x2 = 1 - x 3

15. G1x2 = x 3 - 6x 2 + 10

2

27. G1x2 = 29. g1x2 =

-8 x + 1 2

4x x2 + 1

3

28. F1x2 = 30. g1x2 =

5 x + 1 2

x2 x2 + 1

Exercise Set 2.1

31. f 1x2 = 2x

32. f 1x2 = 1x + 121>3

3

33. g1x2 = 2x + 2x + 5 2

34. F1x2 =

1

213

85. Consider this graph. y

2x 2 + 1

35–68. Check the results of Exercises 1–34 using a calculator. For Exercises 69–84, draw a graph to match the description given. Answers will vary. 69. f 1x2 is increasing over 1- q , 22 and decreasing over 12, q 2.

70. g 1x2 is decreasing over 1- q , - 32 and increasing over 1- 3, q 2.

71. G1x2 is decreasing over 1- q , 42 and 19, q 2 and increasing over 14, 92.

x1 x 2 x 3 x 4 x 5

x 6 x 7 x 8 x 9 x10 x

Explain the idea of a critical value. Then determine which x-values are critical values, and state why. 86. Consider this graph. y

72. F1x2 is increasing over 1- q , 52 and 112, q 2 and decreasing over 15, 122.

73. g 1x2 has a positive derivative over 1- q , - 32 and a negative derivative over 1- 3, q 2.

74. f 1x2 has a negative derivative over 1- q , 12 and a positive derivative over 11, q 2. 75. F1x2 has a negative derivative over 1- q , 22 and 15, 92 and a positive derivative over 12, 52 and 19, q 2.

76. G1x2 has a positive derivative over 1- q , - 22 and 14, 72 and a negative derivative over 1- 2, 42 and 17, q 2.

77. f 1x2 has a positive derivative over 1- q , 32 and 13, 92, a negative derivative over 19, q 2, and a derivative equal to 0 at x = 3.

78. g 1x2 has a negative derivative over 1- q , 52 and 15, 82, a positive derivative over 18, q 2, and a derivative equal to 0 at x = 5.

79. F1x2 has a negative derivative over 1- q , - 12 and a positive derivative over 1- 1, q 2, and F¿1- 12 does not exist.

80. G1x2 has a positive derivative over 1- q , 02 and 13, q 2 and a negative derivative over 10, 32, but neither G¿102 nor G¿132 exists. 81. f 1x2 has a negative derivative over 1- q , - 22 and 11, q 2 and a positive derivative over 1- 2, 12, and f ¿1- 22 = 0, but f ¿112 does not exist.

82. g 1x2 has a positive derivative over 1- q , - 32 and 10, 32, a negative derivative over 1- 3, 02 and 13, q 2, and a derivative equal to 0 at x = - 3 and x = 3, but g¿102 does not exist.

83. H1x2 is increasing over 1- q , q 2, but the derivative does not exist at x = 1.

84. K1x2 is decreasing over 1- q , q 2, but the derivative does not exist at x = 0 and x = 2.

a

b

c

d

e

x

Using the graph and the intervals noted, explain how to relate the concept of the function being increasing or decreasing to the first derivative.

APPLICATIONS Business and Economics 87. Employment. According to the U.S. Bureau of Labor Statistics, the number of professional services employees fluctuated during the period 2000–2009, as modeled by E1t2 = - 28.31t 3 + 381.86t 2 - 1162.07t + 16,905.87, where t is the number of years since 2000 (t = 0 corresponds to 2000) and E is thousands of employees. (Source: www.data.bls.gov.) Find the relative extrema of this function, and sketch the graph. Interpret the meaning of the relative extrema. 88. Advertising. Brody Electronics estimates that it will sell N units of a new toy after spending a thousands of dollars on advertising, where N1a2 = - a 2 + 300a + 6, 0 … a … 300. Find the relative extrema and sketch a graph of the function.

Life and Physical Sciences 89. Temperature during an illness. The temperature of a person during an intestinal illness is given by T1t2 = - 0.1t 2 + 1.2t + 98.6, 0 … t … 12,

where T is the temperature 1°F2 at time t, in days. Find the relative extrema and sketch a graph of the function.

214

CHAPTER 2

•

Applications of Differentiation

90. Solar eclipse. On January 15, 2010, the longest annular solar eclipse until 3040 occurred over Africa and the Indian Ocean (in an annular eclipse, the sun is partially obscured by the moon and looks like a ring). The path of the full eclipse on the earth’s surface is modeled by

f 1x2 = 0.0125x 2 - 1.157x + 22.864, 15 6 x 6 90, where x is the number of degrees of longitude east of the prime meridian and f 1x2 is the number of degrees of latitude north (positive) or south (negative) of the equator. (Source: NASA.) Find the longitude and latitude of the southernmost point at which the full eclipse could be viewed.

93.

f′

y 5 4 3 2 1

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

95.

94.

y

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x

96.

y 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x

1 2 3 4 5

x

y 5 4 3 2 1

f′

f′

f′ 1 2 3 4 5

x

TECHNOLOGY CONNECTION Graph each function. Then estimate any relative extrema. 97. f 1x2 = - x 6 - 4x 5 + 54x 4 + 160x 3 - 641x 2 - 828x + 1200 98. f 1x2 = x 4 + 4x 3 - 36x 2 - 160x + 400 99. f 1x2 = 2 ƒ 4 - x 2 ƒ + 1 3

SYNTHESIS In Exercises 91–96, the graph of a derivative f ¿ is shown. Use the information in each graph to determine where f is increasing or decreasing and the x-values of any extrema. Then sketch a possible graph of f. 91.

92.

y 5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

f′

f′ 1 2 3 4 5

x

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Use your calculator’s absolute-value feature to graph the following functions and determine relative extrema and intervals over which the function is increasing or decreasing. State the x-values at which the derivative does not exist. 101. f 1x2 = ƒ x - 2 ƒ

102. f 1x2 = ƒ 2x - 5 ƒ

y

103. f 1x2 = ƒ x 2 - 1 ƒ

104. f 1x2 = ƒ x 2 - 3x + 2 ƒ 1 2 3 4 5

x

100. f 1x2 = x29 - x 2

105. f 1x2 = ƒ 9 - x 2 ƒ

106. f 1x2 = ƒ - x 2 + 4x - 4 ƒ 107. f 1x2 = ƒ x 3 - 1 ƒ

108. f 1x2 = ƒ x 4 - 2x 2 ƒ

Exercise Set 2.1

Life science: caloric intake and life expectancy. The data in the following table give, for various countries, daily caloric intake, projected life expectancy, and infant mortality. Use the data for Exercises 109 and 110.

Country

Daily Caloric Intake

Life Expectancy at Birth (in years)

Infant Mortality (number of deaths before age 1 per 1000 births)

Argentina

3004

77

13

Australia

3057

82

5

Bolivia

2175

67

46

Canada

3557

81

5

Dominican Republic

2298

74

30

Germany

3491

79

4

Haiti

1835

61

62

Mexico

3265

76

17

United States

3826

78

6

Venezuela

2453

74

17

(Source: U.N. FAO Statistical Yearbook, 2009.)

109. Life expectancy and daily caloric intake. a) Use the regression procedures of Section R.6 to fit a cubic function y = f 1x2 to the data in the table, where x is daily caloric intake and y is life expectancy. Then fit a quartic function and decide which fits best. Explain. b) What is the domain of the function? c) Does the function have any relative extrema? Explain. 110. Infant mortality and daily caloric intake. a) Use the regression procedures of Section R.6 to fit a cubic function y = f 1x2 to the data in the table, where x is daily caloric intake and y is infant mortality. Then fit a quartic function and decide which fits best. Explain. b) What is the domain of the function? c) Does the function have any relative extrema? Explain. 111. Describe a procedure that can be used to select an appropriate viewing window for the functions given in (a) Exercises 1–16 and (b) Exercises 97–100.

Answers to Quick Checks

1. Relative maximum at 1- 3, 482, relative minimum at 13, - 602

y 100 50 −4

−2 − 50

2

− 100

2. Relative minimum at A 2, - 16 3 B

4

x

g(x)  x 3  27x  6

y 4 2 −1

1

2

−2 −4 8 3

h(x)  x 4  x 3

3. There are no extrema.

y 6 4 2 −4

−2

g (x)  3  x 1/3

2 −2

4

x

3 x

215

216

CHAPTER 2

•

Applications of Differentiation

2.2 OBJECTIVES • Classify the relative extrema of a function using the Second-Derivative Test. • Sketch the graph of a continuous function.

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs The “turning” behavior of a graph is called its concavity. The second derivative plays a pivotal role in analyzing the concavity of a function’s graph.

Concavity: Increasing and Decreasing Derivatives The graphs of two functions are shown below. The graph of f is turning up and the graph of g is turning down. Let’s see if we can relate these observations to the functions’ derivatives. Consider first the graph of f. Take a ruler, or straightedge, and draw tangent lines as you move along the curve from left to right. What happens to the slopes of the tangent lines? Do the same for the graph of g. Look at the curvature and decide whether you see a pattern. y

y g f

a

b

x

a

b

f '(x) increasing

g'(x) decreasing

f ''(x) > 0

g''(x) < 0

x

For the graph of f, the slopes of the tangent lines are increasing. That is, f ¿ is increasing over the interval. This can be determined by noting that f –1x2 is positive, since the relationship between f ¿ and f – is like the relationship between f and f ¿. Note also that all the tangent lines for f are below the graph. For the graph of g, the slopes are decreasing. This can be determined by noting that g¿ is decreasing whenever g–1x2 is negative. For g, all tangent lines are above the graph.

DEFINITION Suppose that f is a function whose derivative f ¿ exists at every point in an open interval I. Then f is concave up on I if f ′ is increasing over I.

f is concave down on I if f ′ is decreasing over I.

The following theorem states how the concavity of a function’s graph and the second derivative of the function are related.

THEOREM 4

A Test for Concavity 1. If f –1x2 7 0 on an interval I, then the graph of f is concave up. (f ¿ is increasing, so f is turning up on I.) 2. If f –1x2 6 0 on an interval I, then the graph of f is concave down. (f ¿ is decreasing, so f is turning down on I.)

2.2

•

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

217

Keep in mind that a function can be decreasing and concave up, decreasing and concave down, increasing and concave up, or increasing and concave down. That is, concavity and increasing/decreasing are independent concepts. It is the increasing or decreasing aspect of the derivative that tells us about the function’s concavity. f ¿1x2 6 0

f ¿1x2 7 0

Concave down, Decreasing

Concave down, Increasing

Concave up, Decreasing

Concave up, Increasing

f –1x2 6 0

f –1x2 7 0

TECHNOLOGY CONNECTION

Exploratory Now graph the first derivative

Graph the function f 1x2 = - 13 x 3 + 6x 2 - 11x - 50

f ¿1x2 = - x 2 + 12x - 11 and the second derivative

and its second derivative, f –1x2 = - 2x + 12,

using the viewing window 3- 10, 25, - 100, 1504, with Xscl = 5 and Yscl = 25. Over what intervals is the graph of f concave up? Over what intervals is the graph of f concave down? Over what intervals is the graph of f– positive? Over what intervals is the graph of f– negative? What can you conjecture?

f –1x2 = - 2x + 12

using the viewing window 3- 10, 25, - 200, 504, with Xscl = 5 and Yscl = 25. Over what intervals is the first derivative f ¿ increasing? Over what intervals is the first derivative f ¿ decreasing? Over what intervals is the graph of f– positive? Over what intervals is the graph of f– negative? What can you conjecture?

Classifying Relative Extrema Using Second Derivatives Let’s see how we can use second derivatives to determine whether a function has a relative extremum on an open interval. The following graphs show both types of concavity at a critical value (where f ¿1c2 = 0). When the second derivative is positive (graph is concave up) at the critical value, the critical point is a relative minimum point, and when the second derivative is negative (graph is concave down), the critical point is a relative maximum point.

218

CHAPTER 2

•

Applications of Differentiation y

y

⎫ ⎪ ⎬ f(c) ⎪ ⎭ a

c

b

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ f(c) ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

x

a

f '(c) = 0, f ''(c) > 0, The graph is concave up around c. Therefore, f(c) is a relative minimum.

c

x

b

f '(c) = 0, f ''(c) < 0, The graph is concave down around c. Therefore, f(c) is a relative maximum.

This analysis is summarized in Theorem 5:

THEOREM 5

The Second-Derivative Test for Relative Extrema Suppose that f is differentiable for every x in an open interval 1a, b2 and that there is a critical value c in 1a, b2 for which f ¿1c2 = 0. Then: 1. f1c2 is a relative minimum if f –1c2 7 0. 2. f1c2 is a relative maximum if f –1c2 6 0. For f –1c2 = 0, the First-Derivative Test can be used to determine whether f1x2 is a relative extremum.

Consider the following graphs. In each one, f ¿ and f – are both 0 at c = 2, but the first function has an extremum and the second function does not. When c is a critical value and f –1c2 = 0, an extremum may or may not exist at c. Note too that if f ¿1c2 does not exist and c is a critical value, then f –1c2 also does not exist. Again, an approach other than the Second-Derivative Test must be used to determine whether f1c2 is an extremum.

–1

y

y

10 9 8 7 6 5 4 3 2 1

10 9 8 7 6 5 4 3 2 1

–1

f(x) = (x – 2)4 + 1 f(x) = (x – 2)4 + 1 f '(x) = 4(x – 2) 3 f ''(x) = 12(x – 2) 2 f '(2) = 0 f ''(2) = 0 Relative minimum at c = 2

1

c=2

3

4

x

–1

–1

f(x) = (x – 2)3 + 1

1

c=2

3

f(x) = (x – 2) 3 + 1 f '(x) = 3(x – 2) 2 f ''(x) = 6(x – 2) f '(2) = 0 f ''(2) = 0 No relative extremum

4

x

The second derivative is used to help identify extrema and determine the overall behavior of a graph, as we see in the following examples.

•

2.2

219

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

■ EXAMPLE 1 Find the relative extrema of the function f given by f1x2 = x 3 + 3x 2 - 9x - 13, and sketch the graph.

Solution To find any critical values, we determine f ¿1x2. To determine whether any critical values lead to extrema, we also find f –1x2: f ¿1x2 = 3x 2 + 6x - 9, f –1x2 = 6x + 6. Then we solve f ¿1x2 = 0: 3x 2 + 6x - 9 x 2 + 2x - 3 1x + 321x - 12 x + 3 = 0 x = -3

TECHNOLOGY CONNECTION

Exploratory

= 0 = 0 Dividing both sides by 3 = 0 Factoring or x - 1 = 0 Using the Principle of Zero Products or x = 1.

We next find second coordinates by substituting the critical values in the original function:

Consider the function given by

f1- 32 = 1- 323 + 31- 322 - 91- 32 - 13 = 14; f112 = 1123 + 31122 - 9112 - 13 = - 18.

y = x 3 - 3x 2 - 9x - 1. Use a graph to estimate the relative extrema. Then find the first and second derivatives. Graph both in the same window. Use the ZERO feature to determine where the first derivative is zero. Verify that relative extrema occur at those x-values by checking the sign of the second derivative. Then check your work using the approach of Example 1.

Are the points 1- 3, 142 and 11, - 182 relative extrema? Let’s look at the second derivative. We use the Second-Derivative Test with the critical values - 3 and 1: f –1- 32 = 61- 32 + 6 = - 12 6 0; f –112 = 6112 + 6 = 12 7 0.

Relative maximum Relative minimum

Thus, f1- 32 = 14 is a relative maximum and f112 = - 18 is a relative minimum. We plot both 1- 3, 142 and 11, - 182, including short arcs at each point to indicate the graph’s concavity. Then, by calculating and plotting a few more points, we can make a sketch, as shown below. y

y

40

40

Relative maximum

30 20

(−3, 14)

−4

−3

−2

−1

20

2

3

4

x

−5

−4

−3

−2

1

2

3

(2, −11)

(−1, −2)

−10

(3, 14)

(−2, 9) 10

(−4, 7) 1

−20

30

(−3, 14)

10 −5

f

Relative maximum

(0, −13) (−5, −18)

(1, −18)

−20

(1, −18)

Relative minimum

■ EXAMPLE 2 Find the relative extrema of the function f given by f1x2 = 3x 5 - 20x 3, and sketch the graph.

Relative minimum

4

x

220

CHAPTER 2

•

Applications of Differentiation

Solution We find both the first and second derivatives: f ¿1x2 = 15x 4 - 60x 2, f –1x2 = 60x 3 - 120x. Then we solve f ¿1x2 = 0 to find any critical values: 15x 4 15x 21x 2 15x 21x + 221x 15x 2 = 0 or

60x 2 = 0 - 42 = 0 - 22 = 0 x + 2 = 0

x = 0 or

Factoring

or x - 2 = 0

x = - 2 or

Using the Principle of Zero Products

x = 2.

We next find second coordinates by substituting in the original function: f1- 22 = 31- 225 - 201- 223 = 64; f102 = 31025 - 201023 = 0. f122 = 31225 - 201223 = - 64; All three of these y-values are candidates for relative extrema. We now use the Second-Derivative Test with the numbers - 2, 2, and 0: f –1- 22 = 601- 223 - 1201- 22 = - 240 6 0; f –102 = 601023 - 120102 = 0.

Relative maximum The Second-Derivative Test fails. Use the First-Derivative Test. Relative minimum

f –122 = 601223 - 120122 = 240 7 0;

Quick Check 1 Find the relative extrema of the function g given by g1x2 = 10x 3 - 6x 5, and sketch the graph.

(−3, 14)

20 10

−4 −3 −2

−10 −20

Concave down

(−1, −2) 2

3

(1, −18) Concave up

f(x) = 3x5 − 20x3

80

(−2, 64)

60 40 20 −3

−2

−1

(0, 0)

−20

1

2

3

x

−40 −60 −80

Increasing

Decreasing

(2, −64) Increasing

O Quick Check 1

Look again at the graphs in Examples 1 and 2. The concavity changes from down to up at the point 1- 1, - 22 in Example 1, and the concavity changes from up to down at the point (0, 0) in Example 2. (In fact, the graph in Example 2 has other points where the concavity changes direction. This is addressed in Example 3.)

f

1

y

Points of Inflection

f(x) = x 3 + 3x 2 − 9x − 13 y 30

Thus, f1- 22 = 64 is a relative maximum, and f122 = - 64 is a relative minimum. Since f ¿1- 12 6 0 and f ¿112 6 0, we know that f is decreasing on both 1- 2, 02 and 10, 22. Thus, we know by the FirstDerivative Test that f has no relative extremum at 10, 02. We complete the graph, plotting other points as needed. The extrema are shown in the graph at right.

x

O

2.2

•

221

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

A point of inflection, or an inflection point, is a point across which the direction of concavity changes. For example, in Figs. 1–3, point P is an inflection point. The figures display the sign of f –1x2 to indicate the concavity on either side of P. y

y

–

P

y

–

x0 '' f (x 0) = 0

P

P

+

+ x

x x0 '' f (x 0) does not exist.

FIGURE 2

FIGURE 1

–

+ x0 '' f (x 0) = 0

x

FIGURE 3

As we move to the right along each curve, the concavity changes at P. Since, as we move through P, the sign of f –1x2 changes, either the value of f –1x 02 at P must be 0, as in Figs. 1 and 3, or f –1x 02 must not exist, as in Fig. 2.

THEOREM 6

Finding Points of Inflection If a function f has a point of inflection, it must occur at a point x 0, where f –1x 02 = 0 or f –1x 02 does not exist.

The converse of Theorem 6 is not necessarily true. That is, if f –1x 02 is 0 or does not exist, then there is not necessarily a point of inflection at x 0. There must be a change in the direction of concavity on either side of x 0 for 1x 0, f1x 022 to be a point of inflection. For example, for f1x2 = 1x - 224 + 1 (see the graph on p. 218), we have f –122 = 0, but 12, f1222 is not a point of inflection since the graph of f is concave up to the left and to the right of x = 2. To find candidates for points of inflection, we look for numbers x 0 for which f –1x 02 = 0 or for which f –1x 02 does not exist. Then, if f –1x2 changes sign as x moves through x 0 (see Figs. 1–3), we have a point of inflection at x 0. Theorem 6, about points of inflection, is completely analogous to Theorem 2 about relative extrema. Theorem 2 tells us that relative extrema occur when f ¿1x2 = 0 or f ¿1x2 does not exist. Theorem 6 tells us that points of inflection occur when f –1x2 = 0 or f –1x2 does not exist.

■ EXAMPLE 3 Use the second derivative to determine the point(s) of inflection for the function in Example 2.

Solution The function is f1x2 = 3x 5 - 20x 3, and its second derivative is

f –1x2 = 60x 3 - 120x. We set the second derivative equal to 0 and solve for x: 60x 3 - 120x = 0 60x1x 2 - 22 = 0

60x A x + 22 B A x - 22 B = 0

Factoring out 60x Factoring x 2 - 2 as a difference of squares

x = 0 or x = - 22 or x = 22.

Using the Principle of Zero Products

Next, we check the sign of f –1x2 over the intervals bounded by these three x-values. We are looking for a change in sign from one interval to the next:

222

CHAPTER 2

•

Applications of Differentiation

Interval

- 22

x

⎪

⎪ ⎬

⎪

⎪ ⎭

⎫

⎪ ⎪

⎬ ⎪ ⎪

⎭

⎫

⎪ ⎪

⎬ ⎪ ⎪

⎭

⎫ ⎪ ⎪

Result

⎬

Sign of f –1x2

⎪

⎪

⎪ Test Value

22

0

x = -2

x = -1

x = 1

x = 2

f –1- 22 6 0

f –1- 12 7 0

f –112 6 0

f –122 7 0

Concave down

Concave up

Concave down

Concave up

The graph changes from concave down to concave up at x = - 22, from concave up to concave down at x = 0, and from concave down to concave up at x = 22. Therefore, A - 22, f A - 22 B B , 10, f1022, and A 22, f A 22 B B are points of inflection. Since f A - 22 B = 3 A - 22 B 5 - 20 A - 22 B 3 = 2822, f102 = 31025 - 201023 = 0,

and

f A 22 B = 3 A 22 B 5 - 20 A 22 B 3 = - 2822,

these points are A - 22, 2822 B , 10, 02 , and A 22, - 2822 B , shown in the graph below. y (−√2, 28√2)

−3

Quick Check 2 Determine the points of inflection for the function given by g1x2 = 10x 3 - 6x 5.

−2

80 60 40 20 −1

f

(0, 0)

−20 −40 −60 −80

1

2

3

x

(√2, −28√2)

Concave Concave Concave Concave down up down up

O Quick Check 2 Curve Sketching The first and second derivatives enhance our ability to sketch curves. We use the following strategy:

Strategy for Sketching Graphs* a) Derivatives and domain. Find f ¿1x2 and f –1x2. Note the domain of f. b) Critical values of f. Find the critical values by solving f ¿1x2 = 0 and finding where f ¿1x2 does not exist. These numbers yield candidates for relative maxima or minima. Find the function values at these points. c) Increasing and/or decreasing; relative extrema. Substitute each critical value, x 0, from step (b) into f –1x2. If f –1x 02 6 0, then f1x 02 is a relative maximum and f is increasing to the left of x 0 and decreasing to the right. If f –1x 02 7 0, then f1x 02 is a relative minimum and f is decreasing to the left of x 0 and increasing to the right. d) Inflection points. Determine candidates for inflection points by finding where f –1x2 = 0 or where f –1x2 does not exist. Find the function values at these points. e) Concavity. Use the candidates for inflection points from step (d) to define intervals. Substitute test values into f –1x2 to determine where the graph is concave up 1f –1x2 7 02 and where it is concave down 1f –1x2 6 02. f) Sketch the graph. Sketch the graph using the information from steps (a)–(e), calculating and plotting extra points as needed.

*This strategy is refined further, for rational functions, in Section 2.3.

O

2.2

•

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

223

The examples that follow apply this step-by-step strategy to sketch the graphs of several functions.

■ EXAMPLE 4 Find the relative extrema of the function f given by f1x2 = x 3 - 3x + 2, and sketch the graph.

Solution a) Derivatives and domain. Find f ¿1x2 and f –1x2: f ¿1x2 = 3x 2 - 3, f –1x2 = 6x.

The domain of f (and of any polynomial function) is 1- q , q 2, or the set of all real numbers, which is also written as . b) Critical values of f. Find the critical values by determining where f ¿1x2 does not exist and by solving f ¿1x2 = 0. We know that f ¿1x2 = 3x 2 - 3 exists for all values of x, so the only critical values are where f ¿1x2 is 0: y 3x 2 - 3 3x 2 x2 x

= = = =

0 Setting f ¿1x2 equal to 0 3 1 ;1.

We have f1- 12 = 4 and f112 = 0, so 1- 1, 42 and 11, 02 are on the graph.

6

(−1, 4) 5 4 3 2 1

−6 −5−4−3−2−1 −1

(1, 0) 1 2 3 4 5 6 x

−2 −3 −4 −5 −6

c) Increasing and/or decreasing; relative extrema. Substitute the critical values into f –1x2: y f –1- 12 = 61- 12 = - 6 6 0, so f1- 12 = 4 is a relative maximum, with f increasing on 1- q , - 12 and decreasing on 1- 1, 12. f –112 = 6 # 1 = 6 7 0,

so f112 = 0 is a relative minimum, with f decreasing on 1- 1, 12 and increasing on 11, q 2.

d) Inflection points. Find possible inflection points by finding where f –1x2 does not exist and by solving f –1x2 = 0. We know that f –1x2 = 6x exists for all values of x, so we try to solve f –1x2 = 0: 6x = 0 x = 0.

Setting f –1x2 equal to 0 Dividing both sides by 6

We have f102 = 2, which gives us another point, 10, 22, that lies on the graph.

6

(−1, 4) 5 4 3 2 1 −6 −5−4−3−2−1 −1

(1, 0) 1 2 3 4 5 6 x

−2 −3 −4 −5 −6

y 6

(−1, 4) 5 4 3 (0, 2) 2 1 −6 −5−4−3−2−1 −1 −2 −3 −4 −5 −6

(1, 0) 1 2 3 4 5 6 x

224

CHAPTER 2

•

Applications of Differentiation

e) Concavity. Find the intervals on which f is concave up or concave down, using the point 10, 22 from step (d). From step (c), we can conclude that f is concave down over the interval 1- q , 02 and concave up over 10, q 2. Interval

y

x

Sign of f –1x2

1 2 3 4 5 6 x

−2 −3 −4 −5 −6

Result

⎪ ⎪ ⎪ ⎬

Test Value

⎪⎪ ⎪

⎭

−6 −5−4−3−2−1 −1

0

Point of inflection, (0, 2) (1, 0)

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪⎪ ⎪ ⎪

6 (−1, 4) 5 4 3 2 1

x = -1

x = 1

f –1- 12 6 0

f –112 7 0

f ¿ is decreasing; f is concave down.

f ¿ is increasing; f is concave up.

Change indicates a point of inflection.

f ) Sketch the graph. Sketch the graph using the information in steps (a)–(e). Calculate some extra function values if desired. The graph follows. y

x

f1x2

-3

- 16

-2

0

Relative maximum at (–1, 4)

6

f(x) = x 3 – 3x + 2

5 4

Point of inflection at (0, 2)

3 2

-1

4

0

2

1

0

2

4

3

20

1 –6 –5 –4 –3 – 2 –1

1

2

3

4

5

6

x

–1 –2 –3

Relative minimum at (1, 0)

–4 –5

TECHNOLOGY CONNECTION Check the results of Example 4 using a calculator.

–6

Increasing Concave down

Decreasing

Increasing Concave up

■ EXAMPLE 5 Find the relative maxima and minima of the function f given by f1x2 = x 4 - 2x 2, and sketch the graph.

Solution a) Derivatives and domain. Find f ¿1x2 and f –1x2: f ¿1x2 = 4x 3 - 4x, f –1x2 = 12x 2 - 4. The domain of f is .

2.2 y

•

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

b) Critical values. Since f ¿1x2 = 4x 3 - 4x exists for all values of x, the only critical values are where f ¿1x2 = 0:

2

4x 3 4x1x 2 4x x

1

(0, 0) −2

−1

1 −1

(−1, −1)

225

2

x

(1, −1)

= =

4x 12 0 0

−2

= 0 Setting f ¿1x2 equal to 0 = 0 or x 2 - 1 = 0 or x2 = 1 x = ;1.

We have f102 = 0, f1- 12 = - 1, and f112 = - 1, which gives the points 10, 02, 1- 1, - 12, and 11, - 12 on the graph. c) Increasing and/or decreasing; relative extrema. Substitute the critical values into f –1x2: f –102 = 12 # 02 - 4 = - 4 6 0,

so f102 = 0 is a relative maximum, with f increasing on 1- 1, 02 and decreasing on 10, 12.

y

f –1- 12 = 121- 122 - 4 = 8 7 0,

2

so f1- 12 = - 1 is a relative minimum, with f decreasing on 1- q , - 12 and increasing on 1- 1, 02.

1

(0, 0) −2

−1

1

(−1, −1)

−1

2

f –112 = 12 # 12 - 4 = 8 7 0,

x

so f112 = - 1 is also a relative minimum, with f decreasing on 10, 12 and increasing on 11, q 2. d) Inflection points. Find where f –1x2 does not exist and where f –1x2 = 0. Since f –1x2 exists for all real numbers, we just solve f –1x2 = 0:

(1, −1)

−2

12x 2 - 4 413x 2 - 12 3x 2 - 1 3x 2

= = = =

Setting f –1x2 equal to 0

0 0 0 1 1 x2 = 3 x = ; = ;

1 A3 1 23

.

We have fa

1 23

b = a =

y

fa-

(− 13 , − 59 ) 1 ( 13 , − 59 ) −1

1 −1 −2

23

4

b - 2a

1 23

b

2

1 2 5 = 9 3 9

and

2

−2

1

2

x

5 b = - . 9 23 1

These values give a-

1 23

, -

5 1 5 b and a , - b 9 9 23

as possible inflection points.

226

CHAPTER 2

•

Applications of Differentiation

e) Concavity. Find the intervals on which f is concave up or concave down, using the 1 5 1 5 points a , - b and a , - b, from step (d). From step (c), we can 9 23 23 9 conclude that f is concave up over the intervals A - q , - 1> 23 B and A 1> 23, q B and concave down over the interval A - 1> 23, 1> 23 B .

y

Interval

x = 0

x = 1

f –1- 12 7 0

f –102 6 0

f –112 7 0

f ¿ is increasing; f is concave up.

f ¿ is decreasing; f is concave down.

f ¿ is increasing; f is concave up.

Sign of f –1x2 Result

Change indicates a point of inflection.

TECHNOLOGY CONNECTION

EXERCISE 1. Consider

f 1x2 = x 31x - 223. How many relative extrema do you anticipate finding? Where do you think they will be? Graph f, f ¿, and f – using 3- 1, 3, - 2, 64 as a viewing window. Estimate the relative extrema and the inflection points of f. Then check your work using the analytic methods of Examples 4 and 5.

⎪ ⎪ ⎪ ⎬ ⎪ ⎪

−2

⎭ ⎫ ⎪

x

x

x = -1

Test Value

−1

1> 23

⎪

2

⎬

1

⎪⎪ ⎭

−1

Point of inflection

- 1> 23

⎫ ⎪ ⎪ ⎪ ⎬

−2

⎪

Point of 1 inflection

⎪ ⎪⎪

2

Change indicates a point of inflection.

f) Sketch the graph. Sketch the graph using the information in steps (a)–(e). By solving x 4 - 2x 2 = 0, we can find the x-intercepts easily. They are A - 22, 0 B , 10, 02, and A 22, 0 B . This also aids with graphing. Extra function values can be calculated if desired. The graph is shown below. y

x

f1x2, approximately

-2

8

- 1.5

0.56

-1

-1

- 0.5

- 0.44

0

0

2

⎞– 1 , – 5– ⎞ ⎠ √3 9 ⎠

Decreasing

0.5

- 0.44

1

-1

Quick Check 3

1.5

0.56

Find the relative maxima and minima of the function f given by f1x2 = 1 + 8x 2 - x 4, and sketch the graph.

2

8

1 (0, 0)

(– √ 2, 0 ) –2

f(x) = x 4 – 2x 2

–1

(√2, 0 ) 1

–1 (–1, –1)

(1, –1)

Increasing Decreasing

Concave up

Concave down

x

2 1 , –5 – 9 √3

Increasing

Concave up

O Quick Check 3 ■ EXAMPLE 6 Graph the function f given by f1x2 = 12x - 521>3 + 1.

List the coordinates of any extrema and points of inflection. State where the function is increasing or decreasing, as well as where it is concave up or concave down.

O

2.2

•

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

227

Solution a) Derivatives and domain. Find f ¿1x2 and f –1x2: f ¿1x2 = 13 12x - 52-2>3 # 2 = 23 12x - 52-2>3, or

2 ; 312x - 522>3

f –1x2 = - 49 12x - 52-5>3 # 2 = - 89 12x - 52-5>3, or

-8 . 912x - 525>3

The domain of f is . b) Critical values. Since f ¿1x2 =

2 312x - 522>3

is never 0 (a fraction equals 0 only when its numerator is 0), the only critical value is when f ¿1x2 does not exist. The only time f ¿1x2 does not exist is when its denominator is 0: y 3 2

( 52 , 1)

1 −1

1 −1

2

3

x

312x - 522>3 12x - 522>3 12x - 522 2x - 5 2x x

= = = = = =

f – A 52 B =

-8

0 0 0 0 5

Dividing both sides by 3 Cubing both sides

5 2

We now have f A 52 B = A 2 # 52 - 5 B 1>3 + 1 = 0 + 1 = 1, so the point A 52, 1 B is on the graph. c) Increasing and/or decreasing; relative extrema. Substitute the critical value into f –1x2: 9 A 2 # 52

- 5B

5>3

=

-8 -8 = . 9#0 0

Since f – A B does not exist, the Second-Derivative Test cannot be used at x = 52. Instead, we use the First-Derivative Test, selecting 2 and 3 as test values on either side of 52: 5 2

f ¿122 =

312 # 2 - 522>3

2

=

2 2 2 = # = , 2>3 3 1 3 31- 12

and f ¿132 =

312 # 3 - 522>3

2

=

3 # 12>3

2

=

2

3#1

=

2 . 3

Since f ¿1x2 7 0 on either side of x = 52, we know that f is increasing on both A - q , 52 B and A 52, q B ; thus, f A 52 B = 1 is not an extremum. d) Inflection points. Find where f –1x2 does not exist and where f –1x2 = 0. Since f –1x2 is never 0 (why?), we only need to find where f –1x2 does not exist. Since f –1x2 cannot exist where f ¿1x2 does not exist, we know from step (b) that a possible inflection point is A 52, 1 B . e) Concavity. We check the concavity on either side of x = 52. We choose x = 2 and x = 3 as our test values. f –122 =

-8 -8 = 7 0, 91- 12 912 # 2 - 525>3

228

CHAPTER 2

•

Applications of Differentiation

so f is concave up on A - q , 52 B . f –132 =

-8

912 # 3 - 525>3

so f is concave down on

5 2

x

(

1 1

2

3

)

1

Sign of f –1x2 x

Result

−1

⎪ ⎪ ⎪ ⎪

Test Value 5 , 2

⎬ ⎪ ⎪

Point of inflection

⎪ ⎭ ⎫ ⎪ ⎪ ⎪ ⎬

⎪

⎪ ⎪ ⎪

3

−1

A 52, q B .

-8 6 0, 9#1

Interval

y

2

=

x = 2

x = 3

f –122 7 0

f –132 6 0

f ¿ is increasing; f is concave up.

f ¿ is decreasing; f is concave down.

Change indicates a point of inflection.

f) Sketch the graph. Sketch the graph using the information in steps (a)–(e). By solving 12x - 521>3 + 1 = 0, we can find the x-intercept—it is (2, 0). Extra function values can be calculated, if desired. The graph is shown below. y

x

f1x2, approximately

0

- 0.71

2

1

- 0.44

1

2

0

5 2

1

3

2

4

2.44

5

2.71

f(x) = (2x − 5)1/3 + 1

3

−2

−1

(3, 2)

(

(2, 0) 1

2

5 , 2

)

1

3

4

5

x

−1

Increasing Concave up

Concave down

The following figures illustrate some information concerning the function in Example 1 that can be found from the first and second derivatives of f. The relative extrema are shown in Figs. 4 and 5. In Fig. 5, we see that the x-coordinates of the x-intercepts of f ¿ are the critical values of f. Note that the intervals over which f is increasing or decreasing are those intervals for which f ¿ is positive or negative, respectively.

•

2.2

y

f ′(x) = 3x2 + 6x − 9

40

f

Relative maximum

y

30

40

20

(−3, 14)

Relative maximum

30

10

20

(−3, 14)

−5

−4

−3

−2

−1

10 −5

−4

−3

229

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

−2

1

2

−1

1

2

3

−20

x

4

x

4

(1, −18)

−10

Concave down

3

−10

Relative minimum

−20

f ′(x) > 0 f is increasing

(1, −18)

f(x) = x3 + 3x2 − 9x − 13 Concave up

Relative minimum

f ′(x) < 0 f is decreasing f ′(1) = 0

f ′(−3) = 0

FIGURE 4

f′(x) > 0 f is increasing

FIGURE 5

In Fig. 6, the intervals over which f ¿ is increasing or decreasing are, respectively, those intervals over which f – is positive or negative. And finally, in Fig. 7, we note that when f –1x2 6 0, the graph of f is concave down, and when f –1x2 7 0, the graph of f is concave up. y y

40

40

f

f′

30 30

f″ 20

20 10 10 −5

−4

−3

−2

f ″(x) = 6x + 6

−5

−1

1

2

3

4

−4

x

−3

−2

1

3

4

(−1, −2) −10

−10

−20

−20

f ′ is decreasing

2

f ′ is increasing

f ″(x) < 0 f is concave down

f ″(−1) = 0

FIGURE 6

f ″(x) > 0 f is concave up

f ″(−1) = 0

FIGURE 7

TECHNOLOGY CONNECTION

Using Graphicus to Find Roots, Extrema, and Inflection Points

Graphing a Function and Finding Its Roots, Relative Extrema, and Points of Inflection

Graphicus has the capability of finding roots, relative extrema, and points of inflection. Let’s consider the function of Example 5, given by f 1x2 = x 4 - 2x 2.

After opening Graphicus, touch the blank rectangle at the top of the screen and enter the function as y(x)=x^4-2x^2. Press + in the upper right. You will see the graph (Fig. 1). (continued)

x

230

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•

Applications of Differentiation

Graphing a Function and Finding Its Roots, Relative Extrema, and Points of Inflection (continued)

FIGURE 1

FIGURE 4

FIGURE 2

FIGURE 5

Then touch the fourth icon from the left at the bottom, and you will see the roots highlighted on the graph (Fig. 2). Touch the left-hand root symbol, and a value for one root, in this case, an approximation, - 1.414214, is displayed (Fig. 3). To find the relative extrema, touch the fifth icon from the left; the relative extrema will be highlighted on the graph in a different color. Touch each point to identify the relative minima at 1- 1, - 12, and 11, - 12, and a relative maximum at 10, 02 (Fig. 4). To find points of inflection, touch the sixth icon from the left, and the points of inflection will be highlighted as shown in Fig. 5. Touch these points and the approximations 1- 0.577, - 0.5562 and 10.577, - 0.5562 will be displayed.

Graphing a Function and Its Derivatives Graphicus can graph derivatives. Go back to the original graph of f (Fig. 1). Touch + ; then press Add derivative, and you will see the original graph and the graph of the first derivative as a dashed line of a different color (Fig. 6). Touch + and Add derivative again, and you will see the graph of the original function, with dashed graphs of the first and second derivatives in different colors (Fig. 7). You can toggle between the derivatives by pressing the colored squares above the graphs.

FIGURE 3

FIGURE 6

FIGURE 7

EXERCISES Use Graphicus to graph each function and its first and second derivatives. Then find approximations for roots, relative extrema, and points of inflection. 1. f 1x2 = 2x 3 - x 4

2. f 1x2 = x1200 - x2 3. f 1x2 = x 3 - 6x 2

4. f 1x2 = - 4.32 + 1.44x + 3x 2 - x 3 5. g1x2 = x 24 - x 2 6. g1x2 = 7. f 1x2 =

4x x + 1 2

x 2 - 3x x - 1

8. f 1x2 = ƒ x + 2 ƒ - 3

Exercise Set 2.2

231

TECHNOLOGY CONNECTION

EXERCISES

1. From the graph of f ¿, estimate the critical values of f.

Graph the following:

2. From the graph of f –, estimate the x-values of any

f 1x2 = 3x 5 - 5x 3, f ¿1x2 = 15x 4 - 15x 2,

inflection points of f.

and f –1x2 = 60x 3 - 30x, using the window 3- 3, 3, - 10, 104.

Section Summary • The second derivative f – determines the concavity of the graph of function f. • If f –1x2 7 0 for all x in an open interval I, then the graph of f is concave up over I. • If f –1x2 6 0 for all x in an open interval I, then the graph of f is concave down over I. • If c is a critical value and f –1c2 7 0, then f1c2 is a relative minimum. • If c is a critical value and f –1c2 6 0, then f1c2 is a relative maximum.

• If c is a critical value and f –1c2 = 0, the First-Derivative Test must be used to classify f1c2. • If f –1x 02 = 0 or f –1x 02 does not exist, and there is a change in concavity to the left and to the right of x 0, then the point 1x 0, f1x 022 is called a point of inflection. • Finding relative extrema, intervals over which a function is increasing or decreasing, intervals of upward or downward concavity, and points of inflection is all part of a strategy for accurate curve sketching.

EXERCISE SET

2.2 For each function, find all relative extrema and classify each as a maximum or minimum. Use the Second-Derivative Test where possible. 1. f 1x2 = 5 - x 2

2. f 1x2 = 4 - x 2

5. f 1x2 = - 5x 2 + 8x - 7

6. f 1x2 = - 4x 2 + 3x - 1

3. f 1x2 = x 2 - x

7. f 1x2 = 8x 3 - 6x + 1

4. f 1x2 = x 2 + x - 1

8. f 1x2 = x 3 - 12x - 1

Sketch the graph of each function. List the coordinates of where extrema or points of inflection occur. State where the function is increasing or decreasing, as well as where it is concave up or concave down. 9. f 1x2 = x 3 - 12x

11. f 1x2 = 3x 3 - 36x - 3

10. f 1x2 = x 3 - 27x

12. f 1x2 = 2x 3 - 3x 2 - 36x + 28 13. f 1x2 = 83 x 3 - 2x +

1 3

14. f 1x2 = 80 - 9x 2 - x 3

15. f 1x2 = - x 3 + 3x 2 - 4

16. f 1x2 = - x 3 + 3x - 2

17. f 1x2 = 3x 4 - 16x 3 + 18x 2 (Round results to three decimal places.) 18. f 1x2 = 3x 4 + 4x 3 - 12x 2 + 5 (Round results to three decimal places.) 19. f 1x2 = x 4 - 6x 2 20. f 1x2 = 2x 2 - x 4

21. f 1x2 = x 3 - 2x 2 - 4x + 3 22. f 1x2 = x 3 - 6x 2 + 9x + 1 23. f 1x2 = 3x 4 + 4x 3 24. f 1x2 = x 4 - 2x 3

25. f 1x2 = x 3 - 6x 2 - 135x

26. f 1x2 = x 3 - 3x 2 - 144x - 140 27. f 1x2 = x 4 - 4x 3 + 10 28. f 1x2 = 43 x 3 - 2x 2 + x

29. f 1x2 = x 3 - 6x 2 + 12x - 6

232

•

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Applications of Differentiation

30. f 1x2 = x 3 + 3x + 1

APPLICATIONS

31. f 1x2 = 5x 3 - 3x 5

Business and Economics

32. f 1x2 = 20x - 3x (Round results to three decimal places.) 3

5

Total revenue, cost, and profit. Using the same set of axes, sketch the graphs of the total-revenue, total-cost, and totalprofit functions.

33. f 1x2 = x 213 - x22 (Round results to three decimal places.)

103. R1x2 = 50x - 0.5x 2, C1x2 = 4x + 10

34. f 1x2 = x 211 - x22 (Round results to three decimal places.) 35. f 1x2 = 1x + 12

2>3

37. f 1x2 = 1x - 32

1>3

104. R1x2 = 50x - 0.5x 2, C1x2 = 10x + 3 105. Small business. The percentage of the U.S. national income generated by nonfarm proprietors may be modeled by the function

36. f 1x2 = 1x - 122>3

p1x2 =

- 1

38. f 1x2 = 1x - 221>3 + 3

where x is the number of years since 1970. Sketch the graph of this function for 0 … x … 40.

39. f 1x2 = - 21x - 422>3 + 5 40. f 1x2 = - 31x - 222>3 + 3 41. f 1x2 = x24 - x 2

42. f 1x2 = - x21 - x 2

43. f 1x2 =

44. f 1x2 =

45. f 1x2 =

x x2 + 1 3 x2 + 1

13x 3 - 240x 2 - 2460x + 585,000 , 75,000

46. f 1x2 =

8x x2 + 1 -4 x2 + 1

For Exercises 47–56, sketch a graph that possesses the characteristics listed. Answers may vary. 47. f is increasing and concave up on 1- q , 42, f is increasing and concave down on 14, q 2. 48. f is decreasing and concave up on 1- q , 22, f is decreasing and concave down on12, q 2.

49. f is increasing and concave down on 1- q , 12, f is increasing and concave up on 11, q 2.

50. f is decreasing and concave down on 1- q , 32, f is decreasing and concave up on 13, q 2.

106. Labor force. The percentage of the U.S. civilian labor force aged 45–54 may be modeled by the function f 1x2 = 0.025x 2 - 0.71x + 20.44,

where x is the number of years after 1970. Sketch the graph of this function for 0 … x … 30.

Life and Physical Sciences 107. Coughing velocity. A person coughs when a foreign object is in the windpipe. The velocity of the cough depends on the size of the object. Suppose a person has a windpipe with a 20-mm radius. If a foreign object has a radius r, in millimeters, then the velocity V, in millimeters per second, needed to remove the object by a cough is given by V1r2 = k120r 2 - r 32, 0 … r … 20,

where k is some positive constant. For what size object is the maximum velocity required to remove the object?

51. f is concave down at 11, 52, concave up at 17, - 22, and has an inflection point at 14, 12. 52. f is concave up at 11, - 32, concave down at 18, 72, and has an inflection point at 15, 42. 53. f ¿1- 12 = 0, f –1 - 12 7 0, f 1- 12 = - 5; f ¿172 = 0, f –172 6 0, f 172 = 10; f –132 = 0, and f 132 = 2

54. f ¿1- 32 = 0, f –1 - 32 6 0, f 1- 32 = 8; f ¿192 = 0, f –192 7 0, f 192 = - 6; f –122 = 0, and f 122 = 1

55. f ¿1- 12 = 0, f –1 - 12 7 0, f 1- 12 = - 2; f ¿112 = 0, f –112 7 0, f 112 = - 2; f ¿102 = 0, f –102 6 0, and f 102 = 0

56. f ¿102 = 0, f –102 6 0, f 102 = 5; f ¿122 = 0, f –122 7 0, f 122 = 2; f ¿142 = 0, f –142 6 0, and f 142 = 3 57–102. Check the results of Exercises 1–46 with a graphing calculator.

108. New York temperatures. The average temperature in New York can be approximated by the function T1x2 = 43.5 - 18.4x + 8.57x 2 - 0.996x 3 + 0.0338x 4, where T represents the temperature, in degrees Fahrenheit, x = 1 represents the middle of January, x = 2 represents the middle of February, and so on. (Source: www.WorldClimate.com.)

Exercise Set 2.2

112. Social sciences: three aspects of love. Researchers at Yale University have suggested that the following graphs may represent three different aspects of love.

T T(x) = 43.5 – 18.4x + 8.57x 2 – 0.996x 3 + 0.0338x 4 80 70

PASSION

INTIMACY

COMMITMENT Level

50

Level

60 Level

Temperature (in degrees Fahrenheit)

233

40 Time

Time

Time

30 2

0

4

6 8 Month

10

12

(Source: From “A Triangular Theory of Love,” by R. J. Sternberg, 1986, Psychological Review, 93(2), 119–135. Copyright 1986 by the American Psychological Association, Inc. Reprinted by permission.)

x

Analyze each of these graphs in terms of the concepts you have learned: relative extrema, concavity, increasing, decreasing, and so on. Do you agree with the researchers regarding the shapes of these graphs? Why or why not?

a) Based on the graph, when would you expect the highest temperature to occur in New York? b) Based on the graph, when would you expect the lowest temperature to occur? c) Use the Second-Derivative Test to estimate the points of inflection for the function T1x2. What is the significance of these points?

113. Use calculus to prove that the relative minimum or maximum for any function f for which f 1x2 = ax 2 + bx + c, a Z 0, occurs at x = - b>12a2.

109. Hours of daylight. The number of hours of daylight in Chicago is represented in the graph below. On what dates is the number of hours of daylight changing most rapidly? How can you tell?

114. Use calculus to prove that the point of inflection for any function g given by g1x2 = ax 3 + bx 2 + cx + d,

a Z 0,

Number of hours of daylight

occurs at x = - b>13a2. For Exercises 115–121, assume the function f is differentiable over the interval 1 ˆ , ˆ 2; that is, it is smooth and continuous for all real numbers x and has no corners or vertical tangents. Classify each of the following statements as either true or false. If you choose false, explain why.

18 15 12

115. If f has exactly two critical values at x = a and x = b, where a 6 b, then there must exist exactly one point of inflection at x = c such that a 6 c 6 b. In other words, exactly one point of inflection must exist between any two critical points.

9 6 3 0

60

120

180

240

300

116. If f has exactly two critical values at x = a and x = b, where a 6 b, then there must exist at least one point of inflection at x = c such that a 6 c 6 b. In other words, at least one point of inflection must exist between any two critical points.

360

Number of days since January 1 ( Source: Astronomical Applications Dept., U.S. Naval Observatory.)

117. The function f can have no extrema but can have at least one point of inflection.

SYNTHESIS

118. If the function f has two points of inflection, then there must be a critical value located between those points of inflection.

In each of Exercises 110 and 111, determine which graph is the derivative of the other and explain why. 110.

y

h

4

111. h

y

g

119. The function f can have a point of inflection at a critical value.

100

2 –4

–2

2 –2

4

–2

x g

–1 –100

1

2

x

120. The function f can have a point of inflection at an extreme value. 121. The function f can have exactly one extreme value but no points of inflection.

234

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•

Applications of Differentiation

TECHNOLOGY CONNECTION

a) Use the regression procedures of Section R.6 to fit linear, cubic, and quartic functions y = f 1x2 to the data, where x is age and y is average use per month. Decide which function best fits the data. Explain. b) What is the domain of the function? c) Does the function have any relative extrema? Explain.

Graph each function. Then estimate any relative extrema. Where appropriate, round to three decimal places. 122. f 1x2 = 3x 2>3 - 2x

123. f 1x2 = 4x - 6x 2>3

124. f 1x2 = x 21x - 223 126. f 1x2 = x - 2x 127. f 1x2 = 1x - 12

2>3

125. f 1x2 = x 211 - x23

Answers to Quick Checks

- 1x + 12

1. Relative minimum: - 4 at x = - 1; relative maximum: 4 at x = 1

2>3

128. Social sciences: time spent on home computer. The following data relate the average number of minutes spent per month on a home computer to a person’s age.

Age (in years)

Average Use (in minutes per month)

6.5 14.5 21 29.5 39.5 49.5 55 and up

363 645 1377 1727 1696 2052 2299

y 4

g(x)  10x3  6x5

2 − 1.0 − 0.5 −2

0.5 1.0

x

−4

22 2. 10, 02, A 22 2 , 2.475 B , and A - 2 , - 2.475 B 3. Relative maxima: 17 at x = - 2 and x = 2; relative minimum: 1 at x = 0 y 10 −2

−1 − 10 − 20

1

2

x

f(x)  1  8x 2  x 4

− 30

(Source: Media Matrix; The PC Meter Company.)

2.3 OBJECTIVES • Find limits involving infinity. • Determine the asymptotes of a function’s graph. • Graph rational functions.

Graph Sketching: Asymptotes and Rational Functions Thus far we have considered a strategy for graphing a continuous function using the tools of calculus. We now want to consider some discontinuous functions, most of which are rational functions. Our graphing skills must now allow for discontinuities as well as certain lines called asymptotes. Let’s review the definition of a rational function.

Rational Functions DEFINITION A rational function is a function f that can be described by f1x2 =

P1x2 Q1x2

,

where P1x2 and Q1x2 are polynomials, with Q1x2 not the zero polynomial. The domain of f consists of all inputs x for which Q1x2 Z 0.

2.3

•

Graph Sketching: Asymptotes and Rational Functions

235

Polynomials are themselves a special kind of rational function, since Q1x2 can be 1. Here we are considering graphs of rational functions in which the denominator is not a constant. Before we do so, however, we need to reconsider limits.

Vertical and Horizontal Asymptotes Figure 1 shows the graph of the rational function 1x - 121x + 12 x2 - 1 = . 1x - 221x + 32 x + x - 6 2

Vertical asymptote

y

Vertical asymptote

f1x2 =

6 5 4 3 2

x2 – 1 f(x) = ——–—— x2 + x – 6

⎛0, –1–⎛ ⎝ 6⎝

y = 1 Horizontal asymptote

(–1, 0) – 6 – 5 – 4 –3

–2

–1 (1, 0) –1

3

4

5

6

7

8

9

10 11 x

–2 –3 –4 –5 –6 x = –3

x=2

FIGURE 1

Note that as x gets closer to 2 from the left, the function values get smaller and smaller negatively, approaching - q . As x gets closer to 2 from the right, the function values get larger and larger positively. Thus, lim f1x2 = - q

x : 2-

and

lim f1x2 = q .

x : 2+

For this graph, we can think of the line x = 2 as a “limiting line” called a vertical asymptote. Similarly, the line x = - 3 is another vertical asymptote.

DEFINITION The line x = a is a vertical asymptote if any of the following limit statements is true: lim f1x2 = q ,

x : a-

lim f1x2 = - q ,

x : a-

lim f1x2 = q , or

x : a+

lim f1x2 = - q .

x : a+

The graph of a rational function never crosses a vertical asymptote. If the expression that defines the rational function f is simplified, meaning that it has no common factor other than - 1 or 1, then if a is an input that makes the denominator 0, the line x = a is a vertical asymptote.

236

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•

Applications of Differentiation

For example, f1x2 =

1x - 321x + 32 x2 - 9 = x - 3 x - 3

does not have a vertical asymptote at x = 3, even though 3 is an input that makes the denominator 0. This is because when 1x 2 - 92>1x - 32 is simplified, it has x - 3 as a common factor of the numerator and the denominator. In contrast, g1x2 =

1x + 221x - 22 x2 - 4 = 2 1x - 321x + 42 x + x - 12

is simplified and has x = 3 and x = - 4 as vertical asymptotes. Figure 2 shows the four ways in which a vertical asymptote can occur. The dashed lines represent the asymptotes. They are sketched in for visual assistance only; they are not part of the graphs of the functions. y

y

y

y y = f(x)

y = f(x)

y = f(x) x

x

x

x

y = f(x) x=a lim f(x) = –∞

x=a lim f(x) = ∞ (b)

(a)

x=a lim f(x) = ∞

x → a–

x → a–

x → a+

(c)

x=a lim f(x) = –∞ x → a+

(d)

FIGURE 2

■ EXAMPLE 1 Determine the vertical asymptotes: f1x2 =

Quick Check 1

O

Determine the vertical asymptotes: f1x2 =

1 . x1x - 162 2

3x - 2 . x1x - 521x + 32

Solution The expression is in simplified form. The vertical asymptotes are the lines x = 0, x = 5, and x = - 3.

O Quick Check 1 ■ EXAMPLE 2 Determine the vertical asymptotes of the function given by f1x2 =

x 2 - 2x . x3 - x

Solution We write the expression in simplified form: f1x2 = Quick Check 2 For the function in Example 2, explain why x = 0 does not correspond to a vertical asymptote. What kind of discontinuity occurs at x = 0?

=

x1x - 22 x 2 - 2x = 3 x1x - 121x + 12 x - x x - 2 , x Z 0. 1x - 121x + 12

The expression is now in simplified form. The vertical asymptotes are the lines x = - 1 and x = 1.

O Quick Check 2

O

2.3

TECHNOLOGY CONNECTION Our discussion here allows us to attach the term “vertical asymptote” to those mysterious vertical lines that appear when graphing rational functions in the CONNECTED mode. For example, consider the graph of f 1x2 = 8>1x 2 - 42, using the window 3- 6, 6, - 8, 84. Vertical asymptotes occur at x = - 2 and x = 2. These lines are not part of the graph. 8 x2 – 4

Graph Sketching: Asymptotes and Rational Functions

lim f1x2 = 1 and

lim f1x2 = 1.

x: - q

x: q

The line y = 1 is called a horizontal asymptote.

8

DEFINITION

–6

The line y = b is a horizontal asymptote if either or both of the following limit statements is true:

6

lim f1x2 = b or

lim f1x2 = b.

x: - q –8

EXERCISES Graph each of the following in both DOT and CONNECTED modes. Try to locate the vertical asymptotes visually. Then verify your results using the method of Examples 1 and 2. You may need to try different viewing windows. x 2 + 7x + 10 x 2 + 3x - 28

2. f 1x2 =

x: q

The graph of a rational function may or may not cross a horizontal asymptote. Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. (The degree of a polynomial in one variable is the highest power of that variable.)

Not part of the graph!

1. f 1x2 =

237

Look again at the graph in Fig. 1. Note that function values get closer and closer to 1 as x approaches - q , meaning that f1x2 : 1 as x : - q . Also, function values get closer and closer to 1 as x approaches q , meaning that f1x2 : 1 as x : q . Thus,

Asymptotes

y=

•

x2 + 5

In Figs. 3–5, we see three ways in which horizontal asymptotes can occur.

x 3 - x 2 - 6x

y

y

y y = f(x)

y = f(x) lim f(x) = b

x → –∞

y=b

y=b y=b lim f(x) = b

y = f(x) lim f(x) = b

x→∞

x→ ∞

x

x

x

FIGURE 3

FIGURE 4

FIGURE 5

Horizontal asymptotes are found by determining the limit of a rational function as inputs approach - q or q .

■ EXAMPLE 3 Determine the horizontal asymptote of the function given by f1x2 =

3x - 4 . x

Solution To find the horizontal asymptote, we consider lim f1x2 = lim

x: q

x: q

3x - 4 . x

One way to find such a limit is to use an input–output table, as follows, using progressively larger x-values. Inputs, x Outputs,

3x - 4 x

1

10

50

100

2000

-1

2.6

2.92

2.96

2.998

238

CHAPTER 2

•

Applications of Differentiation

TECHNOLOGY CONNECTION

As the inputs get larger and larger without bound, the outputs get closer and closer to 3. Thus,

EXERCISES

3x - 4 = 3. x x: q

1. Verify the limit

lim

x: q

lim

3x - 4 = 3 x

Another way to find this limit is to use some algebra and the fact that

by using the TABLE feature with larger and larger x-values. 2.92 2.9733 2.984 2.9886 2.9911 2.9927 2.9938

X 500 1500 2500 3500 4500 5500 6500 X = 500

2.992 2.9973 2.9984 2.9989 2.9991 2.9993 2.9994

lim

x: q Y1

2. Graph the function

f 1x2 =

1 b : 0, and more generally, n : 0, x ax

for any positive integer n and any constants a and b, a Z 0. We multiply by 1, using 11>x2 , 11>x2. This amounts to dividing both the numerator and the denominator by x:

Y1

X 50 150 250 350 450 550 650 X = 50

as x : q , we have

3x - 4 x

in DOT mode. Then use TRACE, moving the cursor along the graph from left to right, and observe the behavior of the y-coordinates. For Exercises 3 and 4, consider 2x + 5 lim . x x: q 3. Use the TABLE feature to find

the limit.

3x - 4 3x - 4 # 11>x2 = lim q x x x: 11>x2 3x 4 x x = lim x x: q x 4 3 x = lim x: q 1 4 = lim a3 - b x x: q = 3 - 0 = 3.

In a similar manner, it can be shown that lim f1x2 = 3.

x: - q

The horizontal asymptote is the line y = 3.

■ EXAMPLE 4 Determine the horizontal asymptote of the function given by f1x2 =

4. Graph the function in DOT

mode, and use TRACE to find the limit.

3x 2 + 2x - 4 . 2x 2 - x + 1

Solution As in Example 3, the degree of the numerator is the same as the degree of the denominator. Let’s adapt the algebraic approach used in that example. To do so, we divide the numerator and the denominator by x 2 and find the limit as ƒ x ƒ gets larger and larger: 3x 2 + 2x - 4 f1x2 = = 2x 2 - x + 1

As ƒ x ƒ gets very large, the numerator approaches 3 and the denominator approaches 2. Therefore, the value of the function gets very close to 32. Thus,

Quick Check 3 Determine the horizontal asymptote of the function given by f1x2 =

12x - 121x + 12

13x + 2215x + 62

2 4 - 2 x x . 1 1 2 + 2 x x 3 +

lim f1x2 =

x: -q

The line y = .

3 2

O Quick Check 3

3 2

and

lim f1x2 =

x: q

is a horizontal asymptote.

3 . 2

O

2.3

TECHNOLOGY CONNECTION To see why a horizontal asymptote is determined by the leading terms in the expression’s numerator and denominator, consider the terms 3x 2, 2x, and - 4 in Example 4. If y1 = 3x 2 and y2 = 3x 2 + 2x - 4, a table reveals that as x gets larger, the difference between y2 and y1 becomes less significant. This is because 3x 2 grows far more rapidly than does 2x - 4. X 20 120 220 320 420 520 620 X = 20

Y1 1200 43200 145200 307200 529200 811200 1.15 E6

Y2 1236 43436 145636 307836 530036 812236 1.15 E6

•

Examples 3 and 4 lead to the following result.

When the degree of the numerator is the same as the degree of the denominator, the line y = a>b is a horizontal asymptote, where a is the leading coefficient of the numerator and b is the leading coefficient of the denominator.

■ EXAMPLE 5 Determine the horizontal asymptote: f1x2 =

2x + 3 . x - 2x 2 + 4 3

Solution Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote. To identify that asymptote, we divide both the numerator and denominator by the highest power of x in the denominator, just as in Examples 3 and 4, and find the limits as ƒ x ƒ : q : 2x + 3 f1x2 = 3 = x - 2x 2 + 4

EXERCISES 1. Let y1 = 2x 2 and

y2 = 2x 2 - x + 1. Use a table to show that as x gets large, y1 L y2.

2. Let y1 = 13x 22>12x 22 and

y2 = 13x 2 + 2x - 12> 12x 2 - x + 12. Show that for large x, we have y1 L y2.

2 3 + 3 2 x x . 4 2 + 3 1 x x

As x gets smaller and smaller negatively, ƒ x ƒ gets larger and larger. Similarly, as x gets larger and larger positively, ƒ x ƒ gets larger and larger. Thus, as ƒ x ƒ becomes very large, every expression with a denominator that is a power of x gets ever closer to 0. Thus, the numerator of f1x2 approaches 0 as its denominator approaches 1; hence, the entire expression takes on values ever closer to 0. That is, for x : - q or x : q , we have f1x2 L so

TECHNOLOGY CONNECTION

239

Graph Sketching: Asymptotes and Rational Functions

0 + 0 , 1 - 0 + 0

lim f1x2 = 0 and

x: - q

lim f1x2 = 0,

x: q

and the x-axis, the line y = 0, is a horizontal asymptote.

EXERCISES

1. f 1x2 = 2. f 1x2 =

x2 + 5 x 3 - x 2 - 6x 9x - 7x - 9 4

2

3x + 7x + 9 4

2

3. f 1x2 =

135x 5 - x 2

4. f 1x2 =

3x 2 - 4x + 3

x7 6x 2 + 2x - 5

When the degree of the numerator is less than the degree of the denominator, the x-axis, or the line y = 0, is a horizontal asymptote.

y 10

Slant Asymptotes Some asymptotes are neither vertical nor horizontal. For example, in the graph of x2 - 4 , f1x2 = x - 1 shown at right, as ƒ x ƒ gets larger and larger, the curve gets closer and closer to y = x + 1. The line y = x + 1 is called a slant asymptote, or oblique asymptote. In Example 6, we will see how the line y = x + 1 was determined.

x2 – 4 f(x) = ——– x–1

5

– 3 –2 –1 e t pto sym a e liqu Ob –5

–10

Vertical asymptote

Graph each of the following. Try to locate the horizontal asymptotes using the TABLE and TRACE features. Verify your results using the methods of Examples 3–5.

1

x=1

y=

2

3

x+

1

x

240

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•

Applications of Differentiation

DEFINITION A linear asymptote that is neither vertical nor horizontal is called a slant asymptote, or an oblique asymptote. For any rational function of the form f1x2 = p1x2>q1x2, a slant asymptote occurs when the degree of p1x2 is exactly 1 more than the degree of q1x2. A graph can cross a slant asymptote.

How can we find a slant asymptote? One way is by division.

TECHNOLOGY CONNECTION

EXERCISES Graph each of the following. Try to visually locate the slant asymptotes. Then use the method in Example 6 to find each slant asymptote and graph it along with the original function. 3x 2 - 7x + 8 1. f 1x2 = x - 2 2. f 1x2 =

5x 3 + 2x + 1 x2 - 4

Quick Check 4

O

Find the slant asymptote: 2x 2 + x - 1 g1x2 = . x - 3

■ EXAMPLE 6 Find the slant asymptote: f1x2 =

x2 - 4 . x - 1

Solution When we divide the numerator by the denominator, we obtain a quotient of x + 1 and a remainder of - 3: x + 1 x - 1 x 2 - 4 x2 - x x - 4 x - 1 - 3

f1x2 =

x2 - 4 -3 = 1x + 12 + . x - 1 x - 1

Now we can see that when ƒ x ƒ gets very large, - 3>1x - 12 approaches 0. Thus, for very large ƒ x ƒ , the expression x + 1 is the dominant part of 1x + 12 +

-3 . x - 1

Thus, y = x + 1 is a slant asymptote.

O Quick Check 4 Intercepts If they exist, the x-intercepts of a function occur at those values of x for which y = f1x2 = 0, and they give us points at which the graph crosses the x-axis. If it exists, the y-intercept of a function occurs at the value of y for which x = 0, and it gives us the point at which the graph crosses the y-axis.

■ EXAMPLE 7 Find the intercepts of the function given by TECHNOLOGY CONNECTION

EXERCISES Graph each of the following. Use the ZERO feature and a table in ASK mode to find the x- and y-intercepts. 1. f 1x2 =

x1x - 321x + 52

2. f 1x2 =

x 3 + 2x 2 - 3x

1x + 221x - 42 x2 + 5

f1x2 =

x 3 - x 2 - 6x . x 2 - 3x + 2

Solution We factor the numerator and the denominator: f1x2 =

x1x + 221x - 32 1x - 121x - 22

.

To find the x-intercepts, we solve the equation f1x2 = 0. Such values occur when the numerator is 0 and the denominator is not. Thus, we solve the equation x1x + 221x - 32 = 0. The x-values that make the numerator 0 are 0, - 2, and 3. Since none of these make the denominator 0, they yield the x-intercepts 10, 02, 1- 2, 02, and 13, 02.

2.3

•

Graph Sketching: Asymptotes and Rational Functions

241

To find the y-intercept, we let x = 0: Quick Check 5 Find the intercepts of the function given by h1x2 =

x3 - x . x2 - 4

f102 =

03 - 02 - 6102 02 - 3102 + 2

= 0.

In this case, the y-intercept is also an x-intercept, 10, 02.

O Quick Check 5 Sketching Graphs We can now refine our analytic strategy for graphing.

Strategy for Sketching Graphs a) b) c) d) e)

f)

g)

h)

Intercepts. Find the x-intercept(s) and the y-intercept of the graph. Asymptotes. Find any vertical, horizontal, or slant asymptotes. Derivatives and domain. Find f ¿1x2 and f –1x2. Find the domain of f. Critical values of f. Find any inputs for which f ¿1x2 is not defined or for which f ¿1x2 = 0. Increasing and/or decreasing; relative extrema. Substitute each critical value, x 0, from step (d) into f –1x2. If f –1x 02 6 0, then x 0 yields a relative maximum and f is increasing to the left of x 0 and decreasing to the right. If f –1x 02 7 0, then x 0 yields a relative minimum and f is decreasing to the left of x 0 and increasing to the right. On intervals where no critical value exists, use f ¿ and test values to find where f is increasing or decreasing. Inflection points. Determine candidates for inflection points by finding x-values for which f –1x2 does not exist or for which f –1x2 = 0. Find the function values at these points. If a function value f1x2 does not exist, then the function does not have an inflection point at x. Concavity. Use the values from step (f) as endpoints of intervals. Determine the concavity over each interval by checking to see where f ¿ is increasing—that is, where f –1x2 7 0—and where f ¿ is decreasing—that is, where f –1x2 6 0. Do this by substituting a test value from each interval into f –1x2. Use the results of step (d). Sketch the graph. Use the information from steps (a)–(g) to sketch the graph, plotting extra points as needed.

■ EXAMPLE 8 Sketch the graph of f1x2 =

8 . x - 4 2

Solution a) Intercepts. The x-intercepts occur at values for which the numerator is 0 but the denominator is not. Since in this case the numerator is the constant 8, there are no x-intercepts. To find the y-intercept, we compute f102: f102 =

8 8 = = - 2. -4 0 - 4 2

This gives us one point on the graph, 10, - 22. b) Asymptotes. Vertical: The denominator, x 2 - 4 = 1x + 221x - 22, is 0 for x-values of - 2 and 2. Thus, the graph has the lines x = - 2 and x = 2 as vertical asymptotes. We draw them using dashed lines (they are not part of the actual graph, just guidelines).

O

242

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Horizontal: The degree of the numerator is less than the degree of the denominator, so the x-axis, y = 0, is the horizontal asymptote. Slant: There is no slant asymptote since the degree of the numerator is not 1 more than the degree of the denominator. c) Derivatives and domain. We find f ¿1x2 and f –1x2 using the Quotient Rule: f ¿1x2 =

- 16x 2 1x - 422

f –1x2 =

and

1613x 2 + 42 1x 2 - 423

.

The domain of f is 1- q , - 22 ´ 1- 2, 22 ´ 12, q 2 as determined in step (b). d) Critical values of f. We look for values of x for which f ¿1x2 = 0 or for which f ¿1x2 does not exist. From step (c), we see that f ¿1x2 = 0 for values of x for which - 16x = 0, but the denominator is not 0. The only such number is 0 itself. The derivative f ¿1x2 does not exist at - 2 and 2, but neither value is in the domain of f. Thus, the only critical value is 0. e) Increasing and/or decreasing; relative extrema. We use the undefined values and the critical values to determine the intervals over which f is increasing and the intervals over which f is decreasing. The values to consider are - 2, 0, and 2. Since f –102 =

1613 # 02 + 42 10 - 42 2

3

=

64 6 0, - 64

we know that a relative maximum exists at 10, f1022, or 10, - 22. Thus, f is increasing on the interval 1- 2, 02 and decreasing on 10, 22. Since f –1x2 does not exist for the x-values - 2 and 2, we use f ¿1x2 and test values to see if f is increasing or decreasing on 1- q , - 22 and 12, q 2: Test - 3, f ¿1- 32 = Test 3,

f ¿132 =

- 161- 32

31- 32 - 44 2

2

- 16132

3132 - 44 2

2

=

=

48 7 0, so f is increasing on 1- q , - 22; 25

- 48 6 0, so f is decreasing on 12, q 2. 25

f) Inflection points. We determine candidates for inflection points by finding where f –1x2 does not exist and where f –1x2 = 0. The only values for which f –1x2 does not exist are where x 2 - 4 = 0, or - 2 and 2. Neither value is in the domain of f, so we focus solely on where f –1x2 = 0, or 1613x 2 + 42 = 0. Since 1613x 2 + 42 7 0 for all real numbers x, there are no points of inflection. g) Concavity. Since no values were found in step (f), the only place where concavity could change is on either side of the vertical asymptotes, x = - 2 and x = 2. To determine the concavity, we check to see where f –1x2 is positive or negative. The numbers - 2 and 2 divide the x-axis into three intervals. We choose test values in each interval and make a substitution into f –: Test - 3, f –1- 32 = Test 0,

f –102 =

Test 3,

f –132 =

16331- 322 + 44 31- 322 - 443 16331022 + 44 31022 - 443

16331322 + 44 31322 - 443

7 0;

6 0; 7 0.

We already knew this from step (e).

2.3

•

Interval

-2

x

⎪ ⎪ ⎪ ⎪ ⎬

⎪ ⎪ ⎭

⎫ ⎪ ⎪ ⎪

⎬

⎪ ⎪ ⎪

⎭ ⎫

⎪ ⎪

Result

⎬

⎪ ⎪

⎪ Sign of f –1x2

2

x = -3

x = 0

x = 3

f –1- 32 7 0

f –102 6 0

f –132 7 0

f ¿ is increasing; f is concave up.

f ¿ is decreasing; f is concave down.

f ¿ is increasing; f is concave up.

Test Value

243

Graph Sketching: Asymptotes and Rational Functions

Change does not indicate a point of inflection since f 1- 22 does not exist.

Change does not indicate a point of inflection since f 122 does not exist.

The function is concave up over the intervals 1- q , - 22 and 12, q 2. The function is concave down over the interval 1- 2, 22. h) Sketch the graph. We sketch the graph using the information in the following table, plotting extra points as needed. The graph is shown below. y

x

f1x2 approximately

-5

0.38

-4

0.67

-3

1.6

-1

- 2.67

0

-2

1

- 2.67

3

1.6

4

0.67

5

0.38

8 7 6 5 4 3 2 1 –5

–4

–3

–2

x = –2

–1

8 f(x) = ——– x2 – 4

1 2 3 4 –1 Relative maximum –2 –3 –4 x=2 – 5 (0, –2) –6 –7 –8

5 x

Increasing

Increasing Decreasing

Decreasing

Concave up

Concave down

Concave up

■ EXAMPLE 9 Sketch the graph of the function given by f1x2 =

x2 + 4 . x

Solution a) Intercepts. The equation f1x2 = 0 has no real-number solution. Thus, there are no x-intercepts. The number 0 is not in the domain of the function. Thus, there is no y-intercept. b) Asymptotes. Vertical: Since replacing x with 0 makes the denominator 0, the line x = 0 is a vertical asymptote. Horizontal: The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote.

244

CHAPTER 2

•

Applications of Differentiation

Slant: The degree of the numerator is 1 greater than the degree of the denominator, so there is a slant asymptote. We do the division x x x 2 + 4 x2 4 and express the function in the form f1x2 = x +

4 . x

As ƒ x ƒ gets larger, 4>x approaches 0, so the line y = x is a slant asymptote. c) Derivatives and domain. We find f ¿1x2 and f –1x2: f ¿1x2 = 1 - 4x -2 = 1 f –1x2 = 8x -3 =

4 ; x2

8 . x3

The domain of f is 1- q , 02 ´ 10, q 2, or all real numbers except 0. d) Critical values of f. We see from step (c) that f ¿1x2 is undefined at x = 0, but 0 is not in the domain of f. Thus, to find critical values, we solve f ¿1x2 = 0, looking for solutions other than 0: 1 -

4 = 0 x2 1 =

Setting f ¿1x2 equal to 0

4 x2

x2 = 4 x = ;2.

Multiplying both sides by x 2

Thus, - 2 and 2 are critical values. e) Increasing and/or decreasing; relative extrema. We use the points found in step (d) to find intervals over which f is increasing and intervals over which f is decreasing. The points to consider are - 2, 0, and 2. Since f –1- 22 =

8 = - 1 6 0, 1- 223

we know that a relative maximum exists at 1- 2, f1- 222, or 1- 2, - 42. Thus, f is increasing on 1- q , - 22 and decreasing on 1- 2, 02. Since f –122 =

8 = 1 7 0, 1223

we know that a relative minimum exists at 12, f1222, or 12, 42. Thus, f is decreasing on 10, 22 and increasing on 12, q 2. f) Inflection points. We determine candidates for inflection points by finding where f –1x2 does not exist or where f –1x2 = 0. The only value for which f –1x2 does not exist is 0, but 0 is not in the domain of f. Thus, the only place an inflection point could occur is where f –1x2 = 0: 8 = 0. x3 But this equation has no solution. Thus, there are no points of inflection.

2.3

•

Graph Sketching: Asymptotes and Rational Functions

245

g) Concavity. Since no values were found in step (f), the only place where concavity could change would be on either side of the vertical asymptote x = 0. In step (e), we used the Second-Derivative Test to determine relative extrema. From that work, we know that f is concave down over the interval 1- q , 02 and concave up over 10, q 2. Interval 0

x

f1x2, approximately

-6

- 6.67

-5

- 5.8

-4

-5

-3

- 4.3

-2

-4

-1

-5

- 0.5

- 8.5

0.5

8.5

1

5

2

4

3

4.3

4

5

5

5.8

6

6.67

Sign of f –1x2 Result

⎪ ⎪ ⎪ ⎬

x

⎪⎪ ⎪

⎭

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪⎪ ⎪ ⎪ x = -2

x = 2

f –1- 22 6 0

f –122 7 0

f ¿ is decreasing; f is concave down.

f ¿ is increasing; f is concave up.

Test Value

Change does not indicate a point of inflection since f 102 does not exist.

h) Sketch the graph. We sketch the graph using the preceding information and additional computed values of f, as needed. The graph follows. y

Relative minimum

9 8 7 6 5 4 3 2 1

–9 –8 –7 –6 –5 – 4 –3 –2 –1 –1 –2 y=x –3 –4 –5

(– 2, – 4)

(2, 4) x2 + 4 f(x) = ——– x

1 2 3 4 5 6 7 8 9

x

Relative maximum

–6 –7 –8 –9

Quick Check 6 Sketch the graph of the function given by x2 - 9 . f1x2 = x - 1

Increasing

Decreasing Decreasing

Concave down

Increasing

Concave up

O Quick Check 6 We can apply our analytic strategy for graphing to “building” a rational function that meets certain initial conditions.

O

246

CHAPTER 2

•

Applications of Differentiation

■ EXAMPLE 10 Determine a rational function f (in lowest terms) whose graph has vertical asymptotes at x = - 5 and x = 2 and a horizontal asymptote at y = 2 and for which f112 = 3.

Solution We know that the graph of f has vertical asymptotes at x = - 5 and x = 2, so we can conclude that the denominator must contain the factors x + 5 and x - 2. Writing these as a product, we see that the denominator will have degree 2 (although we do not carry out the multiplication). Since the graph has a horizontal asymptote, the function must have a polynomial of degree 2 in the numerator, and the leading coefficients must form a ratio of 2. Therefore, a reasonable first guess for f is given by f1x2 =

2x 2 . 1x + 521x - 22

This is a first guess.

However, this does not satisfy the requirement that f112 = 3. We can add a constant in the numerator and then use the fact that f112 = 3 to solve for this constant: f1x2 =

2x 2 + B 1x + 521x - 22 21122 + B

3 =

11 + 521 1 - 22

3 =

2 + B -6

Setting x = 1 and f 1x2 = 3

Multiplying both sides by - 6, we have - 18 = 2 + B. Therefore, B = - 20. The rational function is given by f1x2 =

2x 2 - 20 . 1x + 521x - 22

A sketch of its graph serves as a visual check that all of the initial conditions are met. y x = −5

x=2 5

y=2

(1, 3) −5

Quick Check 7

O

Determine a rational function g that has vertical asymptotes at x = - 2 and x = 2 and a horizontal asymptote at y = 3, and for which g112 = - 3.

5

−5

f(x) =

x

2x 2 − 20 (x + 5)(x − 2)

O Quick Check 7

Section Summary • A line x = a is a vertical asymptote if lim- f1x2 = ; q or x:a lim+ f1x2 = ; q . x:a

• A line y = b is a horizontal asymptote if lim f1x2 = b or x: q lim f1x2 = b. x: -q

• A graph may cross a horizontal asymptote but never a vertical asymptote.

• A slant asymptote occurs when the degree of the numerator is 1 greater than the degree of the denominator. Long division of polynomials can be used to determine the equation of the slant asymptote. • Vertical, horizontal, and slant asymptotes can be used as guides for accurate curve sketching. Asymptotes are not a part of a graph but are visual guides only.

Exercise Set 2.3

247

EXERCISE SET

2.3 Determine the vertical asymptote(s) of each function. If none exists, state that fact. 1. f 1x2 = 3. f 1x2 = 5. f 1x2 = 7. f 1x2 = 9. f 1x2 =

2x - 3 x - 5

2. f 1x2 =

3x

4. f 1x2 =

x - 9 2

x + 2 x - 6x + 8x 3

2

x + 6 x + 7x + 6 2

6. f 1x2 = 8. f 1x2 = 10. f 1x2 =

6 x + 36 2

x + 4 x - 2

13. f 1x2 =

6x 8x + 3

12. f 1x2 =

4x

14. f 1x2 =

x - 3x 2

5x x - 25 2

x + 3 x3 - x x + 2 x + 6x + 8 2

7 x + 49 2

15. f 1x2 = 5 17. f 1x2 =

8x 4 - 5x 2 2x + x

6x 2 + x 2x 3x - x

16. f 1x2 = 4 +

3 x

3

3x 2

3

2

19. f 1x2 =

6x 4 + 4x 2 - 7

21. f 1x2 =

2x 3 - 4x + 1

2x 5 - x + 3 4x 3 + 2x - 3

18. f 1x2 =

3x - 1 x

33. f 1x2 = x +

Determine the horizontal asymptote of each function. If none exists, state that fact. 11. f 1x2 =

31. f 1x2 =

2

2 x

6x 3 + 4x 3x - x 2

20. f 1x2 =

4x 3 - 3x + 2

22. f 1x2 =

5x 4 - 2x 3 + x

x 3 + 2x - 4 x5 - x3 + 8

Sketch the graph of each function. Indicate where each function is increasing or decreasing, where any relative extrema occur, where asymptotes occur, where the graph is concave up or concave down, where any points of inflection occur, and where any intercepts occur.

2 x

32. f 1x2 =

34. f 1x2 = x +

35. f 1x2 =

-1

37. f 1x2 =

x x + 2

38. f 1x2 =

-1

40. f 1x2 =

39. f 1x2 =

36. f 1x2 =

x2

x2 + 2

2x + 1 x 9 x

2 x2 x x - 3 1 x2 + 3

41. f 1x2 =

x + 3 (Hint: x 2 - 9 Simplify.)

42. f 1x2 =

43. f 1x2 =

x - 1 x + 2

44. f 1x2 =

x - 2 x + 1

45. f 1x2 =

x2 - 4 x + 3

46. f 1x2 =

x2 - 9 x + 1

47. f 1x2 = 49. f 1x2 = 51. f 1x2 =

x + 1 x 2 - 2x - 3 2x 2 x - 16 2

1 x - 1 2

48. f 1x2 = 50. f 1x2 = 52. f 1x2 =

53. f 1x2 =

x + 1 x

54. f 1x2 =

55. f 1x2 =

x2 - 9 x - 3

56. f 1x2 =

2

x - 1 x2 - 1

x - 3 x 2 + 2x - 15 x2 + x - 2 2x 2 - 2 10 x + 4 2

x3 x2 - 1 x 2 - 16 x + 4

In Exercises 57–62, determine a rational function that meets the given conditions, and sketch its graph. 57. The function f has a vertical asymptote at x = 2, a horizontal asymptote at y = - 2, and f 102 = 0. 58. The function f has a vertical asymptote at x = 0, a horizontal asymptote at y = 3, and f 112 = 2.

5 x

24. f 1x2 =

4 x

25. f 1x2 =

1 x - 5

26. f 1x2 =

-2 x - 5

27. f 1x2 =

1 x + 2

28. f 1x2 =

1 x - 3

61. The function h has vertical asymptotes at x = - 3 and x = 2, a horizontal asymptote at y = 0, and h112 = 2.

29. f 1x2 =

-3 x - 3

30. f 1x2 =

-2 x + 5

62. The function h has vertical asymptotes at x = - 12 and x = 12, a horizontal asymptote at y = 0, and h102 = - 3.

23. f 1x2 = -

59. The function g has vertical asymptotes at x = - 1 and x = 1, a horizontal asymptote at y = 1, and g102 = 2. 60. The function g has vertical asymptotes at x = - 2 and x = 0, a horizontal asymptote at y = - 3, and g112 = 4.

CHAPTER 2

•

Applications of Differentiation

APPLICATIONS

b) When was the purchasing power $0.50? c) Find lim P1x2. x: q

Business and Economics 63. Depreciation. Suppose that the value V of the inventory at Fido’s Pet Supply decreases, or depreciates, with time t, in months, where V1t2 = 50 -

25t 2

. 1t + 222

a) Find V102, V152, V1102, and V1702. b) Find the maximum value of the inventory over the interval 30, q 2. c) Sketch a graph of V. d) Does there seem to be a value below which V1t2 will never fall? Explain. 64. Average cost. The total-cost function for Acme, Inc., to produce x units of a product is given by C1x2 = 3x 2 + 80. a) The average cost is given by A1x2 = C1x2>x. Find A1x2. b) Graph the average cost. c) Find the slant asymptote for the graph of y = A1x2, and interpret its significance. 65. Cost of pollution control. Cities and companies find that the cost of pollution control increases along with the percentage of pollutants to be removed in a situation. Suppose that the cost C, in dollars, of removing p% of the pollutants from a chemical spill is given by C1p2 = a) b) c) d)

48,000 . 100 - p

Find C102, C1202, C1802, and C1902. Find the domain of C. Sketch a graph of C. Can the company or city afford to remove 100% of the pollutants due to this spill? Explain.

66. Total cost and revenue. The total cost and total revenue, in dollars, from producing x couches are given by C1x2 = 5000 + 600x

and R1x2 =

- 12 x 2

+ 1000x.

a) Find the total-profit function, P1x2. b) The average profit is given by A1x2 = P1x2>x. Find A1x2. c) Graph the average profit. d) Find the slant asymptote for the graph of y = A1x2. 67. Purchasing power. Since 1970, the purchasing power of the dollar, as measured by consumer prices, can be modeled by the function P1x2 =

2.632 , 1 + 0.116x

where x is the number of years since 1970. (Source: U.S. Bureau of Economic Analysis.) a) Find P1102, P1202, and P1402.

y 3.0 Purchasing power (in dollars)

248

2.5 2.0 2.632 P(x) = ——— —— 1 + 0.116x

1.5 1.0 0.5 0

5

10

15 20 25 Years since 1970

30

x

Life and Physical Sciences 68. Medication in the bloodstream. After an injection, the amount of a medication A, in cubic centimeters (cc), in the bloodstream decreases with time t, in hours. Suppose that under certain conditions A is given by A0 A1t2 = 2 , t + 1 where A0 is the initial amount of the medication. Assume that an initial amount of 100 cc is injected. a) Find A102, A112, A122, A172, and A1102. b) Find the maximum amount of medication in the bloodstream over the interval 30, q 2. c) Sketch a graph of the function. d) According to this function, does the medication ever completely leave the bloodstream? Explain your answer.

General Interest 69. Baseball: earned-run average. A pitcher’s earned-run average (the average number of runs given up every 9 innings, or 1 game) is given by r E = 9# , n where r is the number of earned runs allowed in n innings. Suppose that we fix the number of earned runs allowed at 4 and let n vary. We get a function given by 4 E1n2 = 9 # . n a) Complete the following table, rounding to two decimal places: Innings Pitched, n Earned-Run Average, E

9

6

3

1

2 3

1 3

Exercise Set 2.3

b) The number of innings pitched n is equivalent to the number of outs that a pitcher is able to get while pitching, divided by 3. For example, if the pitcher gets just 1 out, he is credited with pitching 13 of an inning. Find lim E1n2. Under what circumstances

a) Find all the x-intercepts. b) Find the y-intercept. c) Find all the asymptotes. 87. Graph the function given by

n:0

might this limit be plausible? c) Suppose a pitcher gives up 4 earned runs over two complete games, or 18 innings. Calculate the pitcher’s earned-run average, and interpret this result.

249

f 1x2 =

2x 2 + 3x + 2 . x - 3

a) Estimate lim f 1x2 and lim f 1x2 using the graph x: q x: - q and input–output tables as needed to refine your estimates. b) Describe the outputs of the function over the interval 1- 2, - 12. c) What appears to be the domain of the function? Explain. d) Find lim - f 1x2 and lim + f 1x2. x: - 2

x: - 1

88. Not all asymptotes are linear. Use long division to find an equation for the nonlinear asymptote that is approached by the graph of f 1x2 =

x5 + x - 9 x 3 + 6x

.

Then graph the function and its asymptote. While pitching for the St. Louis Cardinals in 1968, Bob Gibson had an earned-run average of 1.12, a record low.

89. Refer to Fig. 1 on p. 235. The function is given by f 1x2 =

SYNTHESIS 70. Explain why a vertical asymptote is only a guide and is not part of the graph of a function. 71. Using graphs and limits, explain the idea of an asymptote to the graph of a function. Describe three types of asymptotes. Find each limit, if it exists. 72. 74. 76. 78.

lim

x: - q

lim

x: - 2

- 3x 2 + 5 2 - x

x3 + 8

x: - q

lim

- 6x 3 + 7x

x: q

2

- 6x 3 + 7x 2x - 3x - 10 2

7x 5 + x - 9 6x + x

x: - q

x:0

75. lim

x - 4

lim

ƒxƒ x

73. lim

3

77. lim

x:1

79.

2x 2 - 3x - 10

x3 - 1

2x 4 + x x: - q x + 1 lim

x

2

82. f 1x2 =

x 3 + 4x 2 + x - 6

84. f 1x2 =

x 3 + 2x 2 - 3x

x - x - 2 2

x 2 - 25

86. Graph the function f 1x2 =

x2 - 3 . 2x - 4

.

a) Inspect the graph and estimate the coordinates of any extrema. b) Find f ¿ and use it to determine the critical values. (Hint: you will need the quadratic formula.) Round the x-values to the nearest hundredth. c) Graph this function in the window 30, 0.2, 0.16, 0.174. Use TRACE or MAXIMUM to confirm your results from part (b). d) Graph this function in the window 39.8, 10, 0.9519, 0.951954. Use TRACE or MINIMUM to confirm your results from part (b). e) How close were your estimates of part (a)? Would you have been able to identify the relative minimum point without calculus techniques?

81. f 1x2 = 83. f 1x2 =

Answers to Quick Checks 1. The lines x = 0, x = 4, and x = - 4 are vertical asymptotes. 2. The line x = 0 is not a vertical asymptote because lim f 1x2 = 2. There is a deleted point discontinuity at x = 0. x:0

Graph each function using a calculator, iPlot, or Graphicus. 1

x + x - 6

x2 - 1

TECHNOLOGY CONNECTION 80. f 1x2 = x 2 +

x2 - 1 2

x 2x + 1 2

x 3 + 2x 2 - 15x

85. f 1x2 = 2

2 3. The line y = 15 is a horizontal asymptote. 4. The line y = 2x + 7 is a slant asymptote. 5. x-intercepts: 11, 02, 1- 1, 02, 10, 02; y-intercept: 10, 02 3x 2 + 6 y 6. 7. g1x2 = 2 x - 4 x=1 10

x - 5x - 14

5

2

1 - 22 x

−10

−5

5 −5

y=x+1

−10

f(x) =

10

x

x2 − 9 x−1

250

CHAPTER 2

•

Applications of Differentiation

2.4 OBJECTIVES • Find absolute extrema using Maximum–Minimum Principle 1. • Find absolute extrema using Maximum–Minimum Principle 2.

Using Derivatives to Find Absolute Maximum and Minimum Values An extremum may be at the highest or lowest point for a function’s entire graph, in which case it is called an absolute extremum. For example, the parabola given by f1x2 = x 2 has a relative minimum at 10, 02. This is also the lowest point for the entire graph of f, so it is also called the absolute minimum. Relative extrema are useful for graph sketching and understanding the behavior of a function. In many applications, however, we are more concerned with absolute extrema.

Absolute Maximum and Minimum Values A relative minimum may or may not be an absolute minimum, meaning the smallest value of the function over its entire domain. Similarly, a relative maximum may or may not be an absolute maximum, meaning the greatest value of a function over its entire domain. The function in the following graph has relative minima at interior points c1 and c3 of the closed interval 3a, b4. y Absolute maximum at b

Relative maximum at c2

Maxima

Relative minima at c1 and c3

Absolute minimum at c1 Minima

a

c1

c2

c3

b

x

The relative minimum at c1 is also the absolute minimum. On the other hand, the relative maximum at c2 is not the absolute maximum. The absolute maximum occurs at the endpoint b.

DEFINITION Suppose that f is a function with domain I. f1c2 is an absolute minimum if f1c2 … f1x2 for all x in I. f1c2 is an absolute maximum if f1c2 Ú f1x2 for all x in I.

Finding Absolute Maximum and Minimum Values over Closed Intervals We first consider a continuous function for which the domain is a closed interval. Look at the graphs in Figs. 1 and 2 and try to determine where the absolute maxima and minima (extrema) occur for each interval.

2.4

•

Using Derivatives to Find Absolute Maximum and Minimum Values

y

251

y

a

c1

c2

b

x

FIGURE 1

a

c1

c2

b

x

FIGURE 2

Note that each of the functions does indeed have an absolute maximum value and an absolute minimum value. This leads us to the following theorem.

THEOREM

The Extreme-Value Theorem A continuous function f defined over a closed interval 3a, b4 must have an absolute maximum value and an absolute minimum value over 3a, b4.

Look again at the graphs in Figs. 1 and 2 and consider the critical values and the endpoints. In Fig. 1, the graph starts at f1a2 and falls to f1c12. Then it rises from f1c12 to f1c22. From there, it falls to f1b2. In Fig. 2, the graph starts at f(a) and rises to f1c12. Then it falls from f1c12 to f1c22. From there, it rises to f(b). It seems reasonable that whatever the maximum and minimum values are, they occur among the function values f1a2, f1c12, f1c22, and f1b2. This leads us to a procedure for determining absolute extrema.

THEOREM 8

Maximum–Minimum Principle 1 Suppose that f is a continuous function defined over a closed interval 3a, b4. To find the absolute maximum and minimum values over 3a, b4: a) First find f ¿1x2. b) Then determine all critical values in 3a, b4. That is, find all c in 3a, b4 for which f ¿1c2 = 0 or f ¿1c2 does not exist. c) List the values from step (b) and the endpoints of the interval: a, c1, c2, Á , cn, b. d) Evaluate f1x2 for each value in step (c): f1a2, f1c12, f1c22, Á , f1cn2, f1b2.

The largest of these is the absolute maximum of f over 7a, b8. The smallest of these is the absolute minimum of f over 7a, b8.

A reminder: endpoints of a closed interval can be absolute extrema but not relative extrema.

252

•

CHAPTER 2

Applications of Differentiation

■ EXAMPLE 1 Find the absolute maximum and minimum values of f1x2 = x 3 - 3x + 2

over the interval C - 2, 32 D .

Solution Keep in mind that we are considering only the interval C - 2, 32 D . a) Find f ¿1x2: f ¿1x2 = 3x 2 - 3. b) Find the critical values. The derivative exists for all real numbers. Thus, we merely solve f ¿1x2 = 0: 3x 2 - 3 3x 2 x2 x

f(x) = x3 − 3x + 2

y Maximum = 4, occurs at x = −1

−3

−2

−1

1

3 2

2

3

0 3 1 ;1.

c) List the critical values and the endpoints: - 2, - 1, 1, and 32. d) Evaluate f for each value in step (c): f1- 22 = 1- 223 - 31- 22 + 2 = - 8 + 6 + 2 = 0; f1- 12 = 1- 123 - 31- 12 + 2 = - 1 + 3 + 2 = 4; f112 = 1123 - 3112 + 2 = 1- 3 + 2 = 0;

5 4 3 2 1 −1 −2 −3 −4 −5

= = = =

f A 32 B =

x

A 32 B 3 - 3 A 32 B + 2 =

27 8

-

9 2

+ 2 =

Minimum Maximum Minimum

7 8

The largest of these values, 4, is the maximum. It occurs at x = - 1. The smallest of these values is 0. It occurs twice: at x = - 2 and x = 1. Thus, over the interval C - 2, 32 D , the

Minimum = 0, occurs at x = −2 and x = 1

absolute maximum = 4 at x = - 1

A visualization of Example 1

and the absolute minimum = 0 at x = - 2 and x = 1.

Note that an absolute maximum or minimum value can occur at more than one point.

TECHNOLOGY CONNECTION

Finding Absolute Extrema To find the absolute extrema of Example 1, we can use any of the methods described in the Technology Connection on pp. 209–210. In this case, we adapt Methods 3 and 4.

Next, we check function values at these x-values and at the endpoints, using Maximum–Minimum Principle 1 to determine the absolute maximum and minimum values over C - 2, 32 D . X –2 –1 1 1.5

Method 3 Method 3 is selected because there are relative extrema in the interval C - 2, 32 D . This method gives us approximations for the relative extrema. y = x3 – 3x + 2

Maximum X = –1.000002

y = x 3 – 3x + 2

Y=4

Minimum X = .99999689

Y1 0 4 0 .875

Min Max Min

X=

Method 4

Y = 2.9E–11

Example 2 considers the same function as in Example 1, but over a different interval. Because there are no relative extrema, we can use fMax and fMin features from the MATH menu. The minimum and maximum values occur at the endpoints, as the following graphs show. (continued)

2.4 fMin(Y1, X, –3, –1.5)

fMax(Y1, X, –3, –1.5)

–2.999994692

Y1(Ans)

4 2 −2 −4 −6 −8 −10 −12 −14 −16 −18

1

2

3

253

1. Use a graph to estimate the absolute maximum and

minimum values of f 1x2 = x 3 - x 2 - x + 2, first over the interval 3- 2, 14 and then over the interval 3 - 1, 24. Then check your work using the methods of Examples 1 and 2.

3.124979532

1 Maximum = 3−, 8 occurs at y f(x) = x3 − 3x + 2 x = −−32

−3 −2− 3 −1

Using Derivatives to Find Absolute Maximum and Minimum Values

EXERCISE –1.500005458

Y1(Ans) –15.99987261

2

•

■ EXAMPLE 2 Find the absolute maximum and minimum values of f1x2 = x 3 - 3x + 2

x

over the interval C - 3, - 32 D .

Solution As in Example 1, the derivative is 0 at - 1 and 1. But neither - 1 nor 1 is in the interval C - 3, - 32 D , so there are no critical values in this interval. Thus, the maximum and minimum values occur at the endpoints: f1- 32 = 1- 323 - 31- 32 + 2 = - 27 + 9 + 2 = - 16; 3 3 3 3 f a - b = a - b - 3a - b + 2 2 2 2 27 9 25 1 = + + 2 = = 3 . 8 2 8 8

Minimum = −16, occurs at x = −3

A visualization of Example 2

Quick Check 1 Find the absolute maximum and minimum values of the function given in Example 2 over the interval 30, 34.

Minimum

Maximum

Thus, the absolute maximum over the interval C - 3, - 32 D , is 3 18, which occurs at x = - 32, and the absolute minimum over C - 3, - 32 D is - 16, which occurs at x = - 3.

O Quick Check 1 Finding Absolute Maximum and Minimum Values over Other Intervals When there is only one critical value c in I, we may not need to check endpoint values to determine whether the function has an absolute maximum or minimum value at that point.

THEOREM 9

Maximum–Minimum Principle 2

Suppose that f is a function such that f ¿1x2 exists for every x in an interval I and that there is exactly one (critical) value c in I, for which f ¿1c2 = 0. Then f1c2 is the absolute maximum value over I if f –1c2 6 0 or f1c2 is the absolute minimum value over I if f –1c2 7 0.

Theorem 9 holds no matter what the interval I is—whether open, closed, or infinite in length. If f –1c2 = 0, either we must use Maximum–Minimum Principle 1 or we must know more about the behavior of the function over the given interval.

O

254

CHAPTER 2

•

Applications of Differentiation

TECHNOLOGY CONNECTION

■ EXAMPLE 3 Find the absolute maximum and minimum values of f1x2 = 4x - x 2.

Finding Absolute Extrema Let’s do Example 3 graphically, by adapting Methods 1 and 2 of the Technology Connection on pp. 209–210. Strictly speaking, we cannot use the fMin or fMax options of the MATH menu or the MAXIMUM or MINIMUM options from the CALC menu since we do not have a closed interval.

Solution When no interval is specified, we consider the entire domain of the function. In this case, the domain is the set of all real numbers. a) Find f ¿1x2: f ¿1x2 = 4 - 2x. b) Find the critical values. The derivative exists for all real numbers. Thus, we merely solve f ¿1x2 = 0: 4 - 2x = 0 - 2x = - 4 x = 2.

Methods 1 and 2 We create a graph, examine its shape, and use TRACE and/or TABLE. This procedure leads us to see that there is indeed no absolute minimum. We do find an absolute maximum: f 1x2 = 4 at x = 2.

EXERCISE

c) Since there is only one critical value, we can apply Maximum–Minimum Principle 2 using the second derivative: f –1x2 = - 2. The second derivative is constant. Thus, f –122 = - 2, and since this is negative, we have the absolute maximum: f122 = 4 # 2 - 2 2, = 8 - 4 = 4 at x = 2.

1. Use a graph to estimate

the absolute maximum and minimum values of f 1x2 = x 2 - 4x. Then check your work using the method of Example 3.

The function has no minimum, as the graph, shown below, indicates. y 5 Max = 4

f(x) = 4x – x 2

3 2 1 – 5 – 4 –3 –2 –1 –1

1

2

3

4

5 x

■ EXAMPLE 4 Find the absolute maximum and minimum values of f1x2 = 4x - x 2 over the interval 31, 44.

Solution By the reasoning in Example 3, we know that the absolute maximum of f on 1- q , q 2 is f122, or 4. Since 2 is in the interval 31, 44, we know that the absolute maximum of f over 31, 44 will occur at 2. To find the absolute minimum, we need to check the endpoints: f112 = 4 # 1 - 12 = 3

and

4

f142 = 4 # 4 - 4 = 0. 2

Quick Check 2 Find the absolute maximum and minimum values of f1x2 = x 2 - 10x over each interval: a) 30, 64; b) 34, 104.

y

We see from the graph that the minimum is 0. It occurs at x = 4. Thus, the absolute maximum = 4 at x = 2, and the absolute minimum = 0 at x = 4.

O Quick Check 2

Maximum = 4 at x = 2 f(x) = 4x – x 2 over [1, 4]

3 2 1 0

1

2

3

4

Minimum = 0 at x = 4 x

O

2.4

•

Using Derivatives to Find Absolute Maximum and Minimum Values

255

A Strategy for Finding Absolute Maximum and Minimum Values The following general strategy can be used when finding absolute maximum and minimum values of continuous functions.

A Strategy for Finding Absolute Maximum and Minimum Values To find absolute maximum and minimum values of a continuous function over an interval: a) Find f ¿1x2. b) Find the critical values. c) If the interval is closed and there is more than one critical value, use Maximum–Minimum Principle 1. d) If the interval is closed and there is exactly one critical value, use either Maximum–Minimum Principle 1 or Maximum–Minimum Principle 2. If it is easy to find f –1x2, use Maximum–Minimum Principle 2. e) If the interval is not closed, such as 1- q , q 2, 10, q 2, or 1a, b2, and the function has only one critical value, use Maximum–Minimum Principle 2. In such a case, if the function has a maximum, it will have no minimum; and if it has a minimum, it will have no maximum. Finding absolute maximum and minimum values when more than one critical value occurs in an interval that is not closed, such as any of those listed in step (e) above, requires a detailed graph or techniques beyond the scope of this book.

■ EXAMPLE 5 Find the absolute maximum and minimum values of f1x2 = 1x - 223 + 1.

Solution

y 10 9 8 7 6 5 4 3 2 1 –1

–1

a) Find f ¿1x2. f ¿1x2 = 31x - 222.

f(x) = (x – 2)3 + 1

b) Find the critical values. The derivative exists for all real numbers. Thus, we solve f ¿1x2 = 0:

x=2

4

A visualization of Example 5

x

31x - 222 1x - 222 x - 2 x

= = = =

0 0 0 2.

c) Since there is only one critical value and there are no endpoints, we can try to apply Maximum–Minimum Principle 2 using the second derivative: f –1x2 = 61x - 22. We have f –122 = 612 - 22 = 0,

Quick Check 3 Let f1x2 = x n, where n is a positive odd integer. Explain why functions of this form never have an absolute minimum or maximum.

so Maximum–Minimum Principle 2 does not apply. We cannot use Maximum– Minimum Principle 1 because there are no endpoints. But note that f ¿1x2 = 31x - 222 is never negative. Thus, f(x) is increasing everywhere except at x = 2, so there is no maximum and no minimum. For x 6 2, say x = 1, we have f –112 = - 6 6 0. For x 7 2, say x = 3, we have f –132 = 6 7 0. Thus, at x = 2, the function has a point of inflection.

O Quick Check 3

O

256

CHAPTER 2

•

Applications of Differentiation

■ EXAMPLE 6 Find the absolute maximum and minimum values of f1x2 = 5x +

35 x

over the interval 10, q 2.

Solution a) Find f ¿1x2. We first express f1x2 as f1x2 = 5x + 35x -1. Then f ¿1x2 = 5 - 35x -2 35 = 5 - 2. x

b) Find the critical values. Since f ¿1x2 exists for all values of x in 10, q 2, the only critical values are those for which f ¿1x2 = 0: 5 -

35 = 0 x2 5 =

35 x2

5x 2 = 35 Multiplying both sides by x 2, since x Z 0 x2 = 7 x = ; 27 L ;2.646. TECHNOLOGY CONNECTION

Finding Absolute Extrema Let’s do Example 6 using MAXIMUM and MINIMUM from the CALC menu. The shape of the graph leads us to see that there is no absolute maximum, but there is an absolute minimum. Y1 = 5X + (35/ X )

Minimum X = 2.6457515

c) The interval is not closed and is 10, q 2. The only critical value is 27. Therefore, we can apply Maximum–Minimum Principle 2 using the second derivative, f –1x2 = 70x -3 =

70 , x3

to determine whether we have a maximum or a minimum. Since f – A 27 B = Y = 26.457513

[0, 10, 0, 50]

Note that 27 L 2.6458 and 1027 L 26.458,

70

A 27 B 3

7 0,

an absolute minimum occurs at x = 27: Absolute minimum = f A 27 B

= 5 # 27 +

which confirms the analytic solution.

EXERCISE 1. Use a graph to estimate the absolute maximum and mini-

mum values of f 1x2 = 10x + 1>x over the interval 10, q 2. Then check your work using the analytic method of Example 6.

= 527 + = 527 +

35 27 35

# 27

27 27 3527 7

= 527 + 527 = 1027 L 26.458 at x = 27.

Exercise Set 2.4

257

The function has no maximum value, which can happen since the interval 10, q 2 is not closed. y 40 35 30 Min = 10√7 f(x) = 5x + 35 — x

20

Quick Check 4

15

O

Find the absolute maximum and minimum values of

10 5

2x 2 + 18 g1x2 = x

0

over the interval 10, q 2.

1

√7

4

5

6

x

O Quick Check 4

Section Summary • An absolute minimum of a function f is a value f1c2 such that f1c2 … f1x2 for all x in the domain of f. • An absolute maximum of a function f is a value f1c2 such that f1c2 Ú f1x2 for all x in the domain of f. • If the domain of f is a closed interval and f is continuous over that domain, then the Extreme–Value Theorem guarantees the existence of both an absolute minimum and an absolute maximum.

• Endpoints of a closed interval may be absolute extrema, but not relative extrema. • If there is exactly one critical value c such that f ¿1c2 = 0 in the domain of f, then Maximum–Minimum Principle 2 may be used. Otherwise, Maximum–Minimum Principle 1 has to be used.

EXERCISE SET

2.4 1. Fuel economy. According to the U.S. Department of Energy, a vehicle’s fuel economy, in miles per gallon (mpg), decreases rapidly for speeds over 60 mph.

Find the absolute maximum and minimum values of each function over the indicated interval, and indicate the x-values at which they occur.

35

Fuel economy (mpg)

2. Fuel economy. Using the graph in Exercise 1, estimate the absolute maximum and the absolute minimum fuel economy over the interval 330, 704.

30

3. f 1x2 = 5 + x - x 2; 30, 24

25 20

y 6 5 4 3 2 1

15 10 5 0 5

15

25

35

45

55

65

75

Speed (mph) ( Sources: U.S. Dept. of Energy and a study by West, B.H., McGill, R.N., Hodgson, J.W., Sluder, S.S., and Smith, D.E., Oak Ridge National Laboratory, 1999.)

a) Estimate the speed at which the absolute maximum gasoline mileage is obtained. b) Estimate the speed at which the absolute minimum gasoline mileage is obtained. c) What is the mileage obtained at 70 mph?

–3 –2 –1

–1

1

2

3

4

x

258

•

CHAPTER 2

Applications of Differentiation

4. f 1x2 = 4 + x - x 2; 30, 24

18. f 1x2 = 1 + 6x - 3x 2; 30, 44 19. f 1x2 = x 3 - 3x 2; 30, 54

y

20. f 1x2 = x 3 - 3x + 6; 3- 1, 34

5 4 3 2 1 –3 –2 –1

21. f 1x2 = x 3 - 3x; 3- 5, 14

22. f 1x2 = 3x 2 - 2x 3; 3- 5, 14 1

–1 –2

3

2

23. f 1x2 = 1 - x 3; 3- 8, 84

x

4

24. f 1x2 = 2x 3; 3- 10, 104

25. f 1x2 = 12 + 9x - 3x 2 - x 3; 3- 3, 14

5. f 1x2 = x 3 - x 2 - x + 2; 3- 1, 24

26. f 1x2 = x 3 - 6x 2 + 10; 30, 44 27. f 1x2 = x 4 - 2x 3; 3- 2, 24

y

28. f 1x2 = x 3 - x 4; 3- 1, 14

5 4

29. f 1x2 = x 4 - 2x 2 + 5; 3- 2, 24

3

30. f 1x2 = x 4 - 8x 2 + 3; 3- 3, 34

2 1 −3

−2

−1

−1

1

2

31. f 1x2 = 1x + 322>3 - 5; 3- 4, 54

x

3

32. f 1x2 = 1 - x 2>3; 3- 8, 84

−2

6. f 1x2 = x 3 - 12 x 2 - 2x + 5; 3- 2, 14 y 7

33. f 1x2 = x +

1 ; 31, 204 x

34. f 1x2 = x +

4 ; 3- 8, - 14 x

35. f 1x2 =

6 5

x2 x2 + 1

; 3- 2, 24

4

y

3

0.6

2

0.4

1 −2

−1

−1

1

2

0.2

x –2

7. f 1x2 = x 3 - x 2 - x + 3; 3- 1, 04 8. f 1x2 = x + 3

1 2 2x

- 2x + 4; 3- 2, 04

36. f 1x2 =

4x x2 + 1

–1

, 3- 3, 34

9. f 1x2 = 5x - 7; 3 - 2, 34

y

10. f 1x2 = 2x + 4; 3 - 1, 14

2 1

11. f 1x2 = 7 - 4x; 3 - 2, 54

–3 – 2 –1

12. f 1x2 = - 2 - 3x; 3- 10, 104

15. f 1x2 = x - 6x - 3; 3- 1, 54 2

16. f 1x2 = x - 4x + 5; 3- 1, 34 2

17. f 1x2 = 3 - 2x - 5x 2; 3- 3, 34

1

2

3 x

–1

13. f 1x2 = - 5; 3- 1, 14 14. g1x2 = 24; 34, 134

2 x

1

–2

37. f 1x2 = 1x + 121>3; 3- 2, 264 38. f 1x2 = 2x; 38, 644 3

39–48. Check Exercises 3, 5, 9, 13, 19, 23, 33, 35, 37, and 38 with a graphing calculator.

259

Exercise Set 2.4

Find the absolute maximum and minimum values of each function, if they exist, over the indicated interval. Also indicate the x-value at which each extremum occurs. When no interval is specified, use the real line, 1 ˆ , ˆ 2. 49. f 1x2 = 12x - x

2

51. f 1x2 = 2x - 40x + 270 2

52. f 1x2 = 2x 2 - 20x + 340

53. f 1x2 = x - 43 x 3; 10, q 2

54. f 1x2 = 16x - 43 x 3; 10, q 2 55. f 1x2 = x160 - x2

58. f 1x2 =

- 3x; 3- 2, 24 - 5x; 3- 3, 34

1 3 3x

59. f 1x2 = - 0.001x + 4.8x - 60 2

60. f 1x2 = - 0.01x 2 + 1.4x - 30 61. f 1x2 =

+ 6x - 11x - 50; 10, 32

- 13 x 3

2

62. f 1x2 = - x + x 2 + 5x - 1; 10, q 2 3

63. f 1x2 = 15x 2

64. f 1x2 = 4x 65. f 1x2 = 2x +

1 3 2x ;

1 3 2x ;

2

71. f 1x2 = 2x; 30, 84 3

73. f 1x2 = 1x + 123 74. f 1x2 = 1x - 123

75. f 1x2 = 2x - 3; 3- 1, 14

76. f 1x2 = 9 - 5x; 3- 10, 104 77. f 1x2 = 2x - 3; 3- 1, 52 78. f 1x2 = 9 - 5x; 3- 2, 32

56. f 1x2 = x125 - x2 57. f 1x2 =

70. f 1x2 = 2x 4 + x; 3- 1, 14 72. f 1x2 = 2x; 30, 44

50. f 1x2 = 30x - x 2

1 3 3x

69. f 1x2 = 2x 4 - x; 3- 1, 14

30, 304

79. f 1x2 = x 2>3; 3- 1, 14 80. g1x2 = x 2>3

81. f 1x2 = 13 x 3 - x +

2 3

82. f 1x2 = 13 x 3 - 12 x 2 - 2x + 1

83. f 1x2 = 13 x 3 - 2x 2 + x; 30, 44

84. g1x2 = 13 x 3 + 2x 2 + x; 3- 4, 04 y

85. t1x2 = x 4 - 2x 2

2

30, 84

1

72 ; 10, q 2 x

66. f 1x2 = x +

3600 ; 10, q 2 x

67. f 1x2 = x 2 +

432 ; 10, q 2 x

–2

–1

1

x

2

–1 –2

86. f 1x2 = 2x 4 - 4x 2 + 2

y 4

y 3

700 600

2

500

1

400 –2

300

–1

1

2

x

200

87–96. Check Exercises 49, 51, 53, 57, 61, 65, 67, 69, 73, and 85 with a graphing calculator.

100 5

68. f 1x2 = x 2 +

10 15 20 25 30 x

250 ; 10, q 2 x y

APPLICATIONS Business and Economics 97. Monthly productivity. An employee’s monthly productivity M, in number of units produced, is found to be a function of t, the number of years of service. For a certain product, a productivity function is given by

800 600 400

M1t2 = - 2t 2 + 100t + 180, 0 … t … 40.

200 5

10 15 20 25 30 x

Find the maximum productivity and the year in which it is achieved.

260

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98. Advertising. Sound Software estimates that it will sell N units of a program after spending a dollars on advertising, where N1a2 = - a 2 + 300a + 6, 0 … a … 300, and a is in thousands of dollars. Find the maximum number of units that can be sold and the amount that must be spent on advertising in order to achieve that maximum.

where P(t) is the number of barrels, in billions, produced in a year, t years after 1950. (Source: Beyond Oil, by Kenneth S. Deffeyes, p. xii, Hill and Wang, New York, 2005.) According to this model, in what year did worldwide oil production achieve an absolute maximum? What was that maximum? (Hint: Do not solve P¿1t2 = 0 algebraically.)

99. Small business. The percentage of the U.S. national income generated by nonfarm proprietors may be modeled by 13x 3 - 240x 2 - 2460x + 585,000 , 75,000

p1x2 =

where x is the number of years since 1970. (Source: U.S. Census Bureau.) According to this model, in what year from 1970 through 2000 was this percentage a minimum? Calculate the answer, and then check it on the graph. Percentage of U.S. national income generated by nonfarm proprietors

y 8.6 8.4

13x 3 – 240x 2 – 2460x + 585,000 p(x) = —————————————— 75,000

8.2

102. Maximizing profit. Corner Stone Electronics determines that its total weekly profit, in dollars, from the production and sale of x amplifiers is given by

8.0 7.8 7.6

P1x2 =

7.4 7.2 0

5

10 15 20 25 Number of years since 1970

30

x

100. The percentage of the U.S. civilian labor force aged 35–44 may be modeled by f 1x2 = - 0.029x 2 + 0.928x + 19.103,

where x is the number of years since 1980. (Source: U.S. Census Bureau.) According to this model, in what year from 1980 through 2010 was this percentage a maximum? Calculate the answer, and then check it on the graph.

Percentage of U.S. civilian labor force aged 35-44 years

y

1500 x 2 - 6x + 10

.

Find the number of amplifiers, x, for which the total weekly profit is a maximum. Maximizing profit. The total-cost and total-revenue functions for producing x items are 1 C1x2  5000  600x and R1x2   x 2  1000x, 2 where 0 ◊ x ◊ 600. Use these functions for Exercises 103 and 104. 103. a) Find the total-profit function P1x2. b) Find the number of items, x, for which the total profit is a maximum.

24

104. a) The average profit is given by A1x2 = P1x2>x. Find A(x). b) Find the number of items, x, for which the average profit is a maximum.

23

Life and Physical Sciences

26 25

22 21

105. Blood pressure. For a dosage of x cubic centimeters (cc) of a certain drug, the resulting blood pressure B is approximated by

f(x) = –0.029x 2 + 0.928x + 19.103

20 0

B1x2 = 305x 2 - 1830x 3, 0 … x … 0.16. 5

10 15 20 25 Number of years since 1980

30 x

101. Worldwide oil production. One model of worldwide oil production is the function given by P1t2 = 0.000008533t 4 - 0.001685t 3 + 0.090t 2 - 0.687t + 4.00, 0 … t … 90,

Find the maximum blood pressure and the dosage at which it occurs.

SYNTHESIS 106. Explain the usefulness of the second derivative in finding the absolute extrema of a function.

Exercise Set 2.4

For Exercises 107–110, find the absolute maximum and minimum values of each function, and sketch the graph. 107. f 1x2 = b

2x + 1 for - 3 … x … 1, 4 - x 2, for 1 6 x … 2

113. g1x2 = x2x + 3; 3- 3, 34

115. Business: total cost. Certain costs in a business environment can be separated into two components: those that increase with volume and those that decrease with volume. For example, customer service becomes more expensive as its quality increases, but part of the increased cost is offset by fewer customer complaints. A firm has determined that its cost of service, C1x2, in thousands of dollars, is modeled by

1 - x 2, for - 4 … x 6 0, 109. h1x2 = c 1 - x, for 0 … x 6 1, x - 1, for 1 … x … 2 x 2 + 4, for - 2 … x 6 0, 110. F1x2 = c 4 - x, for 0 … x 6 3, 2x - 2, for 3 … x … 67

C1x2 = 12x + 42 + a

111. Consider the piecewise-defined function f defined by: x 2 + 2, for - 2 … x … 0, for 0 6 x 6 4, f 1x2 = c 2, x - 2, for 4 … x … 6.

2 b, x 7 6, x - 6

where x represents the number of “quality units.” Find the number of “quality units” that the firm should use in order to minimize its total cost of service. 116. Let

a) Sketch its graph. b) Identify the absolute maximum. c) How would you describe the absolute minimum? 112. Physical science: dry lake elevation. Dry lakes are common in the Western deserts of the United States. These beds of ancient lakes are notable for having perfectly flat terrain. Rogers Dry Lake in California has been used as a landing site for space shuttle missions in recent years. The graph shows the elevation E, in feet, as a function of the distance x, in miles, from a point west 1x = 02 of Rogers Dry Lake to a point east of the dry lake. (Source: www.mytopo.com.) E

y = 1x - a22 + 1x - b22. For what value of x is y a minimum?

117. Explain the usefulness of the first derivative in finding the absolute extrema of a function.

TECHNOLOGY CONNECTION 118. Business: worldwide oil production. Refer to Exercise 101. In what year was worldwide oil production increasing most rapidly and at what rate was it increasing? 119. Business: U.S. oil production. One model of oil production in the United States is given by

2295 Elevation (in feet)

Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. 114. h1x2 = x21 - x; 30, 14

x 2, for - 2 … x … 0, 108. g1x2 = b 5x, for 0 6 x … 2

2290

P1t2 = 0.0000000219t 4 - 0.0000167t 3 + 0.00155t 2 + 0.002t + 0.22, 0 … t … 110,

2285 2280 2275 2270 2265 0

261

1

2

3 4 5 Distance (in miles)

6

x

where P(t) is the number of barrels of oil, in billions, produced in a year, t years after 1910. (Source: Beyond Oil, by Kenneth S. Deffeyes, p. 41, Hill and Wang, New York, 2005.) a) According to this model, what is the absolute maximum amount of oil produced in the United States and in what year did that production occur? b) According to this model, at what rate was United States oil production declining in 2004 and in 2010? Graph each function over the given interval. Visually estimate where absolute maximum and minimum values occur. Then use the TABLE feature to refine your estimate. 120. f 1x2 = x 2>31x - 52; 121. f 1x2 =

a) What is the maximum elevation? b) How would you describe the minimum elevation?

3 2 1x - 122>3; 4

122. f 1x2 = xa

4 x - 5b ; 2

31, 44

C 12, q B 

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123. Life and physical sciences: contractions during pregnancy. The following table and graph give the pressure of a pregnant woman’s contractions as a function of time. Time, t (in minutes)

Pressure (in millimeters of mercury)

0 1 2 3 4 5 6

10 8 9.5 15 12 14 14.5

2.5 OBJECTIVE • Solve maximum–minimum problems using calculus.

Use a calculator that has the REGRESSION option. a) Fit a linear equation to the data. Predict the pressure of the contractions after 7 min. b) Fit a quartic polynomial to the data. Predict the pressure of the contractions after 7 min. Find the smallest contraction over the interval 30, 104.

Answers to Quick Checks 1. Absolute maximum is 20 at x = 3; absolute minimum is 0 at x = 1. 2. (a) Absolute maximum is 0 at x = 0; absolute minimum is - 25 at x = 5. (b) Absolute maximum is 0 at x = 10; absolute minimum is - 25 at x = 5. 3. The derivative is f ¿1x2 = nx n - 1. If n is odd, n - 1 is even. Thus, nx n - 1 is always positive or zero, never negative. 4. No absolute maximum; absolute minimum is 12 at x = 3.

Maximum–Minimum Problems; Business and Economics Applications An important use of calculus is the solving of maximum–minimum problems, that is, finding the absolute maximum or minimum value of some varying quantity Q and the point at which that maximum or minimum occurs.

■ EXAMPLE 1 Maximizing Area. A hobby store has 20 ft of fencing to fence off a rectangular area for an electric train in one corner of its display room. The two sides up against the wall require no fence. What dimensions of the rectangle will maximize the area? What is the maximum area?

Solution At first glance, we might think that it does not matter what dimensions we use: They will all yield the same area. This is not the case. Let’s first make a drawing and express the area in terms of one variable. If we let x = the length, in feet, of one side and y = the length, in feet, of the other, then, since the sum of the lengths must be 20 ft, we have x + y = 20 and

y = 20 - x.

Thus, the area is given by A = xy = x120 - x2 = 20x - x 2.

20  x

x

•

Maximum–Minimum Problems; Business and Economics Applications

TECHNOLOGY CONNECTION

We are trying to find the maximum value of

EXERCISES 1. Complete this table, using a calculator as needed.

x

y = 20 - x

A = x120 - x2

0 4 6.5

A1x2 = 20x - x 2

Area (in square feet)

2.5

A 100

263

A(x) = 20x – x 2

80 60

40 over the interval 10, 202. We consider the interval 10, 202 20 because x is a length and 5 10 15 20 x cannot be negative or 0. Length (in feet) Since there is only 20 ft of fencing, x cannot be greater than 20. Also, x cannot be 20 because then the length of y would be 0.

a) We first find A¿1x2: A¿1x2 = 20 - 2x. b) This derivative exists for all values of x in 10, 202. Thus, the only critical values are where

8 10

A¿1x2 = 20 - 2x = 0 - 2x = - 20 x = 10.

12 13.2 20 2. Graph A1x2 = x120 - x2 over the interval 30, 204. 3. Estimate the maximum value, and state where it occurs.

Since there is only one critical value, we can use the second derivative to determine whether we have a maximum. Note that A–1x2 = - 2, which is a constant. Thus, A–1102 is negative, so A1102 is a maximum. Now

Quick Check 1 Repeat Example 1 starting with 50 ft of fencing, and again starting with 100 ft of fencing. Do you detect a pattern? If you had n feet of fencing, what would be the dimensions of the maximum area (in terms of n)?

A1102 = 10120 - 102 = 10 # 10 = 100. Thus, the maximum area of 100 ft2 is obtained using 10 ft for the length of one side and 20 - 10, or 10 ft for the other. Note that A152 = 75, A1162 = 64, and A1122 = 96; so length does affect area.

O Quick Check 1 Here is a general strategy for solving maximum–minimum problems. Although it may not guarantee success, it should certainly improve your chances.

A Strategy for Solving Maximum–Minimum Problems 1. Read the problem carefully. If relevant, make a drawing. 2. Make a list of appropriate variables and constants, noting what varies, what stays fixed, and what units are used. Label the measurements on your drawing, if one exists. 3. Translate the problem to an equation involving a quantity Q to be maximized or minimized. Try to represent Q in terms of the variables of step 2. 4. Try to express Q as a function of one variable. Use the procedures developed in Sections 2.1–2.4 to determine the maximum or minimum values and the points at which they occur.

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■ EXAMPLE 2 Maximizing Volume. From a thin piece of cardboard 8 in. by 8 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume?

Solution We might again think at first that it does not matter what the dimensions are, but our experience with Example 1 suggests otherwise. We make a drawing in which x is the length, in inches, of each square to be cut. It is important to note that since the original square is 8 in. by 8 in., after the smaller squares are removed, the lengths of the sides of the box will be 18 - 2x2 in. by 18 - 2x2 in. x

x

x

x 8 − 2x

8 in. 8 − 2x

x x

x

x x

8 − 2x

8 − 2x

8 in.

TECHNOLOGY CONNECTION

After the four small squares are removed and the sides are folded up, the volume V of the resulting box is

EXERCISES

V = l # w # h = 18 - 2x2 # 18 - 2x2 # x,

1. Complete this table to help visualize Example 2.

x

8 - 2x

4x - 32x + 64x 3

2

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 2. Graph V1x2 = 4x 3 - 32x 2 + 64x over the interval 10, 42. 3. Estimate a maximum value, and state where it occurs.

or

V1x2 = 164 - 32x + 4x 22x = 4x 3 - 32x 2 + 64x.

Since 8 - 2x 7 0, this means that x 6 4. Thus, we need to maximize

V1x2 = 4x 3 - 32x 2 + 64x over the interval 10, 42.

To do so, we first find V¿1x2: V¿1x2 = 12x 2 - 64x + 64.

Since V¿1x2 exists for all x in the interval 10, 42, we can set it equal to 0 to find the critical values: V¿1x2 = 12x 2 - 64x + 64 = 0 413x 2 - 16x + 162 = 0 413x - 421x - 42 = 0 3x - 4 = 0 or x - 4 = 0 3x = 4 or x = 43 or

x = 4 x = 4.

The only critical value in 10, 42 is 43. Thus, we can use the second derivative, V–1x2 = 24x - 64, to determine whether we have a maximum. Since V– A 43 B = 24 # 43 - 64 = 32 - 64 6 0,

we know that V A 43 B is a maximum.

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•

Maximum–Minimum Problems; Business and Economics Applications

265

Thus, to maximize the box’s volume, small squares with edges measuring 34 in., or in., should be cut from each corner of the original 8 in. by 8 in. piece of cardboard. When the sides are folded up, the resulting box will have sides of length

1 13 Quick Check 2 Repeat Example 2 starting with a sheet of cardboard measuring 8.5 in. by 11 in. (the size of a typical sheet of paper). Will this box hold 1 liter (L) of liquid? (Hint: 1 L = 1000 cm3 and 1 in3 = 16.38 cm3.)

8 - 2x = 8 - 2 #

4 24 8 16 1 = = = 5 in. 3 3 3 3 3

and a height of 113 in. The maximum volume is 4 4 3 4 2 4 1024 25 3 Va b = 4a b - 32a b + 64 a b = = 37 in . 3 3 3 3 27 27

O Quick Check 2 In manufacturing, minimizing the amount of material used is always preferred, both from a cost standpoint and in terms of efficiency.

■ EXAMPLE 3

Minimizing Material: Surface Area. A manufacturer of food-storage containers makes a cylindrical can with a volume of 500 milliliters 1mL; 1 mL = 1 cm32. What dimensions (height and radius) will minimize the material needed to produce each can, that is, minimize the surface area?

Solution We let h = height of the can and r = radius, both meas-

r

ured in centimeters. The formula for volume of a cylinder is V = pr 2h.

h

Since we know the volume is 500 cm3, this formula allows us to relate h and r, expressing one in terms of the other. It is easier to solve for h in terms of r: pr 2h = 500 500 h = . pr 2

The can is composed of two circular ends, each with an area equal to pr 2, and a side wall that, when laid out flat, is a rectangle with a height h and a length the same as the circumference of the circular ends, or 2pr. Thus, the area of this rectangle is 2prh.

r

l = 2p r h

r

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The total surface area A is the sum of the areas of the two circular ends and the side wall: A = 21pr 22 + 12prh2 500 = 2pr 2 + 2pr a 2 b. pr

Substituting for h.

Simplifying, we have area A as a function of radius r: A1r2 = 2pr 2 +

1000 . r

The nature of this problem situation requires that r 7 0. We differentiate: A¿1r2 = 4pr -

1000 r2

Note that

d 1000 d 1000 a 11000r -12 = - 1000r -2 = - 2 . b = r dr dr r

We set the derivative equal to 0 and solve for r to determine the critical values: 4pr -

1000 = 0 r2 4pr =

1000 r2

4pr 3 = 1000 1000 250 r3 = = p 4p 250 L 4.3 cm. A p 3

The critical value r L 4.3 is the only critical value in the interval 10, q 2. The second derivative is 2000 A–1r2 = 4p + . r3 Evaluating A–1x2 at the critical value, we get a positive value: Quick Check 3 Repeat Example 3 for a cylindrical can with a volume of 1000 cm3. What do you notice about the relationship between the cylinder’s radius and height? Repeat the example again for any other volume. Does the relationship between radius and height still hold? State this relationship.

A–14.32 = 4p +

2000 7 0. 14.323

y Surface area (in square centimeters)

r =

A(r) 400 300

(4.3, 348.73)

200 100 0

2 4 6 8 Radius (in centimeters)

r

The graph is concave up at the critical value, so the critical value indicates a minimum 3 250 point. Thus, the radius should be , or approximately 4.3 cm. The height is A p 500 approximately h = L 8.6 cm, and the minimum total surface area is approxip14.322 mately 348.73 cm2. Assuming that the material used for the side and the ends costs the same, minimizing the surface area will also minimize the cost to produce each can.

O Quick Check 3

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•

Maximum–Minimum Problems; Business and Economics Applications

267

■ EXAMPLE 4

Business: Maximizing Revenue. A stereo manufacturer determines that in order to sell x units of a new stereo, the price per unit, in dollars, must be p1x2 = 1000 - x. The manufacturer also determines that the total cost of producing x units is given by C1x2 = 3000 + 20x. a) b) c) d) e)

Find the total revenue R1x2. Find the total profit P1x2. How many units must the company produce and sell in order to maximize profit? What is the maximum profit? What price per unit must be charged in order to make this maximum profit?

Solution a) R1x2 = Total revenue = 1Number of units2 # 1Price per unit2 # = x p = x11000 - x2 = 1000x - x 2 b) P1x2 = Total revenue - Total cost = R1x2 - C1x2 = 11000x - x 22 - 13000 + 20x2 = - x 2 + 980x - 3000 c) To find the maximum value of P1x2, we first find P¿1x2: P¿1x2 = - 2x + 980. This is defined for all real numbers, so the only critical values will come from solving P¿1x2 = 0: P¿1x2 = - 2x + 980 = 0 - 2x = - 980 x = 490. There is only one critical value. We can therefore try to use the second derivative to determine whether we have an absolute maximum. Note that P–1x2 = - 2, a constant. Thus, P–14902 is negative, and so profit is maximized when 490 units are produced and sold. d) The maximum profit is given by Quick Check 4 Repeat Example 4 with the price function p1x2 = 1750 - 2x and the cost function C1x2 = 2250 + 15x. Round your answers when necessary.

P14902 = - 149022 + 980 # 490 - 3000 = $237,100.

Thus, the stereo manufacturer makes a maximum profit of $237,100 by producing and selling 490 stereos. e) The price per unit needed to make the maximum profit is p = 1000 - 490 = $510.

O Quick Check 4

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y

Maximum profit

Applications of Differentiation C(x)

A

Loss

B2 R(x)

B0

B1

Loss

x0

x1

x2

x

FIGURE 1

y

Let’s take a general look at the total-profit function and its related functions. Figure 1 shows an example of total-cost and total-revenue functions. We can estimate what the maximum profit might be by looking for the widest gap between R1x2 and C1x2, when R1x2 7 C1x2. Points B0 and B2 are break-even points. Figure 2 shows the related total-profit function. Note that when production is too low 16 x 02, there is a loss, perhaps due to high fixed or initial costs and low revenue. When production is too high 17 x 22, there is also a loss, perhaps due to the increased cost of overtime pay or expansion. The business operates at a profit everywhere between x 0 and x 2. Note that maximum profit occurs at a critical value x 1 of P(x). If we assume that P¿1x2 exists for all x in some interval, usually [0, q 2, this critical value occurs at some number x such that P¿1x2 = 0 and

Maximum profit M

P(x) = R(x) − C(x)

P–1x2 6 0.

Since P1x2 = R1x2 - C1x2, it follows that P¿1x2 = R¿1x2 - C¿1x2 and

x0

x1

x2

x

P–1x2 = R–1x2 - C–1x2.

Thus, the maximum profit occurs at some number x such that P¿1x2 = R¿1x2 - C¿1x2 = 0 and

FIGURE 2

P–1x2 = R–1x2 - C–1x2 6 0,

or R¿1x2 = C¿1x2 and

R–1x2 6 C–1x2.

In summary, we have the following theorem.

THEOREM 10 Maximum profit occurs at those x-values for which R¿1x2 = C¿1x2 and

R–1x2 6 C–1x2.*

You can check that the results in parts (c) and (d) of Example 4 can be easily found using Theorem 10.

■ EXAMPLE 5

Business: Determining a Ticket Price. Promoters of international fund-raising concerts must walk a fine line between profit and loss, especially when determining the price to charge for admission to closed-circuit TV showings in local theaters. By keeping records, a theater determines that at an admission price of $26, it averages 1000 people in attendance. For every drop in price of $1, it gains 50 customers. Each customer spends an average of $4 on concessions. What admission price should the theater charge in order to maximize total revenue?

Solution Let x be the number of dollars by which the price of $26 should be decreased. (If x is negative, the price is increased.) We first express the total revenue R as a function of x. Note that the increase in ticket sales is 50x when the price drops x dollars: R1x2 = 1Revenue from tickets2 + 1Revenue from concessions2

or

= = = R1x2 =

1Number of people2 # 1Ticket price2 + 1Number of people2 # 4 11000 + 50x2126 - x2 + 11000 + 50x2 # 4 26,000 - 1000x + 1300x - 50x 2 + 4000 + 200x, - 50x 2 + 500x + 30,000.

*In Section 2.6, the concepts of marginal revenue and marginal cost are introduced, allowing R¿1x2 = C¿1x2 to be regarded as Marginal revenue = Marginal cost.

2.5

•

Maximum–Minimum Problems; Business and Economics Applications

269

To find x such that R1x2 is a maximum, we first find R¿1x2: R¿1x2 = - 100x + 500. This derivative exists for all real numbers x. Thus, the only critical values are where R¿1x2 = 0; so we solve that equation: Quick Check 5 A baseball team charges $30 per ticket and averages 20,000 people in attendance per game. Each person spends an average of $8 on concessions. For every drop of $1 in the ticket price, the attendance rises by 800 people. What ticket price should the team charge to maximize total revenue?

- 100x + 500 = 0 - 100x = - 500 x = 5

This corresponds to lowering the price by $5.

Since this is the only critical value, we can use the second derivative, R–1x2 = - 100, to determine whether we have a maximum. Since R–152 is negative, R152 is a maximum. Therefore, in order to maximize revenue, the theater should charge $26 - $5, or

$21 per ticket.

O Quick Check 5 Minimizing Inventory Costs A retail business outlet needs to be concerned about inventory costs. Suppose, for example, that an appliance store sells 2500 television sets per year. It could operate by ordering all the sets at once. But then the owners would face the carrying costs (insurance, building space, and so on) of storing them all. Thus, they might make several, say 5, smaller orders, so that the largest number they would ever have to store is 500. However, each time they reorder, there are costs for paperwork, delivery charges, labor, and so on. It seems, therefore, that there must be some balance between carrying costs and reorder costs. Let’s see how calculus can help determine what that balance might be. We are trying to minimize the following function:

Total inventory costs = 1Yearly carrying costs2 + 1Yearly reorder costs2.

The lot size x is the largest number ordered each reordering period. If x units are ordered each period, then during that time somewhere between 0 and x units are in stock. To have a representative expression for the amount in stock at any one time in the period, we can use the average, x>2. This represents the average amount held in stock over the course of each time period. Refer to the graphs shown below and on the next page. If the lot size is 2500, then during the period between orders, there are somewhere between 0 and 2500 units in stock. On average, there are 2500>2, or 1250 units in stock. If the lot size is 1250, then during the period between orders, there are somewhere between 0 and 1250 units in stock. On average, there are 1250>2, or 625 units in stock. In general, if the lot size is x, the average inventory is x>2. 2500 ⎧ ⎪ ⎪ ⎪ ⎪ Lot size (inventory) ⎨⎪ ⎪ ⎪ ⎪ ⎩

Average lot size (average inventory) 1250

0

12

Time (in months)

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•

Applications of Differentiation 1250 ⎧ ⎪ ⎪ Lot size ⎨ (inventory) ⎪ ⎪ ⎩

Average lot size (average inventory) 625 0

6

Lot size 625 ⎧ (inventory) ⎨⎩ 0

3

6

12

9

12

Time (in months) Average lot size (average inventory) 312.5 ≈ 313 Time (in months) Average lot size (average inventory)

x⎧ ⎪ ⎪ Lot size ⎨ (inventory) ⎪ ⎪ ⎩ 0

–x– 2

Time (in months)

■ EXAMPLE 6

Business: Minimizing Inventory Costs. A retail appliance store sells 2500 television sets per year. It costs $10 to store one set for a year. To reorder, there is a fixed cost of $20, plus a fee of $9 per set. How many times per year should the store reorder, and in what lot size, to minimize inventory costs?

Solution Let x = the lot size. Inventory costs are given by

C1x2 = 1Yearly carrying costs2 + 1Yearly reorder costs2.

We consider each component of inventory costs separately. a) Yearly carrying costs. The average amount held in stock is x>2, and it costs $10 per set for storage. Thus, Yearly cost # Average number b a b per item of items x = 10 # . 2

Yearly carrying costs = a

b) Yearly reorder costs. We know that x is the lot size, and we let N be the number of reorders each year. Then Nx = 2500, and N = 2500>x. Thus, Cost of each # Number of b a b order reorders 2500 = 120 + 9x2 . x

Yearly reorder costs = a

c) Thus, we have C1x2 = 10 #

x 2500 + 120 + 9x2 x 2 50,000 = 5x + + 22,500 = 5x + 50,000x -1 + 22,500. x

d) To find a minimum value of C over 31, 25004, we first find C¿1x2: C¿1x2 = 5 -

50,000 x2

.

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•

Maximum–Minimum Problems; Business and Economics Applications

271

e) C¿1x2 exists for all x in 31, 25004, so the only critical values are those x-values such that C¿1x2 = 0. We solve C¿1x2 = 0: 5 -

50,000 x2

= 0

5 =

x2

5x 2 = 50,000 x 2 = 10,000 x = ;100.

y C(x) = 5x + –50,000 –——– + 22,500 x

Since there is only one critical value in 31, 25004, that is, x = 100, we can use the second derivative to see whether it yields a maximum or a minimum: 100,000

C–1x2 =

x3

.

C–1x2 is positive for all x in 31, 25004, so we have a minimum at x = 100. Thus, to minimize inventory costs, the store should order 2500>100, or 25, times per year. The lot size is 100 sets.

x

100

50,000

TECHNOLOGY CONNECTION

Exploratory Many calculators can make tables and/or spreadsheets of function values. In reference to Example 6, without using calculus, one might make an estimate of the lot size that will minimize total inventory costs by using a table like the one below. Complete the table, and estimate the solution of Example 6.

EXERCISES

1. Graph C1x2 over the interval 31, 25004.

2. Graphically estimate the minimum value, and note

where it occurs. Does the table confirm the graph?

Number of Reorders, 2500 x

Average Inventory, x 2

Carrying Costs, x 10 # 2

Cost of Each Order, 20 + 9x

Reorder Costs, 2500 120 + 9x2 x

Total Inventory Costs, 2500 x C1x2 = 10 # + 120 + 9x2 x 2

2500

1

1250

$12,500

$22,520

$22,520

$35,020

1250

2

625

6,250

11,270

22,540

500

5

250

2,500

4,520

250

10

125

167

15

84

125

20

100

25

90

28

50

50

Lot Size, x

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•

Applications of Differentiation

What happens in problems like Example 6 if the answer is not a whole number? For those cases, we consider the two whole numbers closest to the answer and substitute them into C(x). The value that yields the smaller C(x) is the lot size.

■ EXAMPLE 7

Business: Minimizing Inventory Costs. Reconsider Example 6, but change the $10 storage cost to $20. How many times per year should the store reorder television sets, and in what lot size, in order to minimize inventory costs?

Solution Comparing this situation with that in Example 6, we find that the inventory cost function becomes C1x2 = 20 #

2500 x + 120 + 9x2 x 2

= 10x +

50,000 + 22,500 = 10x + 50,000x -1 + 22,500. x

Then we find C¿1x2, set it equal to 0, and solve for x: C¿1x2 = 10 -

50,000 x2

= 0

10 =

50,000 x2

10x 2 = 50,000 x 2 = 5000 x = 25000 L 70.7. It is impossible to reorder 70.7 sets each time, so we consider the two numbers closest to 70.7, which are 70 and 71. Since C1702 L $23,914.29 and

it follows that the lot size that will minimize cost is 71, although the difference, $0.06, is not much. (Note: Such a procedure will not work for all functions but will work for the type we are considering here.) The number of times an order should be placed is 2500>71 with a remainder of 15, indicating that 35 orders should be placed. Of those, 35 - 15 = 20 will be for 71 items and 15 will be for 72 items.

Quick Check 6

O

Repeat Example 7 with a storage cost of $30 per set and assuming that the store sells 3000 sets per year.

C1712 L $23,914.23,

O Quick Check 6 The lot size that minimizes total inventory costs is often referred to as the economic ordering quantity. Three assumptions are made in using the preceding method to determine the economic ordering quantity. First, the demand for the product is the same year round. For television sets, this may be reasonable, but for seasonal items such as clothing or skis, this assumption is unrealistic. Second, the time between the placing of an order and its receipt remains consistent throughout the year. Finally, the various costs involved, such as storage, shipping charges, and so on, do not vary. This assumption may not be reasonable in a time of inflation, although variation in these costs can be allowed for by anticipating what they might be and using average costs. Regardless, the model described above is useful, allowing us to analyze a seemingly difficult problem using calculus.

Section Summary • In many real-life applications, we wish to determine the minimum or maximum value of some function modeling a situation.

• Identify a realistic interval for the domain of the input variable. If it is a closed interval, its endpoints should be considered as possible critical values.

273

Exercise Set 2.5

EXERCISE SET

2.5 1. Of all numbers whose sum is 50, find the two that have the maximum product. That is, maximize Q = xy, where x + y = 50. 2. Of all numbers whose sum is 70, find the two that have the maximum product. That is, maximize Q = xy, where x + y = 70. 3. In Exercise 1, can there be a minimum product? Why or why not? 4. In Exercise 2, can there be a minimum product? Why or why not? 5. Of all numbers whose difference is 4, find the two that have the minimum product.

14. Maximizing area. A rancher wants to enclose two rectangular areas near a river, one for sheep and one for cattle. There are 240 yd of fencing available. What is the largest total area that can be enclosed?

x x x

6. Of all numbers whose difference is 6, find the two that have the minimum product. 7. Maximize Q = xy 2, where x and y are positive numbers such that x + y 2 = 1. 8. Maximize Q = xy 2, where x and y are positive numbers such that x + y 2 = 4. 9. Minimize Q = 2x + 3y , where x + y = 5. 2

2

10. Minimize Q = x 2 + 2y 2, where x + y = 3. 11. Maximize Q = xy, where x and y are positive numbers such that 43 x 2 + y = 16. 12. Maximize Q = xy, where x and y are positive numbers such that x + 43 y 2 = 1. 13. Maximizing area. A lifeguard needs to rope off a rectangular swimming area in front of Long Lake Beach, using 180 yd of rope and floats. What dimensions of the rectangle will maximize the area? What is the maximum area? (Note that the shoreline is one side of the rectangle.)

x x

15. Maximizing area. A carpenter is building a rectangular shed with a fixed perimeter of 54 ft. What are the dimensions of the largest shed that can be built? What is its area? 16. Maximizing area. Of all rectangles that have a perimeter of 42 ft, find the dimensions of the one with the largest area. What is its area? 17. Maximizing volume. From a 50-cm-by-50-cm sheet of aluminum, square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? 18. Maximizing volume. From a thin piece of cardboard 20 in. by 20 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? 19. Minimizing surface area. Drum Tight Containers is designing an open-top, square-based, rectangular box that will have a volume of 62.5 in3. What dimensions will minimize surface area? What is the minimum surface area? 20. Minimizing surface area. A soup company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 32 ft3. What dimensions will minimize surface area? What is the minimum surface area?

274

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•

Applications of Differentiation

21. Minimizing surface area. Open Air Waste Management is designing a rectangular construction dumpster that will be twice as long as it is wide and must hold 12 yd3 of debris. Find the dimensions of the dumpster that will minimize its surface area.

x refrigerators, the price per refrigerator must be p = 280 - 0.4x. It also determines that the total cost of producing x refrigerators is given by C1x2 = 5000 + 0.6x 2.

? ?

x

22. Minimizing surface area. Ever Green Gardening is designing a rectangular compost container that will be twice as tall as it is wide and must hold 18 ft3 of composted food scraps. Find the dimensions of the compost container with minimal surface area (include the bottom and top).

APPLICATIONS Business and Economics Maximizing profit. Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that revenue, R(x), and cost, C(x), are in dollars for Exercises 23–26. 23. R1x2 = 50x - 0.5x 2, C1x2 = 4x + 10 24. R1x2 = 50x - 0.5x 2, C1x2 = 10x + 3 25. R1x2 = 2x, C1x2 = 0.01x 2 + 0.6x + 30 26. R1x2 = 5x, C1x2 = 0.001x 2 + 1.2x + 60 27. R1x2 = 9x - 2x 2, C1x2 = x 3 - 3x 2 + 4x + 1; assume that R1x2 and C1x2 are in thousands of dollars, and x is in thousands of units. 28. R1x2 = 100x - x 2, C1x2 = 13 x 3 - 6x 2 + 89x + 100; assume that R1x2 and C1x2 are in thousands of dollars, and x is in thousands of units. 29. Maximizing profit. Raggs, Ltd., a clothing firm, determines that in order to sell x suits, the price per suit must be p = 150 - 0.5x. It also determines that the total cost of producing x suits is given by C1x2 = 4000 + 0.25x 2. a) Find the total revenue, R1x2. b) Find the total profit, P1x2. c) How many suits must the company produce and sell in order to maximize profit? d) What is the maximum profit? e) What price per suit must be charged in order to maximize profit? 30. Maximizing profit. Riverside Appliances is marketing a new refrigerator. It determines that in order to sell

a) Find the total revenue, R1x2. b) Find the total profit, P1x2. c) How many refrigerators must the company produce and sell in order to maximize profit? d) What is the maximum profit? e) What price per refrigerator must be charged in order to maximize profit? 31. Maximizing revenue. A university is trying to determine what price to charge for tickets to football games. At a price of $18 per ticket, attendance averages 40,000 people per game. Every decrease of $3 adds 10,000 people to the average number. Every person at the game spends an average of $4.50 on concessions. What price per ticket should be charged in order to maximize revenue? How many people will attend at that price? 32. Maximizing profit. Gritz-Charlston is a 300-unit luxury hotel. All rooms are occupied when the hotel charges $80 per day for a room. For every increase of x dollars in the daily room rate, there are x rooms vacant. Each occupied room costs $22 per day to service and maintain. What should the hotel charge per day in order to maximize profit? 33. Maximizing yield. An apple farm yields an average of 30 bushels of apples per tree when 20 trees are planted on an acre of ground. Each time 1 more tree is planted per acre, the yield decreases by 1 bushel (bu) per tree as a result of crowding. How many trees should be planted on an acre in order to get the highest yield? 34. Nitrogen prices. During 2001, nitrogen prices fell by 41%. Over the same period of time, nitrogen demand went up by 12%. (Source: Chemical Week.) a) Assuming a linear change in demand, find the demand function, q1x2, by finding the equation of the line that passes through the points 11, 12 and 10.59, 1.122. Here x is the price as a fraction of the January 2001 price, and q1x2 is the demand as a fraction of the demand in January. b) As a percentage of the January 2001 price, what should the price of nitrogen be to maximize revenue? 35. Vanity license plates. According to a pricing model, increasing the fee for vanity license plates by $1 decreases the percentage of a state’s population that will request them by 0.04%. (Source: E. D. Craft, “The demand for vanity (plates): Elasticities, net revenue maximization, and deadweight loss,” Contemporary Economic Policy, Vol. 20, 133–144 (2002).) a) Recently, the fee for vanity license plates in Maryland was $25, and the percentage of the state’s population that had vanity plates was 2.13%. Use this information to construct the demand function, q1x2, for the percentage of Maryland’s population that will request vanity license plates for a fee of x dollars. b) Find the fee, x, that will maximize revenue from vanity plates.

Exercise Set 2.5

36. Maximizing revenue. When a theater owner charges $5 for admission, there is an average attendance of 180 people. For every $0.10 increase in admission, there is a loss of 1 customer from the average number. What admission should be charged in order to maximize revenue? 37. Minimizing costs. A rectangular box with a volume of 320 ft3 is to be constructed with a square base and top. The cost per square foot for the bottom is 15¢, for the top is 10¢, and for the sides is 2.5¢. What dimensions will minimize the cost?

44. Minimizing inventory costs. Repeat Exercise 42 using the same data, but change the reorder costs from an additional $5 per surfboard to $6 per surfboard. 45. Minimizing surface area. A closed-top cylindrical container is to have a volume of 250 in2. What dimensions (radius and height) will minimize the surface area? 46. Minimizing surface area. An open-top cylindrical container is to have a volume of 400 cm2. What dimensions (radius and height) will minimize the surface area? 47. Minimizing cost. Assume that the costs of the materials for making the cylindrical container described in Exercise 45 are $0.005>in2 for the circular base and top and $0.003>in2 for the wall. What dimensions will minimize the cost of materials?

x x

y

x

275

x

38. Maximizing area. A publisher decides that each page in a new book must have an area of 73.125 in2, a 0.75-in. margin at the top and at the bottom of each page, and a 0.5-in. margin on each of the sides. What should the outside dimensions of each page be so that the printed area is a maximum? 39. Minimizing inventory costs. A sporting goods store sells 100 pool tables per year. It costs $20 to store one pool table for a year. To reorder, there is a fixed cost of $40 per shipment plus $16 for each pool table. How many times per year should the store order pool tables, and in what lot size, in order to minimize inventory costs? 40. Minimizing inventory costs. A pro shop in a bowling center sells 200 bowling balls per year. It costs $4 to store one bowling ball for a year. To reorder, there is a fixed cost of $1, plus $0.50 for each bowling ball. How many times per year should the shop order bowling balls, and in what lot size, in order to minimize inventory costs? 41. Minimizing inventory costs. A retail outlet for Boxowitz Calculators sells 720 calculators per year. It costs $2 to store one calculator for a year. To reorder, there is a fixed cost of $5, plus $2.50 for each calculator. How many times per year should the store order calculators, and in what lot size, in order to minimize inventory costs? 42. Minimizing inventory costs. Bon Temps Surf and Scuba Shop sells 360 surfboards per year. It costs $8 to store one surfboard for a year. Each reorder costs $10, plus an additional $5 for each surfboard ordered. How many times per year should the store order surfboards, and in what lot size, in order to minimize inventory costs? 43. Minimizing inventory costs. Repeat Exercise 41 using the same data, but assume yearly sales of 256 calculators with the fixed cost of each reorder set at $4.

48. Minimizing cost. Assume that the costs of the materials for making the cylindrical container described in Exercise 46 are $0.0015>cm2 for the base and $0.008>cm2 for the wall. What dimensions will minimize the cost of materials?

General Interest 49. Maximizing volume. The postal service places a limit of 84 in. on the combined length and girth of (distance around) a package to be sent parcel post. What dimensions of a rectangular box with square cross-section will contain the largest volume that can be mailed? (Hint: There are two different girths.)

x y x

50. Minimizing cost. A rectangular play area is to be fenced off in a person’s yard and is to contain 48 yd2. The next-door neighbor agrees to pay half the cost of the fence on the side of the play area that lies along the property line. What dimensions will minimize the cost of the fence? Half the cost x

x

y

51. Maximizing light. A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 24 ft. What should its

276

CHAPTER 2

•

Applications of Differentiation

dimensions be in order to allow the maximum amount of light to enter through the window?

should the line come to the shore in order to minimize cost? Note that S could very well be B or A. (Hint: The length of CS is 21 + x 2.) Island

x C y

1

B

52. Maximizing light. Repeat Exercise 51, but assume that the semicircle is to be stained glass, which transmits only half as much light as clear glass does.

SYNTHESIS 53. For what positive number is the sum of its reciprocal and five times its square a minimum? 54. For what positive number is the sum of its reciprocal and four times its square a minimum? 55. Business: maximizing profit. The amount of money that customers deposit in a bank in savings accounts is directly proportional to the interest rate that the bank pays on that money. Suppose that a bank was able to turn around and loan out all the money deposited in its savings accounts at an interest rate of 18%. What interest rate should it pay on its savings accounts in order to maximize profit? 56. A 24-in. piece of wire is cut in two pieces. One piece is used to form a circle and the other to form a square. How should the wire be cut so that the sum of the areas is a minimum? A maximum?

S

x

A

4–x 4

58. Life science: flights of homing pigeons. It is known that homing pigeons tend to avoid flying over water in the daytime, perhaps because the downdrafts of air over water make flying difficult. Suppose that a homing pigeon is released on an island at point C, which is 3 mi directly out in the water from a point B on shore. Point B is 8 mi downshore from the pigeon’s home loft at point A. Assume that a pigeon flying over water uses energy at a rate 1.28 times the rate over land. Toward what point S downshore from A should the pigeon fly in order to minimize the total energy required to get to the home loft at A? Assume that Total energy = 1Energy rate over water2 # 1Distance over water2 + 1Energy rate over land2 # 1Distance over land2. Island C

24 x

24 – x

3

B

x

S

8–x

A

Home loft

8

57. Business: minimizing costs. A power line is to be constructed from a power station at point A to an island at point C, which is 1 mi directly out in the water from a point B on the shore. Point B is 4 mi downshore from the power station at A. It costs $5000 per mile to lay the power line under water and $3000 per mile to lay the line under ground. At what point S downshore from A

59. Business: minimizing distance. A road is to be built between two cities C1 and C2, which are on opposite sides of a river of uniform width r. C1 is a units from the river, and C2 is b units from the river, with a … b. A bridge will carry the traffic across the river. Where should the bridge be located in order to minimize the total distance

2.6

between the cities? Give a general solution using the constants a, b, p, and r as shown in the figure.

277

62. Minimize Q = x 3 + 2y 3, where x and y are positive numbers, such that x + y = 1.

64. Business: minimizing inventory costs—a general solution. A store sells Q units of a product per year. It costs a dollars to store one unit for a year. To reorder, there is a fixed cost of b dollars, plus c dollars for each unit. How many times per year should the store reorder, and in what lot size, in order to minimize inventory costs?

a p–x ⎫ ⎪ ⎬ River ⎪ ⎭

r

Marginals and Differentials

63. Minimize Q = 3x + y 3, where x 2 + y 2 = 2.

C1

Bridge x

•

65. Business: minimizing inventory costs. Use the general solution found in Exercise 64 to find how many times per year a store should reorder, and in what lot size, when Q = 2500, a = $10, b = $20, and c = $9.

b C2 p

60. Business: minimizing cost. The total cost, in dollars, of producing x units of a certain product is given by x3 C1x2 = 8x + 20 + . 100 a) Find the average cost, A1x2 = C1x2>x. b) Find C¿1x2 and A¿1x2. c) Find the minimum of A(x) and the value x 0 at which it occurs. Find C¿1x 02. d) Compare A1x 02 and C¿1x 02. 61. Business: minimizing cost. Consider A1x2 = C1x2>x. a) Find A¿1x2 in terms of C¿1x2 and C1x2. b) Show that if A(x) has a minimum, then it will occur at that value of x 0 for which

Answers to Quick Checks 1. With 50 ft of fencing, the dimensions are 25 ft by 25 ft 1625 ft2 area2; with 100 ft of fencing, they are 50 ft by 50 ft 12500 ft2 area2; in general, n feet of fencing gives n>2 ft by n>2 ft 1n 2>4 ft2 area2. 2. The dimensions are approximately 1.585 in. by 5.33 in. by 7.83 in.; the volume is 66.15 in3, or 1083.5 cm3, slightly more than 1 L. 3. r L 5.42 cm, h L 10.84 cm, surface area L 553.58 cm2; the relationship is h = 2r (height equals diameter). 4. (a) R1x2 = 1750x - 2x 2 (b) P1x2 = - 2x 2 + 1735x - 2250 (c) x = 434 units (d) Maximum profit = $374,028 (e) Price per unit = $882.00 5. $23.50 6. x L 63; the store should place 8 orders for 63 sets and 39 orders for 64 sets.

C¿1x 02 = A1x 02 C1x 02 . = x0

This result shows that if average cost can be minimized, such a minimum will occur when marginal cost equals average cost.

2.6 OBJECTIVES • Find marginal cost, revenue, and profit. • Find ¢ y and dy. • Use differentials for approximations.

Marginals and Differentials In this section, we consider ways of using calculus to make linear approximations. Suppose, for example, that a company is considering an increase in production. Usually the company wants at least an approximation of what the resulting changes in cost, revenue, and profit will be.

Marginal Cost, Revenue, and Profit Suppose that a band is producing its own CD and considering an increase in monthly production from 12 cartons to 13. To estimate the resulting increase in cost, it would be reasonable to find the rate at which cost is increasing when 12 cartons are produced and add that to the cost of producing 12 cartons. That is, C1132 L C1122 + C¿1122.

278

CHAPTER 2

•

Applications of Differentiation

Cost of production

C(x) C(12) + C′(12) C(13)

C′(12)

C(12)

1

12

13

x

Number of cartons produced

The number C¿1122 is called the marginal cost at 12. Remember that C¿1122 is the slope of the tangent line at the point 112, C11222. If, for example, this slope is 43, we can regard it as a vertical change of 3 with a horizontal change of 4, or a vertical change of 34 with a horizontal change of 1. It is this latter interpretation that we use for estimating. Graphically, this interpretation can be viewed as shown at the left. Note in the figure that C¿1122 is slightly more than the difference between C1132 and C1122, or C1132 - C1122. For other curves, C¿1122 may be slightly less than C1132 - C1122. Almost always, however, it is simpler to compute C¿1122 than it is to compute C1132 - C1122. Generalizing, we have the following.

DEFINITIONS Let C(x), R(x), and P(x) represent, respectively, the total cost, revenue, and profit from the production and sale of x items. The marginal cost* at x, given by C¿1x2, is the approximate cost of the 1x + 12st item: C¿1x2 L C1x + 12 - C1x2, or C1x + 12 L C1x2 + C¿1x2. The marginal revenue at x, given by R¿1x2, is the approximate revenue from the 1x + 12st item: R¿1x2 L R1x + 12 - R1x2, or R1x + 12 L R1x2 + R¿1x2. The marginal profit at x, given by P¿1x2, is the approximate profit from the 1x + 12st item: P¿1x2 L P1x + 12 - P1x2, or P1x + 12 L P1x2 + P¿1x2. You can confirm that P¿1x2 = R¿1x2 - C¿1x2.

■ EXAMPLE 1 Business: Marginal Cost, Revenue, and Profit. Given C1x2 = 62x 2 + 27,500 and R1x2 = x 3 - 12x 2 + 40x + 10, find each of the following. a) Total profit, P1x2 b) Total cost, revenue, and profit from the production and sale of 50 units of the product c) The marginal cost, revenue, and profit when 50 units are produced and sold

Solution a) Total profit = P1x2 = R1x2 - C1x2 = x 3 - 12x 2 + 40x + 10 - 162x 2 + 27,5002 = x 3 - 74x 2 + 40x - 27,490

*The term “marginal” comes from the Marginalist School of Economic Thought, which originated in Austria for the purpose of applying mathematics and statistics to the study of economics.

2.6

TECHNOLOGY CONNECTION

Business: Marginal Revenue, Cost, and Profit

P1502 = R1502 - C1502 = $97,010 - $182,500 = - $85,490

1. Using the viewing window

30, 100, 0, 20004, graph these total-revenue and total-cost functions:

and

TECHNOLOGY CONNECTION To check the accuracy of R¿1502 as an estimate of R1512 - R1502, let y1 = x 3 - 12x 2 + 40x + 10, y2 = y11x + 12 - y11x2, and y3 = nDeriv 1y1, x, x2. By using TABLE with Indpnt: Ask, we can display a table in which y2 (the difference between y11x + 12 and y11x22 can be compared with y1 ¿1x2. 40 48 50

We could also use P1x2 from part (a). There is a loss of $85,490 when 50 units are produced and sold.

R¿1x2 = 3x 2 - 24x + 40, so R¿1502 = 3 # 502 - 24 # 50 + 40 = $6340.

Then find P1x2 and graph it using the same viewing window. Find R¿1x2, C¿1x2, and P¿1x2, and graph them using 30, 60, 0, 604. Then find R1402, C1402, P1402, R¿1402, C¿1402, and P¿1402. Which marginal function is constant?

Y2 3989 5933 6479

279

c) C¿1x2 = 124x, so C¿1502 = 124 # 50 = $6200. Once 50 units have been made, the approximate cost of the 51st unit (marginal cost) is $6200.

C1x2 = 10x + 3.

X

Marginals and Differentials

b) C1502 = 62 # 502 + 27,500 = $182,500 (the total cost of producing the first 50 units); R1502 = 503 - 12 # 502 + 40 # 50 + 10 = $97,010 (the total revenue from the sale of the first 50 units);

EXERCISE

R1x2 = 50x - 0.5x 2

•

Y3 3880 5800 6340

Once 50 units have been sold, the approximate revenue from the 51st unit (marginal revenue) is $6340.

P¿1x2 = 3x 2 - 148x + 40, so P¿1502 = 3 # 502 - 148 # 50 + 40 = $140.

Once 50 units have been produced and sold, the approximate profit from the sale of the 51st item (marginal profit) is $140. Often, in business, formulas for C1x2, R1x2, and P1x2 are not known, but information may exist about the cost, revenue, and profit trends at a particular value x = a. For example, C1a2 and C¿1a2 may be known, allowing a reasonable prediction to be made about C1a + 12. In a similar manner, predictions can be made for R1a + 12 and P1a + 12. In Example 1, formulas do exist, so it is possible to see how accurate our predictions were. We check C1512 - C1502 and leave the checks of R1512 - R1502 and P1512 - P1502 for you (see the Technology Connection below, at left): C1512 - C1502 = 62 # 512 + 27,500 - 162 # 502 + 27,5002 = 6262, whereas C¿1502 = 6200.

In this case, C¿1502 provides an approximation of C1512 - C1502 that is within 1% of the actual value. Note that marginal cost is different from average cost: Average cost per unit for 50 units = =

C1502 50

EXERCISE 1. Create a table to check the

effectiveness of using P¿1502 to approximate P1512 - P1502.

The number of units, 50

182,500 = $3650, 50

whereas Marginal cost when 50 units are produced = $6200 L cost of the 51st unit.

Differentials and Delta Notation X=

Total cost of 50 units

Just as the marginal cost C¿1x 02 can be used to estimate C1x 0 + 12, the value of the derivative of any continuous function, f ¿1x 02, can be used to estimate values of f1x2 for x-values near x 0. Before we do so, however, we need to develop some notation. Recall the difference quotient f1x + h2 - f1x2 h

,

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CHAPTER 2

•

Applications of Differentiation y

¢x = 1x + h2 - x = h.

f(x)

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

illustrated in the graph at the right. The difference quotient is used to define the derivative of a function at x. The number h is considered to be a change in x. Another notation for such a change is ¢x, read “delta x” and called delta notation. The expression ¢x is not the product of ¢ and x; it is a new type of variable that represents the change in the value of x from a first value to a second. Thus,

x1 = x

⎫ ⎪ ⎬ Δy = f(x + h) – f(x) ⎪ ⎭

Δx = h x2 = x + h = x + Δx

x

If subscripts are used for the first and second values of x, we have ¢x = x 2 - x 1, or

x 2 = x 1 + ¢x.

Note that the value of ¢x can be positive or negative. For example, and

if x 1 = 4 and ¢x = 0.7, then x 2 = 4.7, if x 1 = 4 and ¢x = - 0.7, then x 2 = 3.3.

We generally omit the subscripts and use x and x + ¢x. Now suppose that we have a function given by y = f1x2. A change in x from x to x + ¢x yields a change in y from f(x) to f1x + ¢x2. The change in y is given by ¢y = f1x + ¢x2 - f1x2.

■ EXAMPLE 2 For y = x 2, x = 4, and ¢x = 0.1, find ¢y. Solution We have

¢y = 14 + 0.122 - 4 2 = 14.122 - 4 2 = 16.81 - 16 = 0.81.

■ EXAMPLE 3 For y = x 3, x = 2, and ¢x = - 0.1, find ¢y. Solution We have Quick Check 1 For y = 2x 4 + x, x = 2, and ¢x = - 0.05, find ¢y.

¢y = 32 + 1- 0.1243 - 2 3 = 11.923 - 2 3 = 6.859 - 8 = - 1.141.

O Quick Check 1 Let’s now use calculus to predict function values. If delta notation is used, the difference quotient f1x + h2 - f1x2 h becomes. f1x + ¢x2 - f1x2 ¢x

=

¢y ¢x

.

We can then express the derivative as dy dx

= lim

¢x : 0

¢y ¢x

.

O

2.6

•

Marginals and Differentials

281

Note that the delta notation resembles Leibniz notation (see Section 1.5). For values of ¢x close to 0, we have the approximation dy dx

L

¢y ¢x

, or f ¿1x2 L

¢y ¢x

.

y f

Multiplying both sides of the second expression by ¢x gives us ¢y L f ¿1x2 ¢x.

Actual value Value from the tangent line

⎫ ⎪ ⎬ Δy ⎪ ⎭

⎫ ⎪ ⎬ ⎪ ⎭

We can see this in the graph at the right. From this graph, it seems reasonable to assume that, for small values of ¢x, the y-values on the tangent line can be used to estimate function values on the curve.

m = f '(x)

Δx

x

x

For f, a continuous, differentiable function, and small ¢x, f ¿1x2 L

¢y ¢x

and

¢y L f ¿1x2 # ¢x.

Let’s illustrate this idea by considering the square-root function, f1x2 = 2x. We know how to approximate 227 using a calculator. But suppose we didn’t. We could begin with 225 and use as a change in input ¢x = 2. We would use the corresponding change in y, that is, ¢y L f ¿1x2¢x, to estimate 227.

■ EXAMPLE 4 Approximate 227 using ¢y L f ¿1x2¢x. Solution We first think of the number closest to 27 that is a perfect square. This is 25. What we will do is approximate how y = 2x changes when x changes from 25 to 27. From the box above, we have ¢y L f ¿1x2 # ¢x L 12 x -1>2 # ¢x

f Using y = 2x = x 1>2 as f 1x2

We are interested in ¢y as x changes from 25 to 27, so ¢y L L

1 2225

#2

Replacing x with 25 and ¢x with 2

1 = 0.2. 5

We can now approximate 227:

Quick Check 2 Approximate 298 using ¢y L f ¿1x2 ¢x. To five decimal places, 298 = 9.89949. How close is your approximation?

227 = = L L

225 + ¢y 5 + ¢y 5 + 0.2 5.2.

To five decimal places, 227 = 5.19615. Thus, our approximation is fairly accurate.

O Quick Check 2 Up to now, we have not defined the symbols dy and dx as separate entities, but have treated dy>dx as one symbol. We now define dy and dx. These symbols are called differentials.

O

282

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•

Applications of Differentiation

DEFINITION For y = f1x2, we define dx, called the differential of x, by dx = ¢x and dy, called the differential of y, by dy = f ¿1x2 dx.

We can illustrate dx and dy as shown at the right. Note that dx = ¢x, but dy Z ¢y, though dy L ¢y, for small values of dx.

y f m = f '(x) ⎫ ⎬ dy ⎭

⎫ ⎪ ⎬ Δy ⎪ ⎭

⎫ ⎪ ⎬ ⎪ ⎭

Δx = dx

x

■ EXAMPLE 5 For y = x14 - x23: a) Find dy. b) Find dy when x = 5 and dx = 0.01. c) Compare dy to ¢y.

Solution a) First, we find dy>dx: dy dx

= x3314 - x221- 124 + 14 - x23 = = = =

- 3x14 - x22 + 14 - x23 14 - x223- 3x + 14 - x24 14 - x223- 4x + 44 - 414 - x221x - 12.

Using the Product and Chain Rules Factoring out 14 - x22 Factoring out - 4

Then we solve for dy:

dy = - 414 - x221x - 12 dx.

b) When x = 5 and dx = 0.01,

dy = - 414 - 52215 - 1210.012 = - 41- 12214210.012 = - 0.16.

c) The value dy = - 0.16 is the approximate change in y between x 1 = 5 and x 2 = 5.01 (that is, x 2 = x 1 + dx = 5 + 0.01). The actual change in y is determined by evaluating the function for x 2 and x 1 and subtracting: ¢y = = = =

35.0114 - 5.01234 - 3514 - 5234 35.011- 1.01234 - 351- 1234 35.011- 1.03030124 - 351- 124 - 0.16180801.

x

2.6

•

Marginals and Differentials

283

We see that the approximation dy and the actual change ¢y are reasonably close. It is easier to calculate the approximation since that involves fewer steps, but the trade-off is some loss in accuracy. As long as dx is small, this loss in accuracy is acceptable for many applications. Differentials are often used in applications involving measurements and tolerance. When we measure an object, we accept that our measurements are not exact, and we allow for a small tolerance in our measurements. If x represents a measurement (a length, a weight, a volume, etc.), then dx represents the tolerance. Even a small tolerance for the input can have a significant effect on the output, as the following example shows.

■ EXAMPLE 6 Business: Cost and Tolerance. In preparation for laying new tile, Michelle measures the floor of a large conference room and finds it to be square, measuring 100 ft by 100 ft. Suppose her measurements are accurate to ;6 in. (the tolerance). a) Use a differential to estimate the difference in area (dA) due to the tolerance. b) Compare the result from part (a) with the actual difference in area 1¢A2. c) If each tile covers 1 ft2 and a box of 12 tiles costs $24, how much extra cost should Michelle allot for the potential overage in floor area?

Solution a) The floor is a square, with a presumed measurement of 100 ft per side and a tolerance of ;6 in. = ;0.5 ft . The area A in square feet 1ft22 for a square of side length x ft is A1x2 = x 2. The derivative is dA>dx = 2x, and solving for dA gives the differential of A: dA = 2x dx. To find dA, we substitute x = 100 and dx = ;0.5: dA = 2110021;0.52 = ;100. The value of dA is interpreted as the approximate difference in area due to the inexactness in measuring. Therefore, if Michelle’s measurements are off by half a foot, the total area can differ by approximately ;100 ft2. A small “error” in measurement can lead to quite a large difference in the resulting area. b) The actual difference in area 1¢A2 is calculated directly. We set x 1 = 100 ft, the presumed length measurement and let x 2 represent the length plus or minus the tolerance. If the true length is at the low end, we have x 2 = 99.5 ft, that is, 100 ft minus the tolerance of 0.5 ft. The floor’s area is then 99.52 = 9900.25 ft2. The actual difference in area is ¢A = = = = =

A1x 22 - A1x 12 A199.52 - A11002 99.52 - 1002 9900.25 - 10,000 - 99.75 ft2.

Thus, the actual difference in area is ¢A = - 99.75 ft2, which compares well with the approximate value of dA = - 100 ft2.

284

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•

Applications of Differentiation

If the true length is at the high end, we have x 2 = 100.5 ft. The floor’s area is then 100.52 = 10,100.25 ft2. The actual difference in area is Quick Check 3

O

The four walls of a room measure 10 ft by 10 ft each, with a tolerance of ;0.25 ft. a) Calculate the approximate difference in area, dA, for the four walls. b) Workers will be texturing the four walls using “knockdown” spray. Each bottle of knockdown spray costs $9 and covers 12 ft2. How much extra cost for knockdown spray should the workers allot for the potential overage in wall area?

¢A = = = = =

A1x 22 - A1x 12 A1100.52 - A11002 100.52 - 1002 10,100.25 - 10,000 100.25 ft2.

In this case, the actual difference in area is ¢A = 100.25 ft2, which again compares well with the approximate value of dA = +100 ft2. c) The tiles (each measuring 1 ft2) come 12 to a box. Thus, if the room were exactly 100 ft by 100 ft (an area of 10,000 ft2), Michelle would need 10,000>12 = 833.33 Á , or 834 boxes to cover the floor. To take into account the possibility that the room is larger by 100 ft2, she needs a total of 10,100>12 = 841.67 Á , or 842 boxes of tiles. Therefore, she should buy 8 extra boxes of tiles, meaning an extra cost of 1821242 = $192.

O Quick Check 3 We see that there is an advantage to using a differential to calculate an approximate difference in an output variable. There is less actual calculating, and the result is often quite accurate. Compare the arithmetic steps needed in parts (a) and (b) of Example 6. Even though dA is an approximation, it is accurate enough for Michelle’s needs: it is sufficient for her to know that the area can be off by as much as “about” 100 ft2. Historically, differentials were quite valuable when used to make approximations. However, with the advent of computers and graphing calculators, such use has diminished considerably. The use of marginals remains important in the study of business and economics.

Section Summary • If C1x2 represents the cost for producing x items, then marginal cost C¿1x2 is its derivative, and C¿1x2 L C1x + 12 - C1x2. Thus, the cost to produce the 1x + 12st item can be approximated by C1x + 12 L C1x2 + C¿1x2. • If R1x2 represents the revenue from selling x items, then marginal revenue R¿1x2 is its derivative, and R¿1x2 L R1x + 12 - R1x2. Thus, the revenue from the 1x + 12st item can be approximated by R1x + 12 L R1x2 + R¿1x2. • If P(x) represents profit from selling x items, then marginal profit P¿1x2 is its derivative, and P¿1x2 L P1x + 12 - P1x2. Thus, the profit from the 1x + 12st item can be approximated by P1x + 12 L P1x2 + P¿1x2.

• In general, profit = revenue - cost, or P1x2 = R1x2 - C1x2. • In delta notation, ¢x = 1x + h2 - x = h, and ¢y = f1x + h2 - f1x2. For small values of ¢x, we have ¢y L f ¿1x2, which is equivalent to ¢y L f ¿1x2 ¢x. ¢x dy • The differential of x is dx = ¢x. Since = f ¿1x2, we dx have dy = f ¿1x2 dx. In general, dy L ¢y, and the approximation can be very close for sufficiently small dx.

Exercise Set 2.6

285

EXERCISE SET

2.6 APPLICATIONS Business and Economics 1. Marginal revenue, cost, and profit. Let R1x2, C1x2, and P1x2 be, respectively, the revenue, cost, and profit, in dollars, from the production and sale of x items. If R1x2 = 5x

and C1x2 = 0.001x 2 + 1.2x + 60,

find each of the following. a) P1x2 b) R11002, C11002, and P11002 c) R¿1x2, C¿1x2, and P¿1x2 d) R¿11002, C¿11002, and P¿11002 e) Describe in words the meaning of each quantity in parts (b) and (d). 2. Marginal revenue, cost, and profit. Let R1x2, C1x2, and P1x2 be, respectively, the revenue, cost, and profit, in dollars, from the production and sale of x items. If R1x2 = 50x - 0.5x 2 and C1x2 = 4x + 10, find each of the following. a) P1x2 b) R1202, C1202, and P1202 c) R¿1x2, C¿1x2, and P¿1x2 d) R¿1202, C¿1202, and P¿1202 3. Marginal cost. Suppose that the monthly cost, in dollars, of producing x chairs is C1x2 = 0.001x 3 + 0.07x 2 + 19x + 700, and currently 25 chairs are produced monthly. a) What is the current monthly cost? b) What would be the additional cost of increasing production to 26 chairs monthly? c) What is the marginal cost when x = 25? d) Use marginal cost to estimate the difference in cost between producing 25 and 27 chairs per month. e) Use the answer from part (d) to predict C(27). 4. Marginal cost. Suppose that the daily cost, in dollars, of producing x radios is C1x2 = 0.002x 3 + 0.1x 2 + 42x + 300, and currently 40 radios are produced daily. a) What is the current daily cost? b) What would be the additional daily cost of increasing production to 41 radios daily? c) What is the marginal cost when x = 40? d) Use marginal cost to estimate the daily cost of increasing production to 42 radios daily. 5. Marginal revenue. Pierce Manufacturing determines that the daily revenue, in dollars, from the sale of x lawn chairs is R1x2 = 0.005x 3 + 0.01x 2 + 0.5x.

Currently, Pierce sells 70 lawn chairs daily. a) What is the current daily revenue? b) How much would revenue increase if 73 lawn chairs were sold each day? c) What is the marginal revenue when 70 lawn chairs are sold daily? d) Use the answer from part (c) to estimate R1712, R1722, and R1732. 6. Marginal profit. For Sunshine Motors, the weekly profit, in dollars, of selling x cars is P1x2 = - 0.006x 3 - 0.2x 2 + 900x - 1200, and currently 60 cars are sold weekly. a) What is the current weekly profit? b) How much profit would be lost if the dealership were able to sell only 59 cars weekly? c) What is the marginal profit when x = 60? d) Use marginal profit to estimate the weekly profit if sales increase to 61 cars weekly. 7. Marginal profit. Crawford Computing finds that its weekly profit, in dollars, from the production and sale of x laptop computers is P1x2 = - 0.004x 3 - 0.3x 2 + 600x - 800. Currently Crawford builds and sells 9 laptops weekly. a) What is the current weekly profit? b) How much profit would be lost if production and sales dropped to 8 laptops weekly? c) What is the marginal profit when x = 9? d) Use the answers from parts 1a2–1c2 to estimate the profit resulting from the production and sale of 10 laptops weekly. 8. Marginal revenue. Solano Carriers finds that its monthly revenue, in dollars, from the sale of x carry-on suitcases is R1x2 = 0.007x 3 - 0.5x 2 + 150x. Currently Solano is selling 26 carry-on suitcases monthly. a) What is the current monthly revenue? b) How much would revenue increase if sales increased from 26 to 28 suitcases? c) What is the marginal revenue when 26 suitcases are sold? d) Use the answers from parts (a)–(c) to estimate the revenue resulting from selling 27 suitcases per month. 9. Sales. Let N1x2 be the number of computers sold annually when the price is x dollars per computer. Explain in words what occurs if N110002 = 500,000 and N¿110002 = - 100. 10. Sales. Estimate the number of computers sold in Exercise 9 if the price is raised to $1025.

CHAPTER 2

•

Applications of Differentiation

For Exercises 11–16, assume that C1x2 and R1x2 are in dollars and x is the number of units produced and sold. 11. For the total-cost function C1x2 = 0.01x 2 + 0.6x + 30, find ¢C and C¿1x2 When x = 70 and ¢x = 1. 12. For the total-cost function C1x2 = 0.01x 2 + 1.6x + 100, find ¢C and C¿1x2 when x = 80 and ¢x = 1. 13. For the total-revenue function R1x2 = 2x, find ¢R and R¿1x2 when x = 70 and ¢x = 1. 14. For the total-revenue function R1x2 = 3x, find ¢R and R¿1x2 when x = 80 and ¢x = 1. 15. a) Using C1x2 from Exercise 11 and R1x2 from Exercise 13, find the total profit, P1x2. b) Find ¢P and P¿1x2 when x = 70 and ¢x = 1. 16. a) Using C1x2 from Exercise 12 and R1x2 from Exercise 14, find the total profit, P1x2. b) Find ¢P and P¿1x2 when x = 80 and ¢x = 1. 17. Marginal demand. The demand, D, for a new rollerball pen is given by D = 0.007p 3 - 0.5p 2 + 150p, where p is the price in dollars. a) Find the rate of change of quantity with respect to price, dD>dp. b) How many units will consumers want to buy when the price is $25 per unit? c) Find the rate of change at p = 25, and interpret this result. d) Would you expect dD>dp to be positive or negative? Why? 18. Marginal productivity. An employee’s monthly productivity, M, in number of units produced, is found to be a function of the number of years of service, t. For a certain product, the productivity function is given by M1t2 = - 2t 2 + 100t + 180. a) Find the productivity of an employee after 5 yr, 10 yr, 25 yr, and 45 yr of service. b) Find the marginal productivity. c) Find the marginal productivity at t = 5; t = 10; t = 25; t = 45; and interpret the results. d) Explain how the employee’s marginal productivity might be related to experience and to age. 19. Average cost. The average cost for a company to produce x units of a product is given by the function A1x2 =

13x + 100 . x

Use A¿1x2 to estimate the change in average cost as production goes from 100 units to 101 units.

20. Supply. A supply function for a certain product is given by S1p2 = 0.08p 3 + 2p 2 + 10p + 11, where S1p2 is the number of items produced when the price is p dollars. Use S¿1p2 to estimate how many more units a producer will supply when the price changes from $18.00 per unit to $18.20 per unit. 21. Gross domestic product. The U.S. gross domestic product, in billions of current dollars, may be modeled by the function P1x2 = 567 + x136x 0.6 - 1042, where x is the number of years since 1960. (Source: U.S. Bureau for Economic Analysis.) Use P¿1x2 to estimate how much the gross domestic product increased from 2009 to 2010. 22. Advertising. Norris Inc. finds that it sells N units of a product after spending x thousands of dollars on advertising, where N1x2 = - x 2 + 300x + 6. Use N¿1x2 to estimate how many more units Norris will sell by increasing its advertising expenditure from $100,000 to $101,000. Marginal tax rate. Businesses and individuals are frequently concerned about their marginal tax rate, or the rate at which the next dollar earned is taxed. In progressive taxation, the 80,001st dollar earned is taxed at a higher rate than the 25,001st dollar earned and at a lower rate than the 140,001st dollar earned. Use the graph below, showing the marginal tax rate for 2005, to answer Exercises 23–26. Marginal tax rate (percentage)

286

50% 40 30 20 10

50

100

150

200

250

300

350

400

Annual income (in thousands of dollars) (Source: “Towards Fundamental Tax Reform” by Alan Auerbach and Kevin Hassett, New York Times, 5/5/05, p. C2.)

23. Was the taxation in 2005 progressive? Why or why not? 24. Marcy and Tyrone work for the same marketing agency. Because she is not yet a partner, Marcy’s year-end income is approximately $95,000; Tyrone’s year-end income is approximately $150,000. Suppose that one of them is to receive another $5000 in income for the year. Which one would keep more of that $5000 after taxes? Why? 25. Alan earns $25,000 per year and is considering a second job that would earn him another $2000 annually. At what rate will his tax liability (the amount he must pay in taxes) change if he takes the extra job? Express your answer in tax dollars paid per dollar earned.

Exercise Set 2.6

26. Iris earns $50,000 per year and is considering extra work that would earn her an extra $3000 annually. At what rate will her tax liability grow if she takes the extra work (see Exercise 25)?

287

Joanne is 160 cm tall and weighs 60 kg. Use a differential to estimate how much her surface area changes after her weight decreases by 1 kg.

Find ¢y and f œ 1x2¢x. Round to four and two decimal places, respectively.

56. Healing wound. The circular area of a healing wound is given by A = pr 2, where r is the radius, in centimeters. By approximately how much does the area decrease when the radius is decreased from 2 cm to 1.9 cm? Use 3.14 for p.

28. For y = f 1x2 = x 3, x = 2, and ¢x = 0.01

57. Medical dosage. The function

27. For y = f 1x2 = x 2, x = 2, and ¢x = 0.01

29. For y = f 1x2 = x + x 2, x = 3, and ¢x = 0.04 30. For y = f 1x2 = x - x 2, x = 3, and ¢x = 0.02 31. For y = f 1x2 = 1>x , x = 1, and ¢x = 0.5 2

32. For y = f 1x2 = 1>x, x = 1, and ¢x = 0.2

33. For y = f 1x2 = 3x - 1, x = 4, and ¢x = 2

34. For y = f 1x2 = 2x - 3, x = 8, and ¢x = 0.5 f œ 1x2¢x

Use ¢y « to find a decimal approximation of each radical expression. Round to three decimal places. 35. 226

36. 210

37. 2102

38. 2103

3

40. 228

Find dy. 41. y = 2x + 1

43. y = 12x + 12 3

3>2

5

45. y = 2x + 27

42. y = 23x - 2

44. y = x 312x + 522 46. y =

0.8t + 1000 5t + 4

gives the bodily concentration N1t2, in parts per million, of a dosage of medication after time t, in hours. Use differentials to determine whether the concentration changes more from 1.0 hr to 1.1 hr or from 2.8 hr to 2.9 hr.

General Interest 58. Major League ticket prices. The average ticket price of a major league baseball game can be modeled by the function p1x2 = 0.09x 2 - 0.19x + 9.41, where x is the number of years after 1990. (Source: Major League Baseball.) Use differentials to predict whether ticket price will increase more between 2010 and 2012 or between 2030 and 2031.

3

39. 21005

N1t2 =

x + x + 2 3

x2 + 3

47. y = x 4 - 2x 3 + 5x 2 + 3x - 4 48. y = 17 - x28

49. In Exercise 47, find dy when x = 2 and dx = 0.1. 50. In Exercise 48, find dy when x = 1 and dx = 0.01.

51. For y = 13x - 1025, find dy when x = 4 and dx = 0.03. 52. For y = x 5 - 2x 3 - 7x, find dy when x = 3 and dx = 0.02.

53. For f 1x2 = x 4 - x 2 + 8, use a differential to approximate f 15.12.

54. For f 1x2 = x 3 - 5x + 9, use a differential to approximate f 13.22.

SYNTHESIS Life and Physical Sciences 55. Body surface area. Certain chemotherapy dosages depend on a patient’s surface area. According to the Gehan and George model, S = 0.02235h0.42246w 0.51456, where h is the patient’s height in centimeters, w is his or her weight in kilograms, and S is the approximation to his or her surface area in square meters. (Source: www.halls.md.)

59. Suppose that a rope surrounds the earth at the equator. The rope is lengthened by 10 ft. By about how much is the rope raised above the earth?

Business and Economics 60. Marginal average cost. In Section 1.6, we defined the average cost of producing x units of a product in terms of the total cost C(x) by A1x2 = C1x2>x. Find a general expression for marginal average cost, A¿1x2. 61. Cost and tolerance. A painting firm contracts to paint the exterior of a large water tank in the shape of a half-dome (a hemisphere). The radius of the tank is measured to be 100 ft with a tolerance of ;6 in. ( ; 0.5 ft). (The formula for the surface area of a hemisphere is A = 2pr 2; use 3.14 as an approximation for p.) Each can of paint costs $30 and covers 300 ft2. a) Calculate dA, the approximate difference in the surface area due to the tolerance. b) Assuming the painters cannot bring partial cans of paint to the job, how many extra cans should they bring to cover the extra area they may encounter? c) How much extra should the painters plan to spend on paint to account for the possible extra area? 62. Strategic oil supply. The U.S. Strategic Petroleum Reserve (SPR) stores petroleum in large spherical caverns built into salt deposits along the Gulf of Mexico. (Source: U.S. Department of Energy.) These caverns can be enlarged by filling the void with water, which dissolves the surrounding salt, and then pumping brine out. Suppose a cavern has a radius of 400 ft, which engineers want to enlarge by 5 ft. Use a differential to estimate how much volume will be added to form the enlarged cavern. (The formula for the volume of a sphere is V = 43pr 3; use 3.14 as an approximation for p.)

288

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•

Applications of Differentiation

Marginal revenue. In each of Exercises 63–67, a demand function, p  D1x2, expresses price, in dollars, as a function of the number of items produced and sold. Find the marginal revenue. 63. p = 100 - 2x

64. p = 400 - x

65. p = 500 - x

66. p =

67. p =

4000 + 3 x

OBJECTIVES • Differentiate implicitly. • Solve related-rate problems.

69. Explain the uses of the differential.

Answers to Quick Checks 1. ¢y = - 3.1319875 2. 298 L 99 10 , or 9.9. This is within 0.001 of the actual value of 298. 3. (a) ;20 ft2 (b) $18 (for 2 extra bottles)

3000 + 5 x

2.7

68. Look up “differential” in a book or Web site devoted to math history. In a short paragraph, describe your findings.

Implicit Differentiation and Related Rates* We often write a function with the output variable (usually y) isolated on one side of the equation. For example, if we write y = x 3, we have expressed y as an explicit function of x. Sometimes, with an equation like y 3 + x 2y 5 - x 4 = 27, it may be cumbersome or nearly impossible to isolate the output variable; in such a case, we have an implicit relationship between the variables x and y. Then, we can find the derivative of y with respect to x using a process called implicit differentiation.

Implicit Differentiation Consider the equation y 3 = x. This equation implies that y is a function of x, for if we solve for y, we get 3

y = 2x = x 1>3. We know from our earlier work that

y 2

y3 = x y=

1 −10

−5

5 −1 −2

10

x 1/3 x

dy dx

=

1 -2>3 x . 3

(1)

A method known as implicit differentiation allows us to find dy>dx without solving for y. To do so, we use the Chain Rule, treating y as a function of x, and differentiate both sides of y3 = x with respect to x: d 3 d y = x. dx dx The derivative on the left side is found using the Extended Power Rule: 3y 2

dy dx

= 1.

Remembering that the derivative of y with respect to x is written dy>dx

Finally, we solve for dy>dx by dividing both sides by 3y 2: dy dx

=

1 , or 3y 2

1 -2 y . 3

We can show that this indeed gives us the same answer as equation (1) by replacing y with x 1>3: dy dx

=

1 -2 1 1 y = 1x 1>32-2 = x -2>3. 3 3 3

*This section can be omitted without loss of continuity.

2.7

•

Implicit Differentiation and Related Rates

289

Often, it is difficult or impossible to solve for y and to express dy>dx solely in terms of x. For example, the equation y 3 + x 2y 5 - x 4 = 27 determines y as a function of x, but it would be difficult to solve for y. We can nevertheless find a formula for the derivative of y without solving for y. To do so usually involves d n computing y for various integers n, and hence involves the Extended Power Rule dx in the form dy d n y = ny n - 1 # . dx dx

■ EXAMPLE 1 For y 3 + x 2y 5 - x 4 = 27: TECHNOLOGY CONNECTION

Exploratory Graphicus can be used to graph equations that relate x and y implicity. Press + and then g(x, y)=0 , and enter y 3 + x 2y 3 - x 4 = 27 as y^3+x^21y^32-x^4=27

EXERCISES Graph each equation. 1. y 3 + x 2y 3 - x 4 = 27 2. y 2x + 2x 3y 3 = y + 1

a) Find dy>dx using implicit differentiation. b) Find the slope of the tangent line to the curve at the point 10, 32.

Solution a) We differentiate the term x 2y 5 using the Product Rule. Because y is a function of x, it is critical that dy>dx is included as a factor in the result any time a term involving y is differentiated. When an expression involving just x is differentiated, there is no factor dy>dx. d 3 d Differentiating both 1y + x 2y 5 - x 42 = 1272 sides with respect to x dx dx d 3 d 2 5 d 4 y + x y x = 0 dx dx dx dy dy Using the Extended Power 3y 2 # + x 2 # 5y 4 # + y 5 # 2x - 4x 3 = 0. Rule and the Product Rule dx dx We next isolate those terms with dy>dx as a factor on one side: 3y 2 #

dy dx

+ 5x 2y 4 #

13y 2 + 5x 2y 42

dy dx dy dx dy dx

= 4x 3 - 2xy 5

Adding 4x 3 - 2xy 5 to both sides

= 4x 3 - 2xy 5

Factoring out dy>dx

=

4x 3 - 2xy 5 3y + 5x y 2

2 4

Solving for dy>dx and leaving the answer in terms of x and y

.

b) To find the slope of the tangent line to the curve at 10, 32, we replace x with 0 and y with 3: dy dx

=

4 # 03 - 2 # 0 # 3 5 = 0. 3 # 3 2 + 5 # 02 # 3 4 y (0, 3)

2 1

y3 + x2y5 – x4 = 27

Quick Check 1 For y 2x + 2x 3y 3 = y + 1, find dy>dx using implicit differentiation.

–10

O Quick Check 1

–5

5

10

x

O

290

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•

Applications of Differentiation

It is not uncommon for the expression for dy>dx to contain both variables x and y. When using the derivative to calculate a slope, we must evaluate it at both the x-value and the y-value of the point of tangency. The steps in Example 1 are typical of those used when differentiating implicitly.

To differentiate implicitly: a) Differentiate both sides of the equation with respect to x (or whatever variable you are differentiating with respect to). b) Apply the rules for differentiation (the Power, Product, Quotient, and Chain Rules) as necessary. Any time an expression involving y is differentiated, dy>dx will be a factor in the result. c) Isolate all terms with dy>dx as a factor on one side of the equation. d) If necessary, factor out dy>dx. e) If necessary, divide both sides of the equation to isolate dy>dx.

The demand function for a product (see Section R.5) is often given implicitly.

■ EXAMPLE 2

For the demand equation x = 2200 - p 3, differentiate implicitly

to find dp>dx.

Solution d d x = 2200 - p 3 dx dx dp 1 1200 - p 32-1>2 # 1- 3p 22 # 2 dx 2 - 3p # dp 1 = 3 dx 22200 - p 1 =

22200 - p 3 - 3p

2

=

Using the Extended Power Rule twice

dp dx

Related Rates Suppose that y is a function of x, say y = f1x2, and x is a function of time, t. Since y depends on x and x depends on t, it follows that y depends on t. The Chain Rule gives the following: dy dt

=

dy dx # . dx dt

Thus, the rate of change of y is related to the rate of change of x. Let’s see how this comes up in problems. It helps to keep in mind that any variable can be thought of as a function of time t, even though a specific expression in terms of t may not be given or its rate of change with respect to t may be 0.

■ EXAMPLE 3 Business: Service Area. A restaurant supplier services the restaurants in a circular area in such a way that the radius r is increasing at the rate of 2 mi per year at the moment when r = 5 mi. At that moment, how fast is the area increasing?

2.7

•

Implicit Differentiation and Related Rates

31

Rocky Ripple

37

465

Clermont

70

Indianapolis 5 mi

465

36

Lawrence

65

74

291

Cumberland

40

Beech Grove

70 31

65 465

52 74

Solution The area A and the radius r are always related by the equation for the area of a circle: A = pr 2. We take the derivative of both sides with respect to t:

Quick Check 2 A spherical balloon is deflating, losing 20 cm3 of air per minute. At the moment when the radius of the balloon is 8 cm, how fast is the radius decreasing? 1Hint: V = 43 pr 3.2

dA dr = 2pr # . dt dt

The factor dr>dt results from the Chain Rule and the fact that r is assumed to be a function of t.

At the moment in question, dr>dt = 2 mi>yr and r = 5 mi, so dA mi = 2p15mi2a2 b yr dt 2 mi L 63 square miles per year. = 20p yr

O Quick Check 2 ■ EXAMPLE 4 Business: Rates of Change of Revenue, Cost, and Profit. For Luce Landscaping, the total revenue from the yard maintenance of x homes is given by R1x2 = 1000x - x 2, and the total cost is given by C1x2 = 3000 + 20x. Suppose that Luce is adding 10 homes per day at the moment when the 400th customer is signed. At that moment, what is the rate of change of (a) total revenue, (b) total cost, and (c) total profit?

Solution a)

b)

dR dx dx = 1000 # - 2x # dt dt dt # = 1000 10 - 21400210 = $2000 per day

Differentiating both sides with respect to time Substituting 10 for dx>dt and 400 for x

dC dx = 20 # Differentiating both sides with respect to time dt dt = 201102 = $200 per day

c) Since P = R - C, dP dR dC = dt dt dt = $2000 per day - $200 per day = $1800 per day.

O

292

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•

Applications of Differentiation

Section Summary • If variables x and y are related to one another by an equation but neither variable is isolated on one side of the equation, we say that x and y have an implicit relationship. To find dy>dx without solving such an equation for y, we use implicit differentiation. • Whenever we implicitly differentiate y with respect to x, the factor dy>dx will appear as a result of the Chain Rule.

• To determine the slope of a tangent line at a point on the graph of an implicit relationship, we may need to evaluate the derivative by inserting both the x-value and the y-value of the point of tangency.

EXERCISE SET

2.7 Differentiate implicitly to find dy/dx. Then find the slope of the curve at the given point. 1. x 3 + 2y 3 = 6; 12, - 12

30. 1000 - 300p + 25p 2 = x

3. 2x 2 - 3y 3 = 5; 1- 2, 12

31. Two variable quantities A and B are found to be related by the equation

4. 2x 3 + 4y 2 = - 12; 1- 2, - 12

A 23, 22 B

6. x + y = 1;

1 23 a , b 2 2

2

2

A3 + B 3 = 9. What is the rate of change dA>dt at the moment when A = 2 and dB>dt = 3? 32. Two nonnegative variable quantities G and H are found to be related by the equation

7. 3x 2y 4 = 12; 12, - 12

8. 2x 3y 2 = - 18; 1- 1, 32

G 2 + H 2 = 25.

9. x 3 - x 2y 2 = - 9; 13, - 22

What is the rate of change dH>dt when dG>dt = 3 and G = 0? G = 1? G = 3?

10. x 4 - x 2y 3 = 12; 1- 2, 12 11. xy - x + 2y = 3;

APPLICATIONS

2 a - 5, b 3

Business and Economics

12. xy + y 2 - 2x = 0; 11, - 22

Rates of change of total revenue, cost, and profit. In Exercises 33–36, find the rates of change of total revenue, cost, and profit with respect to time. Assume that R1x2 and C1x2 are in dollars.

13. x 2y - 2x 3 - y 3 + 1 = 0; 12, - 32

14. 4x 3 - y 4 - 3y + 5x + 1 = 0; 11, - 22

33. R1x2 = 50x - 0.5x 2,

Differentiate implicitly to find dy/dx. 15. 2xy + 3 = 0

16. x 2 + 2xy = 3y 2

17. x 2 - y 2 = 16

18. x 2 + y 2 = 25

19. y 5 = x 3

20. y 3 = x 5

21. x y + x y = 11 2 3

3 4

5 3

23. p + p - 3x = 50

24. p + p + 2x = 40

25. xp = 24

26. x p = 108

3

27.

C1x2 = 4x + 10, when x = 30 and dx>dt = 20 units per day 34. R1x2 = 50x - 0.5x 2, C1x2 = 10x + 3,

22. x y + x y = - 19 3 2

For each demand equation in Exercises 23–30, differentiate implicitly to find dp/dx. 3

x 2p + xp + 1 = 1 (Hint: Clear the fraction first.) 2x + p

29. 1p + 421x + 32 = 48

2. 3x 3 - y 2 = 8; 12, 42

5. x 2 - y 2 = 1;

28.

2

3 2

xp = 2 (Hint: Clear the fraction first.) x + p

when x = 10 and dx>dt = 5 units per day 35. R1x2 = 2x, C1x2 = 0.01x 2 + 0.6x + 30, when x = 20 and dx>dt = 8 units per day 36. R1x2 = 280x - 0.4x 2, C1x2 = 5000 + 0.6x 2, when x = 200 and dx>dt = 300 units per day

Exercise Set 2.7

37. Change of sales. Suppose that the price p, in dollars, and number of sales, x, of a certain item follow the equation 5p + 4x + 2px = 60. Suppose also that p and x are both functions of time, measured in days. Find the rate at which x is changing when x = 3, p = 5, and

dp = 1.5. dt

38. Change of revenue. For x and p as described in Exercise 37, find the rate at which the total revenue R = xp is changing when x = 3, p = 5, and

dp = 1.5. dt

Life and Natural Sciences 39. Rate of change of the Arctic ice cap. In a trend that scientists attribute, at least in part, to global warming, the floating cap of sea ice on the Arctic Ocean has been shrinking since 1950. The ice cap always shrinks in summer and grows in winter. Average minimum size of the ice cap, in square miles, can be approximated by

where h is the patient’s height in centimeters, w is the patient’s weight in kilograms, and S is the approximation to the patient’s surface area in square meters. (Source: www.halls.md.) Assume that Tom’s height is a constant 165 cm, but he is on a diet. If he loses 2 kg per month, how fast is his surface area decreasing at the instant he weighs 70 kg? Poiseuille’s Law. The flow of blood in a blood vessel is faster toward the center of the vessel and slower toward the outside. The speed of the blood V, in millimeters per second 1mm/sec2, is given by V

p 1R 2 - r 22, 4Lv

where R is the radius of the blood vessel, r is the distance of the blood from the center of the vessel, and p, L, and v are physical constants related to pressure, length, and viscosity of the blood vessels, respectively. Assume that dV/dt is measured in millimeters per second squared 1mm/sec 22. Use this formula for Exercises 42 and 43.

A = pr 2. In 2005, the radius of the ice cap was approximately 808 mi and was shrinking at a rate of approximately 4.3 mi>yr. (Source: www.gsfc.nasa.gov and the New York Times, 9/29/05.) How fast was the area changing at that time?

293

r R L

42. Assume that r is a constant as well as p, L, and v. a) Find the rate of change dV>dt in terms of R and dR>dt when L = 80 mm, p = 500, and v = 0.003. b) A person goes out into the cold to shovel snow. Cold air has the effect of contracting blood vessels far from the heart. Suppose that a blood vessel contracts at a rate of dR = - 0.0002 mm>sec dt at a place in the blood vessel where the radius R = 0.075 mm. Find the rate of change, dV>dt, at that location.

40. Rate of change of a healing wound. The area of a healing wound is given by A = pr 2. The radius is decreasing at the rate of 1 millimeter per day 1- 1 mm>day2 at the moment when r = 25 mm. How fast is the area decreasing at that moment? 41. Body surface area. Certain chemotherapy dosages depend on a patient’s surface area. According to the Mosteller model, S =

2hw , 60

The flow of blood in a blood vessel can be modeled by Poiseuille’s Law.

294

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•

Applications of Differentiation

43. Assume that r is a constant as well as p, L, and v. a) Find the rate of change dV>dt in terms of R and dR>dt when L = 70 mm, p = 400, and v = 0.003. b) When shoveling snow in cold air, a person with a history of heart trouble can develop angina (chest pains) due to contracting blood vessels. To counteract this, he or she may take a nitroglycerin tablet, which dilates the blood vessels. Suppose that after a nitroglycerin tablet is taken, a blood vessel dilates at a rate of dR = 0.00015 mm>sec dt

47. The volume of a cantaloupe is given by V = 43 pr 3. The radius is growing at the rate of 0.7 cm>week, at a time when the radius is 7.5 cm. How fast is the volume changing at that moment?

SYNTHESIS Differentiate implicitly to find dy/dx. 48. 2x + 2y = 1

at a place in the blood vessel where the radius R = 0.1 mm. Find the rate of change, dV>dt.

General Interest 44. Two cars start from the same point at the same time. One travels north at 25 mph, and the other travels east at 60 mph. How fast is the distance between them increasing at the end of 1 hr? (Hint: D 2 = x 2 + y 2. To find D after 1 hr, solve D 2 = 252 + 602.)

50. y 3 =

49.

x - 1 x + 1

1 x2

+

51. y 2 =

1 y2

= 5

x2 - 1 x2 + 1

52. x 3>2 + y 2>3 = 1

53. 1x - y23 + 1x + y23 = x 5 + y 5 Differentiate implicitly to find d 2y/dx 2. 54. xy + x - 2y = 4

55. y 2 - xy + x 2 = 5

56. x 2 - y 2 = 5

57. x 3 - y 3 = 8

58. Explain the usefulness of implicit differentiation. 59. Look up the word “implicit” in a dictionary. Explain how that definition can be related to the concept of a function that is defined “implicitly.”

D

x

TECHNOLOGY CONNECTION y

45. A ladder 26 ft long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of 5 ft>s, how fast is the height of the top changing (this will be a negative rate) when the lower end is 10 ft from the wall?

Graph each of the following equations. Equations must be solved for y before they can be entered into most calculators. Graphicus does not require that equations be solved for y. 60. x 2 + y 2 = 4 Note: You will probably need to sketch the graph in two parts: y = 24 - x 2 and y = - 24 - x 2. Then graph the tangent line to the graph at the point 1- 1, 232. 61. x 4 = y 2 + x 6 Then graph the tangent line to the graph at the point 1- 0.8, 0.3842. 62. y 4 = y 2 - x 2

26 ft 26 ft

y

x

46. An inner city revitalization zone is a rectangle that is twice as long as it is wide. A diagonal through the region is growing at a rate of 90 m per year at a time when the region is 440 m wide. How fast is the area changing at that point in time?

63. x 3 = y 212 - x2 64. y 2 = x 3

Answers to Quick Checks 1.

dy y 2 + 6x 2y 3 = dx 1 - 2xy - 6x 3y 2

2. Approximately - 0.025 cm>min

CHAPTER 2 SUMMARY KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 2.1 A function is increasing over an open interval I if, for all x in I, f ¿1x2 7 0.

y f′(x4 ) < 0

f′(x2 ) > 0

A function is decreasing over an open interval I if, for all x in I, f ¿1x2 6 0.

f′(x5 ) < 0 f′(x1) > 0 x0

x1

x2

x3 x4

x5 x6

x

f is increasing over the interval 1x 0, x 32 and decreasing over 1x 3, x 62. y

If f is a continuous function, then a critical value is any number c for which f ¿1c2 = 0 or f ¿1c2 does not exist.

f

If f ¿1c2 does not exist, then the graph of f may have a corner or a vertical tangent at 1c, f 1c22.

The ordered pair 1c, f 1c22 is called a critical point.

c1 c2 c3

c4

c5

x

The values c1, c2, c3, c4, and c5 are critical values of f. • • • • • If f is a continuous function, then a relative extremum (maximum or minimum) always occurs at a critical value.

f ¿1c12 does not exist (corner). f ¿1c22 = 0. f ¿1c32 does not exist (vertical tangent). f ¿1c42 = 0. f ¿1c52 = 0. y

f

The converse is not true: a critical value may not correspond to an extremum. c1 c2 c3

c4

c5

x

• The critical point 1c1, f 1c122 is a relative maximum. • The critical point 1c2, f 1c222 is a relative minimum. • The critical point 1c3, f 1c322 is neither a relative maximum nor a relative minimum. • The critical point 1c4, f 1c422 is a relative maximum. • The critical point 1c5, f 1c522 is neither a relative maximum nor a relative minimum. (continued)

Chapter 2 Summary

295

296

C H A P T E R 21

•• Differentiation Applications of Differentiation

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 2.1 (continued ) The First-Derivative Test allows us to classify a critical value as a relative maximum, a relative minimum, or neither.

Find the relative extrema of the function given by f 1x2 =

1 3

x3 -

1 2

x 2 - 20x + 7.

Critical values occur where f ¿1x2 = 0 or f ¿1x2 does not exist. The derivative f ¿1x2 = x 2 - x - 20 exists for all real numbers, so the only critical values occur when f ¿1x2 = 0. Setting x 2 - x - 20 = 0 and solving, we have x = - 4 and x = 5 as the critical values. To apply the First-Derivative Test, we check the sign of f ¿1x2 to the left and the right of each critical value, using test values: x = -5

x = 0

x = 6

Sign of f ¿1x2

f ¿1- 52 7 0

f ¿102 6 0

f ¿162 7 0

Result

f increasing

f decreasing

f increasing

Test Value

Therefore, there is a relative maximum at x = - 4: f 1- 42 = 57 23.There is a relative minimum at x = 5: f 152 = - 63 56.

SECTION 2.2 If the graph of f is smooth and continuous, then the second derivative, f –1x2, determines the concavity of the graph. If f –1x2 7 0 for all x in an open interval I, then the graph of f is concave up over I. If f –1x2 6 0 for all x in an open interval I, then the graph of f is concave down over I.

A point of inflection occurs at 1x 0, f 1x 022 if f –1x 02 = 0 and there is a change in concavity on either side of x 0.

The function given by f 1x2 =

1 3

x3 -

1 2

x 2 - 20x + 7

has the second derivative f –1x2 = 2x - 1. Setting the second derivative equal to 0, we have x 0 = 12. Using test values, we can check the concavity on either side of x 0 = 12:

Test Value Sign of f –1x2 Result

x = 0

x = 1

f –102 6 0

f –112 7 0

f is concave down

f is concave up

Therefore, the function is concave down over the interval A - q , 12 B and concave up over the interval A 12, q B . Since there is a change in concavity on either side of x 0 = 12, we also conclude that the point A 12, - 3 121 B is a 1 1 point of inflection, where f A 2 B = - 3 12.

The Second-Derivative Test can also be used to classify relative extrema:

For the function

If f ¿1c2 = 0 and f –1c2 7 0, then f1c2 is a relative minimum.

evaluating the second derivative, f –1x2 = 2x - 1, at the critical values yields the following conclusions:

If f ¿1c2 = 0 and f –1c2 6 0, then f1c2 is a relative maximum. If f ¿1c2 = 0 and f –1c2 = 0, then the First-Derivative Test must be used to classify f1c2.

f 1x2 =

1 3

x3 -

1 2

x 2 - 20x + 7,

• At x = - 4, we have f –1- 42 6 0. Since f ¿1- 42 = 0 and the graph is concave down, we conclude that there is a relative maximum at x = - 4. • At x = 5, we have f –152 7 0. Since f ¿152 = 0 and the graph is concave up, we conclude that there is a relative minimum at x = 5.

Chapter 2 Summary

KEY TERMS AND CONCEPTS

297

EXAMPLES

SECTION 2.3 A line x = a is a vertical asymptote if lim- f 1x2 = q ,

Consider the function given by f 1x2 =

x:a

lim- f 1x2 = - q ,

x:a

lim+ f 1x2 = q ,

lim f 1x2 = - q .

x : a+

The graph of a rational function never crosses a vertical asymptote

x + x - 6

f 1x2 =

1x + 121x - 12

1x + 321x - 22

lim f 1x2 = q

and

lim f 1x2 = - q

and

x : - 3-

x : 2-

lim f 1x2 = b

x: - q

or

.

This expression is simplified. Therefore, x = - 3 and x = 2 are vertical asymptotes since

and

A line y = b is a horizontal asymptote if

.

Factoring, we have

x:a

or

x2 - 1 2

lim f 1x2 = - q

x : - 3+

lim f 1x2 = q .

x : 2+

Also, y = 1 is a horizontal asymptote since lim f 1x2 = 1 and

x: q

lim f 1x2 = 1.

x: - q

lim f 1x2 = b.

x: q

The graph of a function can cross a horizontal asymptote. An asymptote is usually sketched as a dashed line; it is not part of the graph itself. For a rational function of the form f 1x2 = p 1x2>q1x2, a slant asymptote occurs if the degree of the numerator is 1 greater than the degree of the denominator.

10 x2 + 1 . Long division yields f 1x2 = x - 3 + . As x : q x + 3 x + 3 10 : 0. Therefore, the slant asymptote is or x : - q , the remainder x + 3 y = x - 3. Let f 1x2 =

y

f

5 −10

−5

5

−10

10

x

Slant asymptote: y=x−3

−15

Asymptotes, extrema, x- and y-intercepts, points of inflection, concavity, and intervals of increasing or decreasing are all used in the strategy for accurate graph sketching.

Consider the function given by f 1x2 = 13 x 3 - 12 x 2 - 20x + 7.

• f has a relative maximum point at A - 4, 57 23 B and a relative minimum point at A 5, - 63 65 B . • f has a point of inflection at

A 12, - 3121 B .

(continued)

298

CHAPTER 2

•

Applications of Differentiation

KEY TERMS AND CONCEPTS SECTION 2.3 (continued )

EXAMPLES • f is increasing over the interval 1- q , - 42 and over the interval 15, q 2, decreasing over the interval 1- 4, 52, concave down over the interval A - q , 12 B , and concave up over the interval A 12, q B . • f has a y-intercept at 10, 72. y (–4, 57.66...) 100

−10

−5

−100

Consider the function given by f 1x2 = • • • •

f

5

10

(5, –63.83

x2 - 1 . x2 + x - 6

f has vertical asymptotes x = - 3 and x = 2. f has a horizontal asymptote given by y = 1. f has a y-intercept at A 0, 16 B . f has x-intercepts at 1- 1, 02 and 11, 02. y 5

f

y=1 −5

5

x = –3

x

x=2 −5

SECTION 2.4 If f is continuous over a closed interval 3a, b4, then the Extreme-Value Theorem tells us that f will have both an absolute maximum value and an absolute minimum value over this interval. One or both points may occur at an endpoint of this interval. Maximum–Minimum Principle 1 can be used to determine these absolute extrema: We find all critical values c1, c2, c3, Á , cn, in 3a, b4, then evaluate

y

Absolute maximum at b Relative maximum at c3

Maxima

Minima

Relative minima at c1 and c4

f 1a2, f 1c12, f 1c22, f 1c32, Á , f 1cn2, f 1b2.

The largest of these is the absolute maximum, and the smallest is the absolute minimum.

Absolute minimum at c1 a

c1

c2

c3 c4

b

x

Chapter 2 Summary

KEY TERMS AND CONCEPTS

299

EXAMPLES

SECTION 2.4 (continued ) If f is differentiable for all x in an interval I, and there is exactly one critical value c in I such that f ¿1c2 = 0, then, according to Maximum–Minimum Principle 2, f 1c2 is an absolute minimum if f –1c2 7 0 or an absolute maximum if f –1c2 6 0.

Let f 1x2 = x +

2 2 , for x 7 0. The derivative is f ¿1x2 = 1 - 2 . We solve x x for the critical value: 1 -

2 = 0 x2 x2 = 2 x = ; 22.

The only critical value over the interval where x 7 0 is x = 22. The 4 4 7 0. second derivative is f –1x2 = 3 . We see that f – A 22 B = x A 22 B 3 Therefore, A 22, f A 22 B B is an absolute minimum. y 4

f 2

(√2, f(√2))

2

x

4

SECTION 2.5 Many real-world applications involve maximum–minimum problems.

See Examples 1–7 in Section 2.5 and the problem-solving strategy on p. 263.

SECTION 2.6

• C¿1x2 L C1x + 12 - C1x2, so C1x + 12 L C1x2 + C¿1x2. • R¿1x2 L R1x + 12 - R1x2, so R1x + 12 L R1x2 + R¿1x2. • P¿1x2 L P1x + 12 - P1x2, so P1x + 12 L P1x2 + P¿1x2. Delta notation represents the change in a variable: ¢x = x 2 - x 1 and

¢y = f 1x 22 - f 1x 12.

y y = C(x) Cost

Marginal cost, marginal revenue, and marginal profit are estimates of the cost, revenue, and profit for the 1x + 12st item produced:

C′(x) ≈ C(x + 1) − C(x)

x

x+1 Quantity produced

x

For f 1x2 = x 2, let x 1 = 2 and x 2 = 2.1. Then, ¢x = 2.1 - 2 = 0.1. Since f 1x 12 = f 122 = 4 and f 1x 22 = f 12.12 = 4.41, we have ¢y = f 1x 22 - f 1x 12 = 4.41 - 4 = 0.41.

Since ¢x = 0.1 is small, we can approximate ¢y by ¢y L f ¿1x2 ¢x.

If x 2 = x 1 + h, then ¢x = h.

The derivative is f ¿1x2 = 2x. Therefore,

If ¢x is small, then the derivative can be used to approximate ¢y:

This approximation, 0.4, is very close to the actual difference, 0.41.

¢y L f ¿1x2 # ¢x.

¢y L f ¿122 # 10.12 = 2122 # 10.12 = 0.4.

(continued)

300

CHAPTER 2

•

Applications of Differentiation

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 2.6 (continued ) Differentials allow us to approximate changes in the output variable y given a change in the input variable x: dx = ¢x

3 x, find dy when x = 27 and dx = 2. For y = 2 dy 1 1 Note that . Thus, dy = dx. Evaluating, we have = 2 dx 32 3 x 32 3 x2 dy =

and dy = f ¿1x2 dx. If ¢x is small, then dy L ¢y. In practice, it is often simpler to calculate dy, and it will be very close to the true value of ¢y.

1

321272 3

2

122 =

2 L 0.074. 27

3 29, using the fact This result can be used to approximate the value of 2 3 27 + dy: 3 29 L 2 that 2 3 27 + dy L 3 + 2 3 29 = 2

2 L 3.074. 27

3 29 L 3.074 is very close to the actual value, Thus, the approximation 2 2 3 29 = 3.07231 Á .

SECTION 2.7 dy

if y 5 = x 3 + 7.

If an equation has variables x and y and y is not isolated on one side of the equation, the derivative dy>dx can be found without solving for y by the method of implicit differentiation.

Find

A related rate occurs when the rate of change of one variable (with respect to time) can be calculated in terms of the rate of change (with respect to time) of another variable of which it is a function.

A cube of ice is melting, losing 30 cm3 of its volume (V) per minute. When the side length (x) of the cube is 20 cm, how fast is the side length decreasing?

dx

We differentiate both sides with respect to x, then solve for

dy dx

.

d 5 d 3 d y = x + 7 dx dx dx dy = 3x 2 5y 4 dx dy 3x 2 . = dx 5y 4

Since V = x 3 and both V and x are changing with time, we differentiate each variable with respect to time: dx dV = 3x 2 . dt dt We have x = 20 and

dV = - 30. Evaluating gives dt

- 30 = 312022

dx dt

or 30 dx = = - 0.025 cm>min. dt 312022

CHAPTER 2 REVIEW EXERCISES These review exercises are for test preparation. They can also be used as a practice test. Answers are at the back of the book. The blue bracketed section references tell you what part(s) of the chapter to restudy if your answer is incorrect.

f)

6. A function with one critical value and a second derivative that is always positive

y

x

CONCEPT REINFORCEMENT Match each description in column A with the most appropriate graph in column B. [2.1–2.4]

Column A 1. A function with a relative maximum but no absolute extrema

Column B a)

y

7. A function with a first derivative that is always positive

g)

y

x x

2. A function with both a vertical asymptote and a horizontal asymptote

b)

In Exercises 8–13, classify each statement as either true or false.

y

8. Every continuous function has at least one critical value. [2.1] x

3. A function that is concave up and decreasing

c)

10. If f ¿1c2 = 0 and f –1c2 7 0, then f(c) is a relative minimum. [2.2]

y

11. If f ¿1c2 = 0 and f –1c2 = 0, then f(c) cannot be a relative minimum. [2.2] x

4. A function that is concave up and increasing

d)

12. If the graph of f 1x2 = P1x2>Q1x2 has a horizontal asymptote, then the degree of the polynomial P1x2 must be the same as that of the polynomial Q1x2. [2.3] 13. Absolute extrema of a continuous function f always occur at the endpoints of a closed interval. [2.4]

y

REVIEW EXERCISES x

5. A function with three critical values

9. If a continuous function y = f 1x2 has extrema, they will occur where f ¿1x2 = 0. [2.1]

e)

For each function given, find any extrema, along with the x-value at which they occur. Then sketch a graph of the function. [2.1]

15. f 1x2 = x 4 - 2x 2 + 3

14. f 1x2 = 4 - 3x - x 2

y

16. f 1x2 = x

- 8x

17. f 1x2 = 4 + 1x - 123

x + 1 2

18. f 1x2 = x 3 + x 2 - x + 3

19. f 1x2 = 3x 2>3

20. f 1x2 = 2x 3 - 3x 2 - 12x + 10 21. f 1x2 = x 3 - 3x + 2

Chapter 2 Review Exercises

301

302

CHAPTER 2

•

Applications of Differentiation

Sketch the graph of each function. List any minimum or maximum values and where they occur, as well as any points of inflection. State where the function is increasing or decreasing, as well as where it is concave up or concave down. [2.2] 22. f 1x2 =

1 3 x + 3x 2 + 9x + 2 3

23. f 1x2 = x 2 - 10x + 8

24. f 1x2 = 4x 3 - 6x 2 - 24x + 5 26. f 1x2 = 3x 4 + 2x 3 - 3x 2 + 1 (Round to three decimal places where appropriate.) 27. f 1x2 = 15 x 5 + 34 x 4 - 43 x 3 + 8 (Round to three decimal places where appropriate.) Sketch the graph of each function. Indicate where each function is increasing or decreasing, the coordinates at which relative extrema occur, where any asymptotes occur, where the graph is concave up or concave down, and where any intercepts occur. [2.3] 2x + 5 28. f 1x2 = x + 1

x 29. f 1x2 = x - 2

30. f 1x2 =

31. f 1x2 = -

5 x - 16

x 2 - 2x + 2 32. f 1x2 = x - 1

x + 1 x - x - 2 2

x2 + 3 33. f 1x2 = x

Find the absolute maximum and minimum values of each function, if they exist, over the indicated interval. Indicate the x-value at which each extremum occurs. Where no interval is specified, use the real line. [2.4] 34. f 1x2 = x4 - 2x2 + 3; 30, 34 35. f 1x2 = 8x2 - x3; 3 - 1, 84 36. f 1x2 = x +

42. Business: minimizing inventory cost. A store in California sells 360 hybrid bicycles per year. It costs $8 to store one bicycle for a year. To reorder, there is a fixed cost of $10, plus $2 for each bicycle. How many times per year should the store order bicycles, and in what lot size, in order to minimize inventory costs? [2.5] 43. Business: marginal revenue. Crane Foods determines that its daily revenue, R(x), in dollars, from the sale of x frozen dinners is

25. f 1x2 = x 4 - 2x 2

2

top is $3, and for the sides is $1, what should the dimensions be in order to minimize the cost? [2.5]

50 ; 10, q 2 x

37. f 1x2 = x 4 - 2x 2 + 1 38. Of all numbers whose sum is 60, find the two that have the maximum product. [2.5] 39. Find the minimum value of Q = x 2 - 2y 2, where x - 2y = 1. [2.5] 40. Business: maximizing profit. If R1x2 = 52x - 0.5x 2 and C1x2 = 22x - 1, find the maximum profit and the number of units that must be produced and sold in order to yield this maximum profit. Assume that R1x2 and C1x2 are in dollars. [2.5] 41. Business: minimizing cost. A rectangular box with a square base and a cover is to have a volume of 2500 ft3. If the cost per square foot for the bottom is $2, for the

R1x2 = 4x 3>4. a) What is Crane’s daily revenue when 81 frozen dinners are sold? b) What is Crane’s marginal revenue when 81 frozen dinners are sold? c) Use the answers from parts (a) and (b) to estimate R(82). [2.6] For Exercises 44 and 45, y  f 1x2  2x 3  x. [2.6] 44. Find ¢y and dy, given that x = 1 and ¢x = - 0.05. 45. a) Find dy. b) Find dy when x = - 2 and dx = 0.01. 46. Approximate 283 using ¢y L f ¿1x2 ¢x. [2.6] 47. Physical science: waste storage. The Waste Isolation Pilot Plant (WIPP) in New Mexico consists of large rooms carved into a salt deposit and is used for long-term storage of radioactive waste. (Source: www.wipp.energy.gov.) A new storage room in the shape of a cube with an edge length of 200 ft is to be carved into the salt. Use a differential to estimate the potential difference in the volume of this room if the edge measurements have a tolerance of ;2 ft. [2.6] 48. Differentiate the following implicitly to find dy>dx. Then find the slope of the curve at the given point. [2.7] 2x 3 + 2y 3 = - 9xy; 1- 1, - 22

49. A ladder 25 ft long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of 6 ft>sec, how fast is the height of the top decreasing when the lower end is 7 ft from the wall? [2.7] 50. Business: total revenue, cost, and profit. Find the rates of change, with respect to time, of total revenue, cost, and profit for R1x2 = 120x - 0.5x 2 and

C1x2 = 15x + 6,

when x = 100 and dx>dt = 30 units per day. Assume that R(x) and C(x) are in dollars. [2.7]

SYNTHESIS 51. Find the absolute maximum and minimum values, if they exist, over the indicated interval. [2.4] f 1x2 = 1x - 322>5; 1- q , q 2

Chapter 2 Test

52. Find the absolute maximum and minimum values of the piecewise-defined function given by 2 - x 2, f 1x2 = c 3x - 2, 1x - 422,

for - 2 … x … 1, for 1 6 x 6 2, for 2 … x … 6. [2.4]

53. Differentiate implicitly to find dy>dx:

1x - y24 + 1x + y24 = x 6 + y 6. [2.7]

54. Find the relative maxima and minima of y = x 4 - 8x 3 - 270x 2. [2.1 and 2.2] 55. Determine a rational function f whose graph has a vertical asymptote at x = - 2 and a horizontal asymptote at y = 3 and includes the point 11, 22. [2.4]

TECHNOLOGY CONNECTION Use a calculator to estimate the relative extrema of each function. [2.1 and 2.2] 56. f 1x2 = 3.8x 5 - 18.6x 3

3 ƒ 9 - x2 ƒ - 1 57. f 1x2 = 2 58. Life and physical sciences: incidence of breast cancer. The following table provides data relating the incidence of breast cancer per 100,000 women of various ages. a) Use REGRESSION to fit linear, quadratic, cubic, and quartic functions to the data.

Age

Incidence per 100,000

0 27 32 37 42 47 52 57 62 67 72 77 82 87

0 10 25 60 125 187 224 270 340 408 437 475 460 420

303

(Source: National Cancer Institute.)

b) Which function best fits the data? c) Determine the domain of the function on the basis of the function and the problem situation, and explain. d) Determine the maximum value of the function on the domain. At what age is the incidence of breast cancer the greatest? [2.1 and 2.2] Note: The function used in Exercise 28 of Section R.1 was found in this manner.

CHAPTER 2 TEST Find all relative minimum or maximum values as well as the x-values at which they occur. State where each function is increasing or decreasing. Then sketch a graph of the function. 1. f 1x2 = x 2 - 4x - 5

2. f 1x2 = 4 + 3x - x 3

3. f 1x2 = 1x - 222>3 - 4

4. f 1x2 =

7. f 1x2 = 1x - 223 + 3

14. f 1x2 = x 3 + x 2 - x + 1;

x + 4

15. f 1x2 = - x 2 + 8.6x + 10

6. f 1x2 = 2x 4 - 4x 2 + 1 8. f 1x2 = x29 - x 2

9. f 1x2 =

2 x - 1

10. f 1x2 =

11. f 1x2 =

x2 - 1 x

12. f 1x2 =

13. f 1x2 = x16 - x2

16 2

Sketch a graph of each function. List any extrema, and indicate any asymptotes or points of inflection. 5. f 1x2 = x 3 + x 2 - x + 1

Find the absolute maximum and minimum values, if they exist, of each function over the indicated interval. Where no interval is specified, use the real line.

-8 x - 4 2

x - 3 x + 2

C - 2, 12 D

16. f 1x2 = - 2x + 5; 3 - 1, 14 17. f 1x2 = - 2x + 5 18. f 1x2 = 3x 2 - x - 1 19. f 1x2 = x 2 +

128 ; 10, q 2 x

20. Of all numbers whose difference is 8, find the two that have the minimum product.

304

CHAPTER 2

•

Applications of Differentiation

21. Minimize Q = x 2 + y 2, where x - y = 10. 22. Business: maximum profit. Find the maximum profit and the number of units, x, that must be produced and sold in order to yield the maximum profit. Assume that R(x) and C(x) are the revenue and cost, in dollars, when x units are produced: R1x2 = x 2 + 110x + 60, C1x2 = 1.1x 2 + 10x + 80. 23. Business: minimizing cost. From a thin piece of cardboard 60 in. by 60 in., square corners are cut out so that the sides can be folded up to make an open box. What dimensions will yield a box of maximum volume? What is the maximum volume? 24. Business: minimizing inventory costs. Ironside Sports sells 1225 tennis rackets per year. It costs $2 to store one tennis racket for a year. To reorder, there is a fixed cost of $1, plus $0.50 for each tennis racket. How many times per year should Ironside order tennis rackets, and in what lot size, in order to minimize inventory costs? 25. For y = f 1x2 = x 2 - 3, x = 5, and ¢x = 0.1, find ¢y and f ¿1x2 ¢x. 26. Approximate 250 using ¢y L f ¿1x2 ¢x. 27. For y = 2x 2 + 3: a) Find dy. b) Find dy when x = 2 and dx = 0.01. 28. Differentiate the following implicitly to find dy>dx. Then find the slope of the curve at 11, 22: x 3 + y 3 = 9. 29. A spherical balloon has a radius of 15 cm. Use a differential to find the approximate change in the volume of the balloon if the radius is increased or decreased by 0.5 cm. (The volume of a sphere is V = 43 pr 3. Use 3.14 for p.) 30. A pole 13 ft long leans against a vertical wall. If the lower end is moving away from the wall at the rate of 0.4 ft>sec, how fast is the upper end coming down when the lower end is 12 ft from the wall?

SYNTHESIS 31. Find the absolute maximum and minimum values of the following function, if they exist, over [0, q 2: f 1x2 =

x2 1 + x3

.

32. Business: minimizing average cost. The total cost in dollars of producing x units of a product is given by C1x2 = 100x + 1002x +

2x 3 . 100

How many units should be produced to minimize the average cost?

TECHNOLOGY CONNECTION 33. Use a calculator to estimate any extrema of this function: f 1x2 = 5x 3 - 30x 2 + 45x + 52x.

34. Use a calculator to estimate any extrema of this function: g1x2 = x 5 - x 3. 35. Business: advertising. The business of manufacturing and selling bowling balls is one of frequent changes. Companies introduce new models to the market about every 3 to 4 months. Typically, a new model is created because of advances in technology such as new surface stock or a new way to place weight blocks in a ball. To decide how to best use advertising dollars, companies track the sales in relation to the amount spent on advertising. Suppose that a company has the following data from past sales. Amount Spent on Advertising (in thousands) $

0 50 100 150 200 250 300

Number of Bowling Balls Sold, N 8 13,115 19,780 22,612 20,083 12,430 4

a) Use REGRESSION to fit linear, quadratic, cubic, and quartic functions to the data. b) Determine the domain of the function in part (a) that best fits the data and the problem situation. Justify your answer. c) Determine the maximum value of the function on the domain. How much should the company spend on advertising its next new model in order to maximize the number of bowling balls sold?

Extended Technology Application Maximum Sustainable Harvest In certain situations, biologists are able to determine what is called a reproduction curve. This is a function y = f1P2 such that if P is the population after P years, then f1P2 is the population a year later, at time t + 1. Such a curve is shown below. y Population 1 year later (at t + 1)

Reproduction function, y = f (P) y=P

( P, f (P))

f(P)

P

f(P)

Population at time t

The line y = P is significant because if it ever coincides with the curve y = f1P2, then we know that the population stays the same from year to year. Here the graph of f lies mostly above the line, indicating that the population is increasing. Too many deer in a forest can deplete the food supply and eventually cause the population to decrease for lack of food. In such cases, often with some controversy, hunters are allowed to “harvest” some of the deer. Then with a greater food supply, the remaining deer population may prosper and increase. We know that a population P will grow to a population f1p2 in a year. If this were a population of fur-bearing animals and the population were increasing, then hunters could “harvest” the amount

y

Reproduction function, y = f ( P)

Maximum sustainable harvest, f (P0 ) − P0

P0

P

Let the harvest function H be given by H1P2 = f1P2 - P.

f1P2 - P each year without shrinking the initial population P. If the population were remaining the same or decreasing, then such a harvest would deplete the population. Suppose that we want to know the value of P0 that would allow the harvest to be the largest. If we could determine that P0, we could let the population grow until it reached that level and then begin harvesting year after year the amount f1P02 - P0.

m=1

Then

H¿1P2 = f ¿1P2 - 1.

Now, if we assume that H¿1P2 exists for all values of P and that there is only one critical value, it follows that the maximum sustainable harvest occurs at that value P0 such that H¿1P02 = f ¿1P02 - 1 = 0

and H–1P02 = f –1P02 6 0.

Extended Technology Application

305

306

CHAPTER 2

•

Applications of Differentiation

Or, equivalently, we have the following.

For Exercises 4 and 5, do the following. a) Graph the reproduction curve, the line y  P, and the

harvest function using the same viewing window.

THEOREM

b) Graphically determine the population at which the

The maximum sustainable harvest occurs at P0 such that

c) Find the maximum sustainable harvest.

f ¿1P02 = 1 and

f –1P02 6 0,

and is given by

H1P02 = f1P02 - P0.

maximum sustainable harvest occurs. 4. f1P2 = 402P, where P is measured in thousands.

Assume that this is the reproduction curve for the brown trout population in a large lake.

EXERCISES For Exercises 1–3, do the following. a) Graph the reproduction curve, the line y  P, and the

harvest function using the same viewing window. b) Find the population at which the maximum

sustainable harvest occurs. Use both a graphical solution and a calculus solution. c) Find the maximum sustainable harvest. 1. f1P2 = P110 - P2, where P is measured in

thousands. 2. f1P2 = - 0.025P 2 + 4P, where P is measured in

thousands. This is the reproduction curve in the Hudson Bay area for the snowshoe hare, a fur-bearing animal.

5. f1P2 = 0.237P22000 - P2, where P is measured in

thousands. 6. The table below lists data regarding the reproduction

of a certain animal. a) Use REGRESSION to fit a cubic polynomial to these

data. b) Graph the reproduction

curve, the line y = P, and the harvest function using the same viewing window. c) Graphically determine the population at which the maximum sustainable harvest occurs.

POPULATION, P (in thousands)

3. f1P2 = - 0.01P 2 + 2P, where P is meas-

ured in thousands. This is the reproduction curve in the Hudson Bay area for the lynx, a fur-bearing animal.

POPULATION, f(P), 1 YEAR LATER

10

9.7

20

23.1

30

37.4

40

46.2

50

42.6

Exponential and Logarithmic Functions

3

Chapter Snapshot What You’ll Learn

Why It’s Important

3.1 Exponential Functions 3.2 Logarithmic Functions 3.3 Applications: Uninhibited and Limited Growth Models 3.4 Applications: Decay 3.5 The Derivatives of ax and loga x 3.6 An Economics Application: Elasticity of Demand

In this chapter, we consider two types of functions that are closely related: exponential functions and logarithmic functions. After learning to find derivatives of such functions, we will study applications in the areas of population growth and decay, continuously compounded interest, spread of disease, and carbon dating.

Where It’s Used

COMIC BOOK VALUE BY YEAR V(t)

A 1939 comic book with the first appearance of the “Caped Crusader,” Batman, sold at auction in Dallas in 2010 for $1.075 million. The comic book originally cost 10¢. What will the value of the comic book be in 2020? After what time will the value of the comic book be $30 million?

Value of comic book

MILLION-DOLLAR COMIC BOOK

$1.075 million

V(t) = 0.10 e 0.228t

$0.10

This problem appears as Example 7 in Section 3.3.

81

Number of years since 1939

t

307

308

CHAPTER 3

•

Exponential and Logarithmic Functions

3.1 OBJECTIVES

Exponential Functions Graphs of Exponential Functions Consider the following graph. The rapid rise of the graph indicates that it approximates an exponential function. We now consider such functions and many of their applications.

• Graph exponential functions.

WORLD POPULATION GROWTH

• Differentiate exponential functions.

8 billion* 2020 6 billion 1998 5 billion 1987 4 billion 1974 3 billion 1960 2 billion 1927 1 billion 1804 1800

1900

2000

2100

Year

*Projected (Source: U.S. Census Bureau.)

Let’s review definitions of expressions of the form a x, where x is a rational number. For example, a 2.34

or a 234>100

means “raise a to the 234th power and then take the 100th root A 100 2a 234 B .” What about expressions with irrational exponents, such as 2 22, 2 p, and 2 -23? An irrational number is a number named by an infinite, nonrepeating decimal. Let’s consider 2 p. We know that p is irrational with an infinite, nonrepeating decimal expansion: 3.141592653 Á . This means that p is approached as a limit by the rational numbers 3, 3.1, 3.14, 3.141, 3.1415, Á , so it seems reasonable that 2 p should be approached as a limit by the rational powers 2 3, 2 3.1, 2 3.14, 2 3.141, 2 3.1415, Á .

TECHNOLOGY CONNECTION We can approximate 2 p using the exponential key: 2 U p [. We obtain

x

EXERCISES Approximate. 1. 5 3. 7

-22

2. 5

8, 8.574188, 8.815241, 8.821353, 8.824411, Á . In general, a x is approximated by the values of a r for rational numbers r near x; a is the limit of a r as r approaches x through rational values. Thus, for a 7 0, the usual laws of exponents, such as

2 p L 8.824977827.

p

Estimating each power with a calculator, we get the following:

23

4. 18-p

a x # a y = a x + y,

a x , a y = a x - y,

1a x2y = a xy, and

a -x =

1 , ax

can be applied to real number exponents. Moreover, the function so obtained, f1x2 = a x, is continuous.

3.1

•

309

Exponential Functions

DEFINITION An exponential function f is given by f1x2 = a x, where x is any real number, a 7 0, and a Z 1. The number a is called the base.

The following are examples of exponential functions: f1x2 = 2 x,

f1x2 =

A 12 B x,

f1x2 = 10.42x.

Note that in contrast to power functions like y = x 2 and y = x 3, an exponential function has the variable in the exponent, not as the base. Exponential functions have countless applications. For now, however, let’s consider their graphs.

■ EXAMPLE 1 Graph: y = f1x2 = 2 x. Solution First, we find some function values. Note that 2 x is always positive: x = 0,

y = 2 0 = 1;

1 , 2 x = 1, x = 2, x = 3,

y = 2 1>2 = 22 L 1.4;

x =

9

y = 2 1 = 2; y = 2 2 = 4; y = 2 3 = 8; 1 x = - 1, y = 2 -1 = ; 2

Quick Check 1 For f1x2 = 3x, complete this table of function values. x

This part of the curve increases without bound. y

x = - 2, y = 2

-2

8 7

5 4 3

1 1 = 2 = . 4 2

2 1

This part of the curve comes very close to the x-axis, but does not touch or cross it. The x-axis is a horizontal asymptote.

f1x2 = 3x

f(x) = 2x

6

–4 –3 – 2 –1

1

2

3

4

x

–1

0 1

x

0

1 2

1

2

3

-1

-2

2

y = f1x2 = 2x

1

1.4

2

4

8

1 2

1 4

3 -1 -2 -3

Graph f1x2 = 3x.

Next, we plot the points and connect them with a smooth curve, as shown above. The graph is continuous, increasing without bound, and concave up. We see too that the x-axis is a horizontal asymptote (see Section 2.3), that is, lim f1x2 = 0 and

x: -q

lim f1x2 = q .

x: q

O Quick Check 1 ■ EXAMPLE 2 Graph: y = g1x2 = A 12 B x. Solution First, we note that y = g1x2 =

A 12 B x

= 12 -12x = 2 -x.

O

310

CHAPTER 3

•

Exponential and Logarithmic Functions

This will ease our work in calculating function values: y = 2 -0 = 1;

x = 0, x =

1 , 2

y = 2 -1>2 = =

Quick Check 2 1 x For g1x2 = a b , complete 3 this table of function values.

O

x

1

L

22

y

1 2 1>2

9 8 7

1 L 0.7; 1.4

5

1 = ; 2

-1

x = 1,

y = 2

x = 2,

y = 2 -2 =

4 x g(x) = ⎛⎝12–⎛⎝ = 2 –x

x

0

3 2

1 ; 4

1

x = - 1, y = 2 -1 - 12 = 2; x = - 2, y = 2 -1 - 22 = 4; x = - 3, y = 2 -1 - 32 = 8.

1 x g1x2 = a b 3

f(x) = 2x

6

–4 –3 – 2 –1

1

2

4

x

0

1 2

1

2

-1

-2

-3

1

0.7

1 2

1 4

2

4

8

y = f1x2 =

1 2

A 12 B x

Next, we plot these points and connect them with a smooth curve, as shown by the red curve in the figure. The graph is continuous, decreasing, and concave up. We see too that the x-axis is a horizontal asymptote, that is,

3 -1

lim g1x2 = 0 and

x: q

-2

lim g1x2 = q .

x: - q

The graph of f1x2 = 2 x of Example 1 is shown as a blue curve, for comparison. Note that the graph of y = g1x2 is the reflection of the graph of y = f1x2 across the y-axis. Thus, we expect the graphs of y = A 12 B x = 2 -x and y = 2 x to be symmetric with respect to the y-axis.

-3

1 x Graph g1x2 = a b . 3

O Quick Check 2

TECHNOLOGY CONNECTION

t

0.004 # 2 t

Start

0

0.004 # 2 0, or 0.004

Step 1

1

0.004 # 2 1, or 0.008

Step 2

2

0.004 # 2 2, or 0.016

Step 3

3

Step 4

4

Step 5

5

Exploratory Exercise: Growth 8 12-in.-by-11-in.

Take a sheet of paper and cut it into two equal pieces. Then cut these again to obtain four equal pieces. Then cut these to get eight equal pieces, and so on, performing five cutting steps.

Start

3

–1

Step 1

Step 2

Step 3

a) Place all the pieces in a stack and measure the thickness.

c) Graph the function f 1t2 = 0.004122t.

b) A piece of paper is typically 0.004 in. thick. Check the measurement in part (a) by completing the table.

d) Compute the thickness of the stack (in miles) after 25 steps.

•

3.1

TECHNOLOGY CONNECTION

Exploratory Use a calculator or iPlot or Graphicus to check the graphs of the functions in Examples 1 and 2. Then graph

y

1. The function given by f1x2 = a x, with a 7 1, is a positive, increasing, continuous function. As x gets smaller, a x approaches 0. The graph is concave up, and the x-axis is the horizontal asymptote.

(2, a 2)

f(x) = a x, a>1

f 1x2 = 3x and g1x2 =

311

Exponential Functions

A B

(–1, a –1)

1 x 3

–4 –3 –2 –1

and look for patterns.

2. The function given by f1x2 = a x, with 0 6 a 6 1, is a positive, decreasing, continuous function. As x gets larger, a x approaches 0. The graph is concave up, and the x-axis is the horizontal asymptote.

(1, a 1) (0, 1) 1

2

3

4

x

y f(x) = a x, 0h has a limit as h approaches 0 and that its approximate value is 0.7; thus, g¿1x2 L 10.722 x.

In other words, the derivative is a constant times 2 x. Similarly, for t1x2 = 3x, t¿1x2 = 3x # lim

3h - 1 . h:0 h

h

3h - 1 h

0.5 0.25 0.175 0.0625 0.03125 0.00111 0.000001

1.4641 1.2643 1.2113 1.1372 1.1177 1.0993 1.0986

Substituting 3 for a

Again, we can find an approximation for the limit that does not depend on the value of x at which we are evaluating the derivative. Consider the table at left. Again, it seems reasonable to conclude that 13h - 12>h has a limit as h approaches 0. This time, the approximate value is 1.1; thus, t¿1x2 L 11.123x.

In other words, the derivative is a constant times 3x. Let’s now analyze what we have done. We proved that if f1x2 = a x, then f ¿1x2 = a x # lim

ah - 1 . h:0 h

ah - 1 . h:0 h

Consider lim

1. For a = 2, TECHNOLOGY CONNECTION

ah - 1 2h - 1 = lim L 0.6931. h:0 h:0 h h lim

2. For a = 3,

EXERCISE Using TBLSET in Ask mode, produce the tables shown above.

ah - 1 3h - 1 = lim L 1.0986. h:0 h:0 h h lim

It seems reasonable to conclude that, for some choice of a between 2 and 3, we have

h

11 + h21>h

0.5

2.25

0.1

2.5937

0.01

2.7048

0.001

2.7169

- 0.01

2.732

- 0.001

2.7196

0.0001

2.7181

ah - 1 = 1. h:0 h lim

To find that a, it suffices to look for a value such that ah - 1 = 1. h

(1)

Multiplying both sides by h and then adding 1 to both sides, we have a h = 1 + h. Raising both sides to the power 1>h, we have a = 11 + h21>h.

(2)

Since equations (1) and (2) are equivalent, it follows that ah - 1 = lim 1 and h:0 h:0 h lim

lim a = lim 11 + h21>h

h:0

h:0

•

3.1

313

Exponential Functions

are also equivalent. Thus, ah - 1 = 1 and h:0 h lim

a = lim 11 + h21>h h:0

ah - 1 = 1, h:0 h we must have a = lim 11 + h21>h. This last equation gives us the special number we are are equivalent. We conclude that, for there to be a number a for which lim h:0

searching for.* The number is named e, in honor of Leonhard Euler (pronounced “Oiler”), the great Swiss mathematician (1707–1783) who did groundbreaking work with it. TECHNOLOGY CONNECTION

Exploratory

Graph f 1x2 = e x using the viewing window 3- 5, 5, - 1, 104. Trace to a point and note the y-value. Then find dy>dx at that point, using the m key, and compare y and dy>dx. Repeat this process for three other values of x. What do you observe? Using iPlot, graph f 1x2 = e x, as a first function in red. Do not turn on Derivate. Graph f 1x2 = e x, as a second function in blue, and turn on Derivate. What happens? Explain. Using Graphicus, graph y = e x. Then use the tangent line feature to move along the curve, noting the x-values, the y-values, and the values of 0y>0x, which are the same as dy>dx in this case. What happens? Explain.

DEFINITION

e = lim 11 + h21>h L 2.718281828459 h:0

We call e the natural base.

It follows that, for the exponential function f1x2 = e x, f ¿1x2 = e x # lim

eh - 1 h:0 h = ex # 1 = e x.

That is, the derivative of e x is e x. We have shown that if f1x2 = e x, it follows that f ¿1x2 = e x.

THEOREM 1 The derivative of the function f given by f1x2 = e x is itself: f ¿1x2 = f1x2,

or

d x e = e x. dx

Theorem 1 says that for the function f1x2 = e x, the derivative at x (the slope of the tangent line) is the same as the function value at x. That is, on the graph of y = e x, at the point 10, 12, the slope is m = 1; at the point 11, e2, the slope is m = e; at the point 12, e 22, the slope is m = e 2, and so on. The function y = e x is the only exponential function for which this correlation between the function and its derivative is true.

y 8

f(2) = f′(2) ≈ 7.39

7 6 5

f(x) = e x = f′(x)

4 3

f(1) = f ′(1) ≈ 2.72

2 1 −2

−1

f(0) = f ′(0) = 1 1

2

3

4

5

*This derivation is based on one presented in Appendix 4 of e: The Story of a Number, by Eli Maor (Princeton University Press, 1998).

6

x

314

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•

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TECHNOLOGY CONNECTION

Exploratory Check the results of Example 3 by entering each function as y1 and letting y2 = nDeriv1y1, x, x2. Then enter the derivatives from Example 3 as y3 and use graphs or a table to compare y2 and y3. Using iPlot, graph f 1x2 = 3e x in red. Turn on Derivate. Now suppose the derivative was mistakenly found to be f ¿1x2 = 3e x - 1. Graph this incorrect function in some other color. What happens? Explain. Then describe a procedure for checking the results of Example 3 using iPlot. Using Graphicus, graph g1x2 = x 2e x as a first function. Then touch + and choose Add derivative. Suppose the derivative was mistakenly found to be g¿1x2 = 2xe x. Graph this incorrect function. What happens? Explain. Then describe a procedure for checking the results of Example 3 using Graphicus.

In Section 3.5, we will develop a formula for the derivative of the more general exponential function given by y = a x.

Finding Derivatives of Functions Involving e We can use Theorem 1 in combination with other theorems derived earlier to differentiate a variety of functions.

■ EXAMPLE 3 Find dy>dx: a) y = 3e x; b) y = x 2e x; c) y =

ex . x3

Solution a)

b)

d d 13e x2 = 3 e x dx dx = 3e x

d 3c # f 1x24 = c # f ¿1x2. dx

d 2 x 1x e 2 = x 2 # e x + e x # 2x dx = e x1x 2 + 2x2, or xe x1x + 22 x

c)

Recall that

d e x a 3b = dx x = =

3

#e

x

- e x6

x

# 3x

Using the Product Rule Factoring

2

x 2e x1x - 32 x6

e x1x - 32 x4

Using the Quotient Rule

Factoring

Simplifying

Suppose that we have a more complicated function in the exponent, as in h1x2 = e x

2

- 5x

.

This is a composition of functions. For such a function, we have h1x2 = g1f1x22 = e f1x2, where g1x2 = e x and f1x2 = x 2 - 5x. Now g¿1x2 = e x. Then by the Chain Rule (Section 1.7), we have h¿1x2 = g¿1f1x22 # f ¿1x2 = e f 1x2 # f ¿1x2.

For the case above, f1x2 = x 2 - 5x, so f ¿1x2 = 2x - 5. Then h¿1x2 = g¿1f1x22 # f ¿1x2

= e f 1x2 # f ¿1x2 2 = e x - 5x12x - 52.

Quick Check 3 Differentiate: a) y = 6e x; b) y = x 3e x; ex c) y = 2 . x

The next theorem, which we have proven using the Chain Rule, allows us to find derivatives of functions like the one above.

O Quick Check 3

O

3.1

•

Exponential Functions

315

THEOREM 2 The derivative of e to some power is the product of e to that power and the derivative of the power: d f 1x2 e = e f 1x2 # f ¿1x2 dx or d u du e = eu # . dx dx

The following gives us a way to remember this rule. 2

h1x2 = e x

- 5x

Rewrite the original function. 2

h¿1x2 = e x

- 5x

12x - 52

Multiply by the derivative of the exponent.

■ EXAMPLE 4 Differentiate each of the following with respect to x: b) y = e -x

2

a) y = e 8x;

+ 4x - 7

;

c) e 2x

2

-3

.

Solution a)

d 8x e = e 8x # 8, or 8e 8x dx

b)

d -x2 + 4x - 7 2 e = e -x + 4x - 71- 2x + 42, or dx

c)

d 2x2 - 3 d 1x2 - 321>2 e = e dx dx = e 1x

2

Quick Check 4

3

b) g1x2 = e x

2

2

= ;

+5

.

2

+ 4x - 7

Using the Chain Rule twice

-3

2

+ 8x

c) h1x2 = e 2x

1x 2 - 32-1>2 # 2x

# x # 1x 2 - 32-1>2 e 2x - 3 # x xe 2x - 3

= e 2x

Differentiate: a) f1x2 = e -4x;

#

- 321>2 1 2

- 21x - 22e -x

2x 2 - 3

2

,

or

2x 2 - 3

O Quick Check 4 Graphs of ex, ex, and 1  ekx Now that we know how to find the derivative of f1x2 = e x, let’s look at the graph of f1x2 = e x from the standpoint of calculus concepts and the curve-sketching techniques discussed in Section 2.2.

O

316

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Exponential and Logarithmic Functions

■ EXAMPLE 5 Graph: f1x2 = e x. Analyze the graph using calculus. Solution We simply find some function values using a calculator, plot the points, and sketch the graph as shown below. y

x

f1x2

-2

0.135

-1

0.368

8

0

7 6 5

f(x) = e x

4

1

3

1

2.718

2

7.389

2 1 – 2 –1 0

1

x

2

We analyze the graph using calculus as follows. a) Derivatives. Since f1x2 = e x, it follows that f ¿1x2 = e x, so f –1x2 = e x. b) Critical values of f. Since f ¿1x2 = e x 7 0 for all real numbers x, we know that the derivative exists for all real numbers and there is no solution of the equation f ¿1x2 = 0. There are no critical values and therefore no maximum or minimum values. c) Increasing. We have f ¿1x2 = e x 7 0 for all real numbers x, so the function f is increasing over the entire real line, 1- q , q 2. d) Inflection points. We have f –1x2 = e x 7 0 for all real numbers x, so the equation f –1x2 = 0 has no solution and there are no points of inflection. e) Concavity. Since f –1x2 = e x 7 0 for all real numbers x, the function f ¿ is increasing and the graph is concave up over the entire real line.

■ EXAMPLE 6 Graph: g1x2 = e -x. Analyze the graph using calculus. Solution First, we find some function values, plot the points, and sketch the graph. y

x

g1x2

-2

7.389

-1

2.718

8

0

7 6 5 4

1

3

1

0.368

2

0.135

g(x) =

e –x

2 1

– 2 –1 0

We can then analyze the graph using calculus as follows. a) Derivatives. Since g1x2 = e -x, we have g¿1x2 = e -x1- 12 = - e -x,

so

g–1x2 = - e -x1- 12 = e -x.

1

2

x

•

3.1

Exponential Functions

317

b) Critical values of g. Since e -x = 1>e x 7 0, we have g¿1x2 = - e -x 6 0 for all real numbers x. Thus, the derivative exists for all real numbers, and the equation g¿1x2 = 0 has no solution. There are no critical values and therefore no maximum or minimum values. c) Decreasing. Since the derivative g¿1x2 = - e -x 6 0 for all real numbers x, the function g is decreasing over the entire real line. d) Inflection points. We have g–1x2 = e -x 7 0, so the equation g–1x2 = 0 has no solution and there are no points of inflection. e) Concavity. We also know that since g–1x2 = e -x 7 0 for all real numbers x, the function g¿ is increasing and the graph is concave up over the entire real line. Functions of the type f1x2 = 1 - e -kx, with x Ú 0, have important applications.

■ EXAMPLE 7 Graph: h1x2 = 1 - e -2x, with x Ú 0. Analyze the graph using calculus.

Solution First, we find some function values, plot the points, and sketch the graph. y

h1x2

x 0

0

0.5

0.63212

1

0.86466

2

0.98168

3

0.99752

4

0.99966

5

0.99995

1

h(x) = 1 − e−2x

0.5

0

0.5 1

2

3

4

x

We can analyze the graph using calculus as follows. a) Derivatives. Since h1x2 = 1 - e -2x, h¿1x2 = - e -2x1- 22 = 2e -2x

and

h–1x2 = 2e -2x1- 22 = - 4e -2x.

b) Critical values. Since e -2x = 1>e 2x 7 0, we have 2e -2x 7 0. Thus, h¿1x2 = 0 has no solution, and since h¿1x2 exists for all x 7 0, it follows that there are no critical values on the interval 10, q 2 c) Increasing. Since 2e -2x 7 0 for all real numbers x, we know that h is increasing over the interval 30, q 2. d) Inflection points. Since h–1x2 = - 4e -2x 6 0, we know that the equation h–1x2 = 0 has no solution; thus, there are no points of inflection. e) Concavity. Since h–1x2 = - 4e -2x 6 0, we know that h¿ is decreasing and the graph is concave down over the interval 10, q 2.

318

CHAPTER 3

•

Exponential and Logarithmic Functions

Quick Check 5

O

Graph each function. Then determine critical values, intervals over which the function is increasing or decreasing, inflection points, and the concavity. a) f1x2 = 2e -x; b) g1x2 = 2e x; c) h1x2 = 1 - e -x.

In general, for k 7 0, the graph of h1x2 = 1 - e -kx is increasing, which we expect since h¿1x2 = ke -kx is always positive. Note that h1x2 approaches 1 as x approaches q . A word of caution! Functions of the type a x (for example, 2 x, 3x, and e x) are different from functions of the type x a (for example, x 2, x 3, x 1>2). For a x, the variable is in the exponent. For x a, the variable is in the base. The derivative of a x is not xa x - 1. In particular, we have the following: d x e Z xe x - 1, but dx

d x e = e x. dx

O Quick Check 5 ■ EXAMPLE 8 Business: Worker Efficiency. It is reasonable for a manufacturer to expect the daily output of a new worker to start out slow and continue to increase over time, but then tend to level off, never exceeding a certain amount. A firm manufactures 5G smart phones and determines that after working t days, the efficiency, in number of phones produced per day, of most workers can be modeled by the function N1t2 = 80 - 70e -0.13t.

a) b) c) d)

Find N102, N112, N152, N1102, N1202, and N1302. Graph N1t2. Find N¿1t2 and interpret this derivative in terms of rate of change. What number of phones seems to determine where worker efficiency levels off?

Solution a) We make a table of input–output values. t

0

1

5

N1t2

10

18.5

43.5

10

20

30

60.9

74.8

78.6

b) Using these values and/or a graphing calculator, we obtain the graph. Number of phones produced per day

N(t) 80 70 60

N(t) = 80 – 70e – 0.13 t

50 40 30 20 10 10

20

30

40

t

Time (in days)

Quick Check 6 Business. Repeat Example 8 for the efficiency function N1t2 = 80 - 60e -0.12t.

c) N¿1t2 = - 70e -0.13t1- 0.132 = 9.1e -0.13t; after t days, the rate of change of number of phones produced per day is given by 9.1e -0.13t. d) Examining the graph and expanding the table of function values, it seems that worker efficiency levels off at no more than 80 phones produced per day.

O Quick Check 6

O

Exercise Set 3.1

319

Section Summary • The exponential function f1x2 = e x, where e L 2.71828, has the derivative f ¿1x2 = e x. That is, the slope of a tangent line to the graph of y = e x is the same as the function value at x.

• The graph of f1x2 = e x is an increasing function with no critical values, no maximum or minimum values, and no points of inflection. The graph is concave up, with lim f1x2 = q

lim f1x2 = 0.

and

x: q

x: -q

• Calculus is rich in applications of exponential functions.

EXERCISE SET

3.1 Graph. 1. y = 4

2. y = 5

x

3. y = 10.252

4. y = 10.22

x

5. f 1x2 =

7. g1x2 =

x x

A B

6. f 1x2 =

3 x 2

A B

2 x 3

8. g1x2 =

9. f 1x2 = 12.52

A B

4 x 3

A B

3 x 4

x

Differentiate.

11. f 1x2 = e -x

12. f 1x2 = e x

13. g1x2 = e

14. g1x2 = e

3x

15. f 1x2 = 6e x

18. F1x2 = e -4x

19. G1x2 = 2e 4x

20. g1x2 = 3e 5x

23. g1x2 = 25. F1x2 =

22. G1x2 = - 7e 24. f 1x2 =

1 -5x 2e 2 - 23 e x

27. G1x2 = 7 + 3e

29. f 1x2 = x - 2e 5

26. g1x2 =

33. F1x2 =

-x

1 -4x 3e

28. F1x2 = 4 - e

6x

30. G1x2 = x 3 - 5e 2x

e 2x x4

2x

32. f 1x2 = x 7e 4x 34. g1x2 =

x6

38. f 1x2 =

x4

39. f 1x2 = e -x

2

+ 7x

41. f 1x2 = e -x >2 2

43. y = e 2x - 7

51. y = 1 - e -kx

52. y = 1 - e -mx

53. g1x2 = 14x 2 + 3x2e x

- 7x

54. g1x2 = 15x 2 - 8x2e x

- 4x

55. f 1x2 = e 2x

56. g1x2 = e -2x

57. g1x2 = e 11>22x

58. f 1x2 = e 11>32x

59. f 1x2 =

60. g1x2 =

1 2

e -x

61. F1x2 = - e 11>32x

1 3

e -x

62. G1x2 = - e 11>22x

64. f 1x2 = 3 - e -x, for x Ú 0 65–74. For each function given in Exercises 55–64, graph the function and its first and second derivatives using a calculator, iPlot, or Graphicus.

76. Find the slope of the line tangent to the graph of f 1x2 = 2e -3x at the point 10, 22.

36. f 1x2 = 1x 2 - 2x + 22e x 37. f 1x2 =

50. y = 1 - e -3x

75. Find the slope of the line tangent to the graph of f 1x2 = e x at the point 10, 12.

e 3x

35. f 1x2 = 1x 2 + 3x - 92e x

ex

49. y = 1 - e -x

63. g1x2 = 211 - e -x2, for x Ú 0

3 - 45 e x

5x

31. g1x2 = x 5e 2x

48. y = e x + x 3 - xe x

Graph each function. Then determine critical values, inflection points, intervals over which the function is increasing or decreasing, and the concavity.

2x

17. F1x2 = e -7x 21. f 1x2 = - 3e

47. y = xe -2x + e -x + x 3

2

16. f 1x2 = 4e x

-x

46. y = 2e x + 1

2

10. f 1x2 = 11.22

x

45. y = 2e x - 1

77. Find an equation of the line tangent to the graph of G1x2 = e -x at the point 10, 12.

ex x5

40. f 1x2 = e -x

2

+ 8x

42. f 1x2 = e x >2 2

44. y = e 2x - 4

78. Find an equation of the line tangent to the graph of f 1x2 = e 2x at the point 10, 12. 79. and 80. For each of Exercises 77 and 78, graph the function and the tangent line using a calculator, iPlot, or Graphicus.

320

CHAPTER 3

•

Exponential and Logarithmic Functions

APPLICATIONS

c) C¿142 (Round to the nearest thousand.) d) Find lim C1t2 and lim C¿1t2. Why do you think t: q t: q the company’s costs tend to level off as time passes?

Business and Economics 81. U.S. exports. U.S. exports of goods are increasing exponentially. The value of the exports, t years after 2009, can be approximated by V1t2 = 1.6e 0.046t, where t = 0 corresponds to 2009 and V is in billions of dollars. (Source: U.S. Commerce Department.) a) Estimate the value of U.S. exports in 2009 and 2020. b) What is the doubling time for the value of U.S. exports?

84. Marginal cost. A company’s total cost, in millions of dollars, is given by C1t2 = 200 - 40e -t, where t is the time in years since the start-up date. C(t)

C(t) = 200 – 40e –t

200 150 100 50 –2

2

4

6

8

10 t

Find each of the following. a) The marginal cost C¿1t2 b) C¿102 c) C¿152 (Round to the nearest thousand.) d) Find lim C1t2 and lim C¿1t2. Why do you think t: q t: q the company’s costs tend to level off as time passes? 85. Marginal demand. At a price of x dollars, the demand, in thousands of units, for a certain music player is given by the demand function 82. Organic food. More Americans are buying organic fruit and vegetables and products made with organic ingredients. The amount A1t2, in billions of dollars, spent on organic food and beverages t years after 1995 can be approximated by A1t2 = 2.43e 0.18t. (Source: Nutrition Business Journal, 2004.) a) Estimate the amount that Americans spent on organic food and beverages in 2009. b) Estimate the rate at which spending on organic food and beverages was growing in 2006. 83. Marginal cost. A company’s total cost, in millions of dollars, is given by C1t2 = 100 - 50e -t, where t is the time in years since the start-up date. C(t)

q = 240e -0.003x. a) How many music players will be bought at a price of $250? Round to the nearest thousand. b) Graph the demand function for 0 … x … 400. c) Find the marginal demand, q¿1x2. d) Interpret the meaning of the derivative. 86. Marginal supply. At a price of x dollars, the supply function for the music player in Exercise 85 is given by q = 75e 0.004x, where q is in thousands of units. a) How many music players will be supplied at a price of $250? Round to the nearest thousand. b) Graph the supply function for 0 … x … 400. c) Find the marginal supply, q¿1x2. d) Interpret the meaning of the derivative.

Life and Physical Sciences

100

C(t) = 100 – 50e –t

80

87. Medication concentration. The concentration C, in parts per million, of a medication in the body t hours after ingestion is given by the function

60 40

C1t2 = 10t 2e -t.

20

a) Find the concentration after 0 hr, 1 hr, 2 hr, 3 hr, and 10 hr. b) Sketch a graph of the function for 0 … t … 10. c) Find the rate of change of the concentration, C¿1t2. d) Find the maximum value of the concentration and the time at which it occurs. e) Interpret the meaning of the derivative.

5

10

Find each of the following. a) The marginal cost, C¿1t2 b) C¿102

15

20

t

321

Exercise Set 3.1

TECHNOLOGY CONNECTION

Social Sciences 88. Ebbinghaus learning model. Suppose that you are given the task of learning 100% of a block of knowledge. Human nature is such that we retain only a percentage P of knowledge t weeks after we have learned it. The Ebbinghaus learning model asserts that P is given by P1t2 = Q + 1100 - Q2e -kt,

where Q is the percentage that we would never forget and k is a constant that depends on the knowledge learned. Suppose that Q = 40 and k = 0.7. a) Find the percentage retained after 0 weeks, 1 week, 2 weeks, 6 weeks, and 10 weeks. b) Find lim P1t2. t: q

c) Sketch a graph of P. d) Find the rate of change of P with respect to time t. e) Interpret the meaning of the derivative.

Use a graphing calculator (or iPlot or Graphicus) to graph each function in Exercises 107 and 108, and find all relative extrema. 108. f 1x2 = e -x

107. f 1x2 = x 2e -x

2

For each of the functions in Exercises 109–112, graph f, f ¿, and f –. 109. f 1x2 = e x

110. f 1x2 = e -x

112. f 1x2 = 1000e -0.08x

111. f 1x2 = 2e 0.3x 113. Graph f 1x2 = a 1 +

1 x b . x

Use the TABLE feature and very large values of x to confirm that e is approached as a limit.

SYNTHESIS

Answers to Quick Checks

Differentiate.

1 1 1 1. 1, 3, 9, 27, , , 3 9 27

89. y = 1e 3x + 125 91. y = 93. y =

e 3t - e 7t e

4t

x2 + 1

95. f 1x2 = e 2x + 2e x 97. f 1x2 = e

2

3

92. y = 2e

3t

y

y

+ t

8

8

6

6 4

4

ex

x>2

90. y = 1e x - 224

1 1 1 2. 1, , , , 3, 9, 27 3 9 27

# 2x - 1

e x - e -x 99. f1x2 = x e + e -x

ex 94. y = 1 - ex 96. f 1x2 =

1 + e 1>x ex

98. f 1x2 =

-x

xe

1 + x2

100. f 1x2 = e

f(x)  3 x

2 −2

−1

1

2

−2

x

3. (a) 6e x; (b) x 2e x 1x + 32; (c) 4. (a) - 4e -4x; (b) e x

3

5. (a)

+ 8x

−1

1

x3

13x 2 + 82; (c)

8

8

6

6

101. For f 1t2 = 11 + t21>t, we have e = lim f 1t2. Find f 112, t:0

f 10.52, f 10.22, f 10.12, and f 10.0012.

102. For g1t2 = t 1>1t - 12, we have e = lim g1t2. Find g10.52, t:1

g10.92, g10.992, g10.9992, and g10.99982. 103. Find the maximum value of f 1x2 = x 2e -x over 30, 44. 104. Find the minimum value of f 1x2 = xe x over 3- 2, 04. 105. A student made the following error on a test: d x e = xe x - 1. dx Identify the error and explain how to correct it. 106. Describe the differences in the graphs of f 1x2 = 3x and g1x2 = x 3.

2 −2

−1

x

xe 2x

2

+5

2x 2 + 5

y

4

4

Exercises 101 and 102 each give an expression for e. Find the function values that are approximations for e. Round to five decimal places.

2

e x 1x - 22

(b)

y

ex

1 x g(x)  () 3

2

f(x)  2ex 1

2

x

No critical values Decreasing on 1- q , q 2 No inflection points Concave up on 1- q , q 2 (c) y

2 −2

−1

g(x)  2e x 1

2

x

No critical values Increasing on 1- q , q 2 No inflection points Concave up on 1- q , q 2

1

0.8 0.6

h(x)  1  ex

0.4 0.2 2

4

6

8 10 x

No critical values Increasing on 1- q , q 2 No inflection points Concave down on 1- q , q 2 6. (a) 20, 26.8, 47.1, 61.9, 74.6, 78.4 (b) (c) N¿1t2 = 7.2e -0.12t; after t days, the rate of change of number of phones produced per day is given by 7.2e -0.12t. (d) 80 phones produced per day

N(t) 80 70 60 50 40 30 20 10

N(t)  80  60e0.12t

10

20

30

40

t

322

CHAPTER 3

•

Exponential and Logarithmic Functions

3.2 OBJECTIVES • Convert between exponential and logarithmic equations. • Solve exponential equations. • Solve problems involving exponential and logarithmic functions. • Differentiate functions involving natural logarithms.

Logarithmic Functions Logarithmic Functions and Their Graphs Suppose that we want to solve the equation 10y = 1000. We are trying to find the power of 10 that will give 1000. Since 103 = 1000, the answer is 3. The number 3 is called “the logarithm, base 10, of 1000.”

DEFINITION A logarithm is defined as follows: log a x = y

means

a y = x, a 7 0, a Z 1.

The number log a x is the power y to which we raise a to get x. The number a is called the logarithmic base. We read log a x as “the logarithm, base a, of x.” For logarithms with base 10, log 10 x is the power y such that 10y = x. Therefore, a logarithm can be thought of as an exponent. We can convert from a logarithmic equation to an exponential equation, and conversely, as follows. Logarithmic Equation

Exponential Equation

log a M = N

aN = M

log 10 100 = 2 1 log 5 25 = -2 log 49 7 = 12

102 = 100 1 5-2 = 25 491>2 = 7

In order to graph a logarithmic equation, we can graph its equivalent exponential equation.

■ EXAMPLE 1 Graph: y = log 2 x. Solution We first write the equivalent exponential equation: 2 y = x. TECHNOLOGY CONNECTION

Exploratory

Graph f 1x2 = 2 x. Then use the TABLE feature to find the coordinates of points on the graph. How can each ordered pair help you make a hand drawing of a graph of g1x2 = log 2 x?

We select values for y and find the corresponding values of 2 y. Then we plot points, remembering that x is still the first coordinate, and connect the points with a smooth curve. x, or 2 y

y

y

1

0

2

1

y = log 2 x (or x = 2y )

3 2 1

4

2

8

3

1 2

-1

1 4

-2 1 Select y. 2 Compute x.

1 –1 –2

2

3

4

5

6

7

8

x

•

3.2

TECHNOLOGY CONNECTION

Graphing Logarithmic Functions To graph y = log 2 x, we first graph y1 = 2 x. We next select the Drawlnv option from the DRAW menu and then the Y-VARS option from the VARS menu, followed by 1, 1, and [ to draw the inverse of y1. Both graphs are drawn together.

Logarithmic Functions

323

The graphs of f1x2 = 2 x and g1x2 = log 2 x are shown below on the same set of axes. Note that we can obtain the graph of g by reflecting the graph of f across the line y = x. Functions whose graphs can be obtained in this manner are known as inverses of each other. y

f(x) = 2x

7 6 5

(2, 4)

4 3

y1 = 2x, y2 = log 2 x

(4, 2) g(x) = log 2 x

2

10

1 –6 –5 –4 –3 – 2 –1

0 1

2

3

4

5

6

7

x

–1 –10

10

–2 –3

y=x

–10

To use Graphicus to graph y = log 2 x, press + and New x(y). Then enter 2 y as 2^y and press Done. The screen will display the graph of the equation x = 2 y, which by the definition of logarithms is also the graph of y = log 2 x.

EXERCISES Graph. 1. y = log 3 x 2. y = log 5 x

–4 –5

Although we do not develop inverses in detail here, it is important to note that they “undo” each other. For example,

and

f132 = 2 3 = 8,

The input 3 gives the output 8.

g182 = log 2 8 = 3.

The input 8 gets us back to 3.

Basic Properties of Logarithms The following are some basic properties of logarithms. The proofs of P1–P3 follow from properties of exponents and are outlined in Exercises 107–109 at the end of this section. Properties P4–P6 follow directly from the definition of a logarithm, and a proof of P7 is outlined in Exercise 110.

3. f 1x2 = log e x

4. f 1x2 = log 10 x

THEOREM 3

Properties of Logarithms For any positive numbers M, N, a, and b, with a, b Z 1, and any real number k: P1. log a 1MN2 = log a M + log a N M P2. log a = log a M - log a N N P3. log a 1M k2 = k # log a M P4. log a a = 1 P5. log a 1a k2 = k P6. log a 1 = 0 P7. log b M =

log a M log a b

(The change-of-base formula)

Let’s illustrate these properties.

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Exponential and Logarithmic Functions

■ EXAMPLE 2 Given log a 2 = 0.301 and find each of the following:

a) log a 6;

d) log a 13;

f) log a 12a2;

Solution a) log a 6 = = = =

e) log a 2a; log a 12 # 32 log a 2 + log a 3 0.301 + 0.477 0.778

c) log a 81 = = = =

log a 34 4 log a 3 410.4772 1.908

c) log a 81; h) log a 5.

By P2

By P3

d) log a 13 = log a 1 - log a 3 = 0 - 0.477 = - 0.477

e) log a 2a = log a 1a 1>22 =

Given log b 2 = 0.356 and log b 5 = 0.827, find each of the following: a) log b 10; b) log b 25; c) log b 52; d) log b 16; e) log b 5b; f) log b 2b.

b) log a 23; log a 3 g) ; log a 2

By P1

b) log a 23 = log a 2 - log a 3 = 0.301 - 0.477 = - 0.176

Quick Check 1

log a 3 = 0.477,

By P2 By P6 1 2

f) log a 12a2 = log a 2 + log a a = 0.301 + 1 = 1.301

By P5 By P1 By P4

log a 3 0.477 = L 1.58 log a 2 0.301 We simply divided and used none of the properties. h) There is no way to find log a 5 using the properties of logarithms 1log a 5 Z log a 2 + log a 32. g)

O Quick Check 1 Common Logarithms The number log 10 x is the common logarithm of x and is abbreviated log x; that is:

DEFINITION For any positive number x, log x = log 10 x.

Thus, when we write “log x” with no base indicated, base 10 is understood. Note the following comparison of common logarithms and powers of 10.

O

3.2

1000 100 10 1 0.1 0.01 0.001

= = = = = = =

103 102 101 100 10-1 10-2 10-3

The common logarithms at the right follow from the powers at the left.

•

Logarithmic Functions

log 1000 log 100 log 10 log 1 log 0.1 log 0.01 log 0.001

= = = = = = =

325

3 2 1 0 -1 -2 -3

Since log 100 = 2 and log 1000 = 3, it seems reasonable that log 500 is somewhere between 2 and 3. Tables of logarithms were originally used for such approximations, but since the advent of the calculator, such tables are rarely used. Using a calculator with a W key, we find that log 500 L 2.6990. Before calculators became so readily available, common logarithms were used extensively to do certain computations. In fact, computation is the reason logarithms were developed. Since standard notation for numbers is based on 10, it was logical to use base-10, or common, logarithms for computations. Today, computations with common logarithms are mainly of historical interest; the logarithmic functions, base e, are far more important.

Natural Logarithms The number e, which is approximately 2.718282, was developed in Section 3.1, and has extensive application in many fields. The number log e x is the natural logarithm of x and is abbreviated ln x.

DEFINITION For any positive number x, ln x = log e x.

The following basic properties of natural logarithms parallel those given earlier for logarithms in general. TECHNOLOGY CONNECTION

THEOREM 4

To enter y = log 10 x, the key labeled W can be used.

Properties of Natural Logarithms P1. ln 1MN2 = ln M + ln N P5. ln 1e k2 = k

EXERCISES

P2. ln

1. Graph f 1x2 = 10x, y = x, and

g1x2 = log 10 x using the same set of axes. Then find f 132, f 10.6992, g152, and g110002.

M = ln M - ln N N

P3. ln 1a k2 = k # ln a

P6. ln 1 = 0 P7. log b M =

P4. ln e = 1

2. Use the W key and P7

of Theorem 4 to graph y = log 2 x.

Let’s illustrate the properties of Theorem 4.

ln M ln b

and

ln M =

log M log e

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Exponential and Logarithmic Functions

■ EXAMPLE 3 Given ln 2 = 0.6931 and ln 3 = 1.0986, find each of the following:

a) ln 6; b) ln 81; c) ln 13; d) ln 12e 52; e) log 2 3.

Solution

a) ln 6 = ln 12 # 32 = ln 2 + ln 3 By P1 = 0.6931 + 1.0986 = 1.7917

b) ln 81 = = = =

ln 1342 4 ln 3 By P3 411.09862 4.3944

c) ln 13 = ln 1 - ln 3 = 0 - 1.0986 = - 1.0986

Quick Check 2 Given ln 2 = 0.6931 and ln 5 = 1.6094, find each of the following: a) ln 10; b) ln 52; c) ln 25; d) ln 32; e) ln 5e 2; f) log 5 2.

By P2 By P6

d) ln 12e 52 = ln 2 + ln 1e 52 = 0.6931 + 5 = 5.6931 e) log 2 3 =

By P1 By P5

ln 3 1.0986 = L 1.5851 ln 2 0.6931

By P7

O Quick Check 2 Finding Natural Logarithms Using a Calculator You should have a calculator with an X key. You can find natural logarithms directly using this key.

■ EXAMPLE 4 Approximate each of the following to six decimal places: a) ln 5.24;

b) ln 0.001278.

Solution We use a calculator with an X key. a) ln 5.24 L 1.656321

b) ln 0.001278 L - 6.662459

Exponential Equations If an equation contains a variable in an exponent, the equation is exponential. We can use logarithms to manipulate or solve exponential equations.

■ EXAMPLE 5 Solve e t = 40 for t. Solution We have ln e t t t t

= = L L

ln 40 ln 40 3.688879 3.7

Taking the natural logarithm on both sides By P5; remember that ln e t means log e e t. Using a calculator

Note that this is an approximation for t even though an equals sign is often used.

O

3.2

•

Logarithmic Functions

327

■ EXAMPLE 6 Solve e -0.04t = 0.05 for t. Solution We have ln e -0.04t = ln 0.05 - 0.04t = ln 0.05 ln 0.05 t = - 0.04 - 2.995732 t L - 0.04 t L 75.

Taking the natural logarithm on both sides By P5

Using a calculator

In Example 6, we rounded ln 0.05 to - 2.995732 in an intermediate step. When using a calculator, you should find ln 0.05 - 0.04 by keying in X.05)d:.04, pressing [, and rounding at the end. Answers at the back of this book have been found in this manner. Remember, the number of places in a table or on a calculator affects the accuracy of the answer. Usually, your answer should agree with that in the Answers section to at least three digits.

Quick Check 3

O

Solve each equation for t: a) e t = 80; b) e -0.08t = 0.25.

O Quick Check 3

TECHNOLOGY CONNECTION

Solving Exponential Equations

Method 2: The ZERO Feature

Let’s solve the equation of Example 5, e t = 40, graphically.

We change the variable to x and get a 0 on one side of the equation: e x - 40 = 0. Then we graph y = e x - 40 in the window 3- 1, 8, - 10, 104.

Method 1: The INTERSECT Feature

10

We change the variable to x and consider the system of equations y1 = e x and y2 = 40. We graph the equations in the window 3 - 1, 8, - 10, 704 to see the curvature and point of intersection. y1 = e x,

–1

y2 = 40

8 Zero X = 3.6888795

70

Y=0

–10

Intersection –1 X = 3.6888795 Y = 40 –10

Using the ZERO option from the CALC menu, we see that the x-intercept is about 13.7, 02, so 3.7 is the solution of e t = 40. 8

Then we use the INTERSECT option from the CALC menu to find the point of intersection, about 13.7, 402. The x-coordinate, 3.7, is the solution of e t = 40.

EXERCISES Solve graphically using a calculator, iPlot, or Graphicus. 1. e t = 1000

2. e -x = 60

3. e -0.04t = 0.05

4. e 0.23x = 41,378

5. 15e 0.2x = 34,785.13

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Exponential and Logarithmic Functions

Graphs of Natural Logarithmic Functions There are two ways in which we might graph y = f1x2 = ln x. One is to graph the equivalent equation x = e y by selecting values for y and calculating the corresponding values of e y. We then plot points, remembering that x is still the first coordinate. y

x, or e y

y

0.1

-2

0.4

-1

1.0

0

2.7

1

7.4

2

20.1

3

y=x

7

g(x) = e x

6 5 4 3 2

f(x) = ln x

1 – 2 –1

1

2

3

4

5

6

7

x

–1 –2

1 Select y. 2 Compute x.

TECHNOLOGY CONNECTION

Graphing Logarithmic Functions

The graph above shows the graph of g1x2 = e x for comparison. Note again that the functions are inverses of each other. That is, the graph of y = ln x, or x = e y, is a reflection, or mirror image, across the line y = x of the graph of y = e x. Any ordered pair 1a, b2 on the graph of g yields an ordered pair 1b, a2 on f. Note too that lim+ ln x = - q x:0 and the y-axis is a vertical asymptote. y The second method of graphing y = ln x is to f (x) = ln x use a calculator to find function values. For example, 2 (2, 0.7) when x = 2, then y = ln 2 L 0.6931 L 0.7. This 1 gives the pair 12, 0.72 shown on the graph. 1

2

3

4

5

6

7

8

x

–1

To graph y = ln x, you can use the X key and enter the function as y1 = ln 1x2.

–2

The following properties can be observed from the graph.

THEOREM 5

ln x exists only for positive numbers x. The domain is 10, q 2. ln x 6 0 for 0 6 x 6 1. ln x = 0 when x = 1. ln x 7 0 for x 7 1. The function given by f1x2 = ln x is always increasing. The range is the entire real line, 1- q , q 2, or the set of real numbers, .

Derivatives of Natural Logarithmic Functions Let’s find the derivative of f1x2 = ln x.

(1)

3.2

•

Logarithmic Functions

329

We first write its equivalent exponential equation: e f 1x2 = x.

ln x = log e x = f 1x2, so e f 1x2 = x, by the definition of logarithms.

(2)

Now we differentiate on both sides of this equation: d f 1x2 d e = x dx dx e f 1x2 # f ¿1x2 = 1 By the Chain Rule x # f ¿1x2 = 1 Substituting x for e f 1x2 from equation (2) 1 f ¿1x2 = . x

TECHNOLOGY CONNECTION

Exploratory Use Graphicus to graph y = ln x. Then touch + and choose Add derivative. Using the tangent line feature, move along the curve, noting the x-values, the y-values, and the values of dy>dx, to verify dy 1 = . that x dx

Thus, we have the following.

THEOREM 6 For any positive number x, d 1 ln x = . x dx

To visualize the meaning of Theorem 6, look back at the graph of f1x2 = ln x on p. 328. Take a small ruler or a credit card and place it as if its edge were a tangent line. Start on the left and move the ruler or card along the curve toward the right, noting how the tangent lines flatten out. Think about the slopes of these tangent lines. The slopes approach 0 as a limit, though they never actually become 0. This is consistent with the formula d 1 1 ln x = , because lim = 0. x x: q x dx Theorem 6 asserts that to find the slope of the tangent line at x for the function f1x2 = ln x, we need only take the reciprocal of x. This is true only for positive values of x, since ln x is defined only for positive numbers. (For negative numbers x, this derivative formula becomes d 1 ln ƒ x ƒ = , x dx but we will seldom consider such a case in this text.) Let’s find some derivatives.

■ EXAMPLE 7 Differentiate: a) y = 3 ln x; b) y = x 2 ln x + 5x; c) y =

ln x . x3

Solution a)

b)

d d 13 ln x2 = 3 ln x dx dx 3 = x d 2 1 1x ln x + 5x2 = x 2 # + ln x # 2x + 5 x dx = x + 2x # ln x + 5

Using the Product Rule on x 2 ln x Simplifying

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Exponential and Logarithmic Functions

x 3 # 11>x2 - 1ln x213x 22 d ln x c) = dx x 3 x6 = Quick Check 4 = =

x 2 - 3x 2 ln x x6

x 211 - 3 ln x2

Factoring

x6 1 - 3 ln x x4

Simplifying

O Quick Check 4 Suppose that we want to differentiate a more complicated function that is of the form h1x2 = ln f1x2, such as h1x2 = ln 1x 2 - 8x2.

This can be regarded as h1x2 = g1f1x22,

where

g1x2 = ln x and

f1x2 = x 2 - 8x.

Now g¿1x2 = 1>x, so by the Chain Rule (Section 1.7), we have h¿1x2 = g¿1f1x22 # f ¿1x2 1 # = f ¿1x2. f1x2

For the above case, f1x2 = x 2 - 8x, so f ¿1x2 = 2x - 8. Then h¿1x2 =

1 # 12x - 82 = 2x2 - 8 . x - 8x x - 8x 2

The following rule, which we have proven using the Chain Rule, allows us to find derivatives of functions like the one above.

THEOREM 7 The derivative of the natural logarithm of a function is the derivative of the function divided by the function: f ¿1x2 d 1 # ln f1x2 = f ¿1x2 = , dx f1x2 f1x2 or d 1 du ln u = # . u dx dx

The following gives us a way of remembering this rule.

⎫ ⎪ ⎬ ⎪ ⎭

h1x2 = ln 1x 2 - 8x2 1

1 Differentiate the “inside” function.

⎫ ⎪ ⎬ ⎪ ⎭

Differentiate: a) y = 5 ln x; b) y = x 3 ln x + 4x; ln x c) y = 2 . x

By the Quotient Rule

h¿1x2 =

⎫ ⎪ ⎬ ⎪ ⎭

2x - 8 x 2 - 8x

2

2 Divide by the “inside” function.

O

3.2

•

331

Logarithmic Functions

■ EXAMPLE 8 Differentiate: a) y = ln 13x2; b) y = ln 1x 2 - 52; c) f1x2 = ln 1ln x2; d) f1x2 = ln a

x3 + 4 b. x

Solution TECHNOLOGY CONNECTION

Exploratory To check part (a) of Example 8, we let y1 = ln 13x2, y2 = nDeriv1y1, x, x2, and y3 = 1>x. Either GRAPH or TABLE can then be used to show that y2 = y3. Use this approach to check parts (b), (c), and (d) of Example 8. (Or use Graphicus to check those results.)

a) If y = ln 13x2, then dy dx

=

3 1 = . x 3x

Note that we could have done this using the fact that ln 1MN2 = ln M + ln N: ln 13x2 = ln 3 + ln x;

then, since ln 3 is a constant, we have 1 d d d 1 = . ln 13x2 = ln 3 + ln x = 0 + x x dx dx dx

b) If y = ln 1x 2 - 52, then dy dx

=

2x . x - 5 2

c) If f1x2 = ln 1ln x2, then f ¿1x2 =

1 #1 1 # d 1 ln x = . = ln x dx ln x x x ln x

d) If f1x2 = ln a

x3 + 4 M = ln M - ln N, we have b, then, since ln x N

d 3ln 1x 3 + 42 - ln x4 dx 3x 2 1 = 3 x x + 4

f ¿1x2 =

=

Quick Check 5 Differentiate: a) y = ln 5x; b) y = ln 13x 2 + 42; c) y = ln 1ln 5x2; x5 - 2 d) y = ln a b. x

=

3x 2 # x 1 # x3 + 4 x x3 + 4 x3 + 4 x 13x 22x - 1x 3 + 42 x1x + 42 3

3x 3 - x 3 - 4 2x 3 - 4 = = . x1x 3 + 42 x1x 3 + 42

By P2; this avoids use of the Quotient Rule.

Finding a common denominator

⎫ ⎪ ⎪ ⎬ Simplifying ⎪ ⎪ ⎭

O Quick Check 5 Applications ■ EXAMPLE 9 Social Science: Forgetting. In a psychological experiment, students were shown a set of nonsense syllables, such as POK, RIZ, DEQ, and so on, and asked to recall them every minute thereafter. The percentage R1t2 who retained the syllables after t minutes was found to be given by the logarithmic learning model R1t2 = 80 - 27 ln t, for

t Ú 1.

O

332

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•

Exponential and Logarithmic Functions

Percentage of students retaining syllables

TECHNOLOGY CONNECTION

Exploratory Graph y = 80 - 27 ln x, from Example 9, using the viewing window 31, 14, - 1, 1004. Trace along the graph. Describe the meaning of each coordinate in an ordered pair.

R(t) 100 90 80 70 60 50 40 30 20 10 0

1

2

3

4

5

6

7

8

9

t

10 11 12 13 14

Time (in minutes)

a) What percentage of students retained the syllables after 1 min? b) Find R¿122, and explain what it represents.

Solution

a) R112 = 80 - 27 # ln 1 = 80 - 27 # 0 = 80%

b)

27 d 1 180 - 27 ln t2 = 0 - 27 # = - , t t dt so

R¿122 = -

27 = - 13.5. 2

This result indicates that 2 min after students have been shown the syllables, the percentage of them who remember the syllables is shrinking at the rate of 13.5% per minute.

■ EXAMPLE 10 Business: An Advertising Model. A company begins a radio advertising campaign in New York City to market a new product. The percentage of the “target market” that buys a product is normally a function of the duration of the advertising campaign. The radio station estimates this percentage, as a decimal, by using f1t2 = 1 - e -0.04t for this type of product, where t is the number of days of the campaign. The target market is approximately 1,000,000 people and the price per unit is $0.50. If the campaign costs $1000 per day, how long should it last in order to maximize profit?

Solution Modeling the percentage of the target market that buys the product, expressed as a decimal, by using f1t2 = 1 - e -0.04t is justified if we graph f. The function increases from 0 (0%) toward 1 (100%). The longer the advertising campaign, the larger the percentage of the market that has bought the product. y 1.0 0.5

25

50

75

100

t

Number of days

The total-profit function, here expressed in terms of time t, is given by Profit = Revenue - Cost P1t2 = R1t2 - C1t2.

3.2

•

Logarithmic Functions

333

We find R1t2 and C1t2:

R1t2 = 1Percentage buying2 # 1Target market2 # 1Price per unit2 = 11 - e -0.04t211,000,000210.52 = 500,000 - 500,000e -0.04t,

and P $400,000

P(t) = 500,000 – 500,000e –0.04t – 1000t

C1t2 = 1Advertising costs per day2 # 1Number of days2 = 1000t. Next, we find P1t2 and take its derivative:

350,000

P1t2 = = P¿1t2 = =

300,000 250,000 200,000

R1t2 - C1t2 500,000 - 500,000e -0.04t - 1000t, - 500,000e -0.04t1- 0.042 - 1000 20,000e -0.04t - 1000.

150,000

We then set the first derivative equal to 0 and solve:

100,000

20,000e -0.04t - 1000 = 0 20,000e -0.04t = 1000 1000 e -0.04t = = 0.05 20,000 ln e -0.04t = ln 0.05 - 0.04t = ln 0.05 ln 0.05 t = - 0.04 t L 75.

50,000 100 ≈75

200

300

400

500

t

TECHNOLOGY CONNECTION

Exploratory Graph the function P in Example 10, and verify that there is maximum profit if the length of the advertising campaign is about 75 days.

We have only one critical value, so we can use the second derivative to determine whether we have a maximum: P–1t2 = 20,000e -0.04t1- 0.042 = - 800e -0.04t.

Quick Check 6

O

Business: An Advertising Model. Repeat Example 10 using the function f1t2 = 1 - e -0.08t and assuming that the campaign costs $2000 per day.

Since exponential functions are positive, e -0.04t 7 0 for all numbers t. Thus, since - 800e -0.04t 6 0 for all t, we have P–1752 6 0, and we have a maximum. The advertising campaign should run for 75 days in order to maximize profit.

O Quick Check 6

Section Summary • A logarithmic function y = log a x is defined by a y = x, for a 7 0 and a Z 1. • The common logarithmic function g is defined by g1x2 = log 10 x = log x, for x 7 0. • The natural logarithm function f is defined by f1x2 = log e x = ln x, where e L 2.71828. The derivative 1 of f is f ¿1x2 = , for x 7 0. The slope of a tangent line to x the graph of f at x is found by taking the reciprocal of the input x.

• The graph of f1x2 = ln x is an increasing function with no critical values, no maximum or minimum values, and no points of inflection. The domain is 10, q 2. The range is 1- q , q 2, or . The graph is concave down, with lim f1x2 = q

x: q

and

lim f1x2 = - q .

x:0

• Properties of logarithms are described in Theorems 3 and 4. • Calculus is rich in applications of natural logarithmic functions.

334

•

CHAPTER 3

Exponential and Logarithmic Functions

EXERCISE SET

3.2 Write an equivalent exponential equation. 1. log 2 8 = 3

2. log 3 81 = 4

1 3

3. log 8 2 =

57. y =

4. log 27 3 =

1 3

5. log a K = J

6. log a J = K

7. - log 10 h = p

8. - log b V = w

Write an equivalent logarithmic equation. 9. e M = b

12. 103 = 1000

13. 10-1 = 0.1

14. 10-2 = 0.01

15. M p = V

16. Qn = T 18. log b 15

19. log b 15

20. log b 2b 3

21. log b 15b2

25. ln 54

26. ln 15

27. ln 15e2 29. ln 2e

32. ln 45

e 33. ln a b 5

4 34. ln a b e

36. ln 99,999

37. ln 0.0182

38. ln 0.00087

39. ln 8100

40. ln 0.011

x2 - 7 b x

64. f 1x2 = ln a

x2 + 5 b x

66. g1x2 = e 2x ln x

65. g1x2 = e x ln x 2 67. f 1x2 = ln 1e x + 12

(Hint: Use the Extended Power Rule.)

70. g1x2 = 1ln x23

71. f 1x2 = ln 1ln 18x22

72. f 1x2 = ln 1ln 13x22

74. g1x2 = ln 12x2 # ln 17x2 75. Find the equation of the line tangent to the graph of y = 1x 2 - x2 ln 16x2 at x = 2. 76. Find the equation of the line tangent to the graph of y = e 3x # ln 14x2 at x = 1. 77. Find the equation of the line tangent to the graph of y = 1ln x22 at x = 3.

Find each logarithm. Round to six decimal places. 35. ln 5894

A = ln A - ln B.b B

73. g1x2 = ln 15x2 # ln 13x2

30. ln 2e 8

31. ln 14

x4 2

69. g1x2 = 1ln x24

28. ln 14e2

6

60. y = ln

a Hint: ln

x5

68. f 1x2 = ln 1e x - 22

Given ln 4 = 1.3863 and ln 5 = 1.6094, find each value. Do not use a calculator. 24. ln 80

x2 4

63. f 1x2 = ln a

22. log b 75

23. ln 20

59. y = ln

ln x

62. y = ln 17x 2 + 5x + 22

Given log b 3  1.099 and log b 5  1.609, find each value. 17. log b 53

x

58. y =

4

61. y = ln 13x 2 + 2x - 12

10. e t = p

11. 102 = 100

ln x

78. Find the equation of the line tangent to the graph of y = ln 14x 2 - 72 at x = 2.

APPLICATIONS

Solve for t.

Business and Economics

41. e t = 80

42. e t = 10

43. e 2t = 1000

44. e 3t = 900

45. e -t = 0.1

46. e -t = 0.01

N1a2 = 2000 + 500 ln a, a Ú 1,

47. e -0.02t = 0.06

48. e 0.07t = 2

where N1a2 is the number of units sold and a is the amount spent on advertising, in thousands of dollars. a) How many units were sold after spending $1000 on advertising? b) Find N¿1a2 and N¿1102. c) Find the maximum and minimum values, if they exist. d) Find lim N¿1a2. Discuss whether it makes sense to a: q to continue to spend more and more dollars on advertising.

79. Advertising. A model for consumers’ response to advertising is given by

Differentiate. 49. y = - 8 ln x 51. y = x ln x 4

50. y = - 9 ln x 1 2

53. f 1x2 = ln 16x2

x

2

55. g1x2 = x ln 17x2 2

52. y = x 6 ln x -

1 4

54. f 1x2 = ln 19x2

x4

56. g1x2 = x 5 ln 13x2

Exercise Set 3.2

80. Advertising. A model for consumers’ response to advertising is given by N1a2 = 1000 + 200 ln a, a Ú 1, where N1a2 is the number of units sold and a is the amount spent on advertising, in thousands of dollars. a) How many units were sold after spending $1000 on advertising? b) Find N¿1a2 and N¿1102. c) Find the maximum and minimum values of N, if they exist. d) Find N¿1a2. Discuss lim N¿1a2. Does it make sense a: q to spend more and more dollars on advertising? Why or why not? 81. An advertising model. Solve Example 10 given that the advertising campaign costs $2000 per day. 82. An advertising model. Solve Example 10 given that the advertising campaign costs $4000 per day. 83. Growth of a stock. The value, V1t2, in dollars, of a stock t months after it is purchased is modeled by V1t2 = 5811 - e -1.1t2 + 20.

Life and Physical Sciences 86. Acceptance of a new medicine. The percentage P of doctors who prescribe a certain new medicine is P1t2 = 10011 - e -0.2t2,

where t is the time, in months. a) Find P(1) and P(6). b) Find P¿1t2. c) How many months will it take for 90% of doctors to prescribe the new medicine? d) Find lim P1t2, and discuss its meaning. t: q

Social Sciences 87. Forgetting. Students in a botany class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t months, the average score S1t2, as a percentage, was found to be a) What was the average score when the students initially took the test? b) What was the average score after 4 months? c) What was the average score after 24 months? d) What percentage of their original answers did the students retain after 2 years (24 months)? e) Find S¿1t2. f) Find the maximum value, if one exists. g) Find lim S1t2, and discuss its meaning.

70 60 50

30

b) Find P¿11502, and explain what this number represents. c) How many thousands of mechanical pencils should be sold to maximize profit?

S1t2 = 68 - 20 ln 1t + 12, t Ú 0.

V

40

335

V(t) = 58(1 – e –1.1t) + 20

t: q

20 10 5

10

15

t

a) Find V112 and V1122. b) Find V¿1t2. c) After how many months will the value of the stock first reach $75? d) Find lim V1t2. Discuss the value of the stock over a t: q long period of time. Is this trend typical? 84. Marginal revenue. The demand for a new computer game can be modeled by p1x2 = 53.5 - 8 ln x, where p1x2 is the price consumers will pay, in dollars, and x is the number of games sold, in thousands. Recall that total revenue is given by R1x2 = x # p1x2. a) Find R1x2. b) Find the marginal revenue, R¿1x2. c) Is there any price at which revenue will be maximized? Why or why not? 85. Marginal profit. The profit, in thousands of dollars, from the sale of x thousand mechanical pencils, can be estimated by P1x2 = 2x - 0.3x ln x. a) Find the marginal profit, P¿1x2.

88. Forgetting. Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t months, the average score S1t2, as a percentage, was found to be given by S1t2 = 78 - 15 ln 1t + 12, t Ú 0.

a) What was the average score when they initially took the test, t = 0? b) What was the average score after 4 months? c) What was the average score after 24 months? d) What percentage of their original answers did the students retain after 2 years (24 months)? e) Find S¿1t2. f) Find the maximum and minimum values, if they exist. g) Find lim S1t2 and discuss its meaning. t: q

89. Walking speed. Bornstein and Bornstein found in a study that the average walking speed v, in feet per second, of a person living in a city of population p, in thousands, is v1p2 = 0.37 ln p + 0.05. (Source: M. H. Bornstein and H. G. Bornstein, “The Pace of Life,” Nature, Vol. 259, pp. 557–559 (1976).) a) The population of Seattle is 571,000 1p = 5712. What is the average walking speed of a person living in Seattle? b) The population of New York is 8,100,000. What is the average walking speed of a person living in New York? c) Find v¿1p2. d) Interpret v¿1p2 found in part (c).

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90. Hullian learning model. A keyboarder learns to type W words per minute after t weeks of practice, where W is given by W1t2 = 10011 - e -0.3t2.

109. Proof of P3 of Theorem 3. M = a X, so M k = 1a X2k = a Xk. Thus, log a M k = Xk = k # log a M.

W(t) 100

W 80 60 40 20 5

10

15

t

a) Find W112 and W182. b) Find W¿1t2. c) After how many weeks will the keyboarder’s speed be 95 words per minute? d) Find lim W1t2, and discuss its meaning. t: q

111. Find lim

ln 11 + h2

h:0

SYNTHESIS Differentiate.

92. f 1x2 = ln 1x 3 + 125

93. f 1t2 = ln 1t 2 - t27

94. g1x2 = 3ln 1x + 5244

95. f 1x2 = ln 3ln 1ln 13x224

96. f 1t2 = ln 31t + 321t - 124 3

h

.

112. For any k 7 0, ln 1kx2 = ln k + ln x. Use this fact to show graphically why

91. Solve P = P0e kt for t.

97. f 1t2 = ln

110. Proof of P7 of Theorem 3. Let log b M = R. Then b R = M, and log a 1b R2 = log a M. Thus, R # log a b = log a M, log a M . and R = log a b It follows that log a M . log b M = log a b

2

1 - t 1 + t

99. f 1x2 = log 5 x

101. y = ln 25 + x 2

98. y = ln

x

5

18x + 522

100. f 1x2 = log 7 x 102. f 1t2 =

ln t 2 t

d 1 d ln 1kx2 = ln x = . x dx dx

TECHNOLOGY CONNECTION 113. Use natural logarithms to determine which is larger, e p or pe. (Hint: y = ln x is an increasing function.) e 114. Find 2 e. Compare it to other expressions of the type x 2x, with x 7 0. What can you conclude?

Use input–output tables to find each limit.

2

115. lim ln x x:1

116. lim ln x x: q

1 1 103. f 1x2 = x 5 aln x - b 5 5

Graph each function f and its derivative f œ . Use a graphing calculator, iPlot, or Graphicus.

xn+1 1 104. y = aln x b n + 1 n + 1

117. f 1x2 = ln x

118. f 1x2 = x ln x

119. f 1x2 = x 2 ln x

120. f 1x2 =

105. f 1x2 = ln

1 + 2x 1 - 2x

106. f 1x2 = ln 1ln x2

108. Proof of P2 of Theorem 3. M = a X and N = a Y, M aX so = Y = a X - Y. N a M Thus, log a = X - Y N = log a M - log aN.

x2

3

To prove Properties P1, P2, P3, and P7 of Theorem 3, let X  log a M and Y  log a N, and give reasons for the steps listed in Exercises 107–110. 107. Proof of P1 of Theorem 3. M = a X and N = a Y, so MN = a X # a Y = a X + Y. Thus, log a 1MN2 = X + Y = log a M + log a N.

ln x

Find the minimum value of each function. Use a graphing calculator, iPlot, or Graphicus. 121. f 1x2 = x ln x

122. f 1x2 = x 2 ln x

Answers to Quick Checks 1. (a) 1.183; (b) - 0.471; (c) 0.471; (d) 1.424; (e) 1.827; (f) 12 2. (a) 2.3025; (b) 0.9163; (c) - 0.9163; (d) 3.4655; (e) 3.6094; (f) 0.4307 3. (a) t L 4.3820; 5 (b) t L 17.329 4. (a) ; (b) x 2 + 3x 2 ln x + 4; x 1 1 - 2 ln x 6x 1 (c) 5. (a) ; (b) 2 ; (c) ; x x1ln 5x2 x3 3x + 4 4x 5 + 2 (d) 6. The advertising campaign should run for x1x 5 - 22 about 37 days to maximize profit.

3.3

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•

Exponential Growth

• Find functions that satisfy dP>dt = kP.

Consider the function

• Solve application problems using exponential growth and limited growth models.

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OBJECTIVES

• Convert between growth rate and doubling time.

Applications: Uninhibited and Limited Growth Models

f1x2 = 2e 3x. Differentiating, we get

f ¿1x2 = 2e 3x # 3 = f1x2 # 3.

y

Graphically, this says that the derivative, or slope of the tangent line, is simply the constant 3 times the function value.

m = f′(x) = 3 ⋅ f(x)

Quick Check 1 Differentiate f1x2 = 5e 4x. Then express f ¿1x2 in terms of f1x2.

f(x)

O Quick Check 1

x

x

Although we do not prove it here, the exponential function f1x2 = ce kx is the only function for which the derivative is a constant times the function itself.

THEOREM 8 A function y = f1x2 satisfies the equation dy dx

= ky

or

f ¿1x2 = k # f1x2

or

f1x2 = ce kx

if and only if y = ce kx for some constant c.

■ EXAMPLE 1 Find the general form of the function that satisfies the equation dA = 5A. dt

Solution The function is A = ce 5t, or A1t2 = ce 5t, where c is an arbitrary constant. As a check, note that

A¿1t2 = ce 5t # 5 = 5 # A1t2.

■ EXAMPLE 2 Find the general form of the function that satisfies the equation dP = kP. dt

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Quick Check 2

Solution The function is P = ce kt, or P1t2 = ce kt, where c is an arbitrary constant.

Find the general form of the function that satisfies the equation

Check:

O

dN = kN. dt

dP = ce kt # k = kP. dt

O Quick Check 2 Whereas the solution of an algebraic equation is a number, the solutions of the equations in Examples 1 and 2 are functions. For example, the solution of 2x + 5 = 11 is the number 3, and the solution of the equation dP>dt = kP is the function P1t2 = ce kt. An equation like dP>dt = kP, which includes a derivative and which has a function as a solution, is called a differential equation.

■ EXAMPLE 3 Solve the differential equation f ¿1z2 = k # f1z2.

Solution The solution is f1z2 = ce kz. Check: f ¿1z2 = ce kz # k = f1z2 # k. We will discuss differential equations in more depth in Section 5.7.

Uninhibited Population Growth The equation dP = kP or P¿1t2 = kP1t2, with k 7 0, dt is the basic model of uninhibited (unrestrained) population growth, whether the population is comprised of humans, bacteria in a culture, or dollars invested with interest compounded continuously. In the absence of inhibiting or stimulating factors, a population normally reproduces at a rate proportional to its size, and this is exactly what dP>dt = kP says. The only function that satisfies this differential equation is given by P1t2 = ce kt, where t is time and k is the rate expressed in decimal notation. Note that P102 = ce k 0 = ce 0 = c # 1 = c,

#

so c represents the initial population, which we denote P0: P1t2 = P0e kt.

The graph of P1t2 = P0e kt, for k 7 0, shows how uninhibited growth produces a “population explosion.”

y

P(t) = P0ekt, for k  0

What will the world population be in 2020?

P0 t

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339

The constant k is called the rate of exponential growth, or simply the growth rate. This is not the rate of change of the population size, which varies according to dP = kP, dt but the constant by which P must be multiplied in order to get the instantaneous rate of change at any point in time. It is similar to the daily interest rate paid by a bank. If the daily interest rate is 0.07>365, then any given balance P is growing at the rate of 0.07>365 # P dollars per day. Because of the compounding, after 1 year, the interest earned will exceed 7% of P. When interest is compounded continuously, the interest rate is a true exponential growth rate. A detailed explanation of this is presented at the end of this section.

■ EXAMPLE 4 Business: Interest Compounded Continuously. Suppose that an amount P0, in dollars, is invested in the Von Neumann Hi-Yield Fund, with interest compounded continuously at 7% per year. That is, the balance P grows at the rate given by dP = 0.07P. dt a) Find the function that satisfies the equation. Write it in terms of P0 and 0.07. b) Suppose that $100 is invested. What is the balance after 1 yr? c) In what period of time will an investment of $100 double itself?

Solution a) P1t2 = P0e 0.07t Note that P102 = P0. 0.07112 b) P112 = 100e = 100e 0.07 L 10011.0725082

It is best to skip this step when using a calculator.

L $107.25 c) We are looking for a number T such that P1T2 = $200. The number T is called the doubling time. To find T, we solve the equation

#

200 = 100e 0.07 T 2 = e 0.07T. We use natural logarithms to solve this equation:

Quick Check 3 Business: Interest Compounded Continuously. Repeat Example 4 for interest compounded continuously at 4% per year.

ln 2 = ln e 0.07T ln 2 = 0.07T ln 2 = T 0.07

Finding the natural logarithm of both sides By P5: ln e k = k

9.9 L T. Thus, $100 will double itself in approximately 9.9 yr.

O Quick Check 3 To find a general expression relating growth rate k and doubling time T, we solve the following: 2P0 2 ln 2 ln 2

= = = =

P0e kT e kT Dividing by P0 kT ln e kT.

Note that this relationship between k and T does not depend on P0. We now have the following theorem.

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THEOREM 9 The growth rate k and the doubling time T are related by kT = ln 2 L 0.693147, or

k =

ln 2 0.693147 L , T T

and

T =

ln 2 0.693147 L . k k

■ EXAMPLE 5 Business: Facebook Membership. The social-networking Web site Facebook connects people with other members they designate as friends. Membership in Facebook has been doubling every 6 months. What is the exponential growth rate of Facebook membership, as a percentage?

Solution We have Quick Check 4 Business: Internet Use. Worldwide use of the Internet is increasing at an exponential rate, with traffic doubling every 100 days. What is the exponential growth rate of Internet use?

k =

ln 2 0.693147 L T 6 months L 0.116 #

If possible, enter the calculation as 1ln 22>6 without approximating the logarithmic value.

1 . month

The exponential growth rate of Facebook membership is 11.6% per month.

O Quick Check 4 The Rule of 70 The relationship between doubling time T and interest rate k is the basis of a rule often used in business, called the Rule of 70. To estimate how long it takes to double your money, divide 70 by the rate of return: T =

ln 2 0.693147 100 # 0.693147 L = k k 100 k 69.3147 70 Remember that k is the interest rate L L . written as a decimal. 100k 100k

■ EXAMPLE 6 Life Science: World Population Growth. The world population was approximately 6.0400 billion at the beginning of 2000. It has been estimated that the population is growing exponentially at the rate of 0.016, or 1.6% per year. (How was this estimate determined? The answer is in the model we develop in the following Technology Connection.) Thus, dP = 0.016P, dt where t is the time, in years, after 2000. (Source: U.S. Census Bureau.) a) Find the function that satisfies the equation. Assume that P0 = 6.0400 and k = 0.016. b) Estimate the world population at the beginning of 2020 1t = 202. c) After what period of time will the population be double that in 2000?

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Solution

Quick Check 5

a) P1t2 = 6.0400e 0.016t b) P1202 = 6.0400e 0.0161202 = 6.0400e 0.32 L 8.3179 billion ln 2 ln 2 c) T = = = 43.3 yr k 0.016

O

Life Science: Population Growth in China. In 2006, the population of China was 1.314 billion, and the exponential growth rate was 0.6% per year. Thus,

Thus, according to this model, the population in 2000 will double itself by 2043. (No wonder environmentalists are alarmed!)

dP = 0.006P, dt

Under ideal conditions, the growth rate of this rapidly growing population of rabbits might be 11.7% per day. When will this population of rabbits double?

where t is the time, in years, after 2006. (Source: Time Almanac, 2007.) a) Find the function that satisfies the equation. Assume that P0 = 1.314 and k = 0.006. b) Estimate the population of China at the beginning of 2020. c) After what period of time will the population be double that in 2006?

O Quick Check 5

TECHNOLOGY CONNECTION

Exponential Models Using Regression Projecting World Population Growth The table below shows data regarding world population growth. A graph illustrating these data, along with the projected population in 2020, appeared in Section 3.1.

Year

World Population (in billions)

1927 1960 1974 1987 1998

2 3 4 5 6

How was the population projected for 2020? The graph shows a rapidly growing population that can be modeled with an exponential function. We carry out the regression procedure very much as we did in Section R.6, but here we choose ExpReg rather than LinReg. EDI T CALC TESTS 7 QuartReg 8:LinReg(a+bx) 9:LnReg 0:ExpReg A:PwrReg B:Logistic C:SinReg

ExpReg y = a*bx a = 1.488241E –13 b = 1.01579058

Note that this gives us an exponential model of the type y = a # b x, where y is the population, in billions, in year x. y = 11.488241 # 10-13211.015790582x.

(1)

The base here is not e, but we can make a conversion to an exponential function, base e, using the fact that b = e ln b and then multiplying exponents: b x = 11.015790582x = 1e ln 1.015790582x = e 1ln 1.015790582x

L e 0.0156672059x. We can now write equation (1) as

y = 11.488241 # 10-132e 0.0156672059x.

(2)

The advantage of this form is that we see the growth rate. Here the world population growth rate is about 0.016, or 1.6%. To find world population in 2008, we can substitute 2008 for x in either equation (1) or (2). We choose equation (2): y = 11.488241 # 10-132e 0.0156672059120082 L 6.8465 billion.

EXERCISES Use equation (1) or equation (2) to estimate world population in each year. 1. 2020

2. 2050

3. 2060

4. 2080

(continued)

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Exponential Models Using Regression (continued) Projecting College Costs

EXERCISES

For Exercises 5 and 6, use the data regarding projected college costs (tuition and room and board) listed in the table below.

School Year, x

Costs of Attending a Public 4-year College or University (2006–2007 dollars)

1998–1999, 0

9,959

1999–2000, 1

9,978

2000–2001, 2

10,089

2001–2002, 3

10,535

2002–2003, 4

10,971

2003–2004, 5

11,709

2004–2005, 6

12,168

2005–2006, 7

12,421

2006–2007, 8

12,797

2007–2008, 9

12,944

5. Use REGRESSION to fit an exponential function

y = a # b x to the data. Let 1998–1999 be represented by x = 0 and let y = the cost, in dollars. Then convert that formula to an exponential function, base e, and determine the exponential growth rate.

6. Use either of the exponential functions found in

Exercise 5 to estimate college costs in the 2014–2015, 2017–2018, and 2039–2040 school years.

(Source: National Center for Education Statistics, Annual Digest of Education Statistics: 2008.)

In the preceding Technology Connection, we used regression to create an exponential model. There is another way to create such a model if regression is not an option. As shown in Example 7, two representative data points are sufficient to determine P0 and k in P1t2 = P0e kt.

■ EXAMPLE 7 Business: Batman Comic Book. A 1939 comic book with the first appearance of the “Caped Crusader,” Batman, sold at auction in Dallas in 2010 for a record $1.075 million. The comic book originally cost 10¢ (or $0.10). Using two representative data points (0, $0.10) and (71, $1,075,000), we can find an exponential function that models the increasing value of the comic book. The modeling assumption is that the value V of the comic book has grown exponentially, as given by dV = kV. dt (Source: Heritage Auction Galleries.) a) Find the function that satisfies this equation. Assume that V0 = $0.10. b) Estimate the value of the comic book in 2020. c) What is the doubling time for the value of the comic book? d) In what year will the value of the comic book be $30 million, assuming there is no change in the growth rate?

Solution a) Because of the modeling assumption, we have V1t2 = V0e kt. Since V0 = $0.10, it follows that V1t2 = 0.10e kt.

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343

We have made use of the data point (0, $0.10). Next, we use the data point (71, $1,075,000) to determine k. We solve V1t2 = 0.10e kt, or 1,075,000 = 0.10k1712 for k, using natural logarithms: 1,075,000 1,075,000 0.10 10,750,000 ln 10,750,000 ln 10,750,000

= 0.10e k1712 = 0.10e 71k = e 71k

Dividing to simplify

= e 71k = ln e 71k = 71k

Finding the logarithm of both sides By P5: ln e k = k

16.190416 L k 71

Skip this step when using a calculator.

0.228 L k.

Quick Check 6 Business: Batman Comic Book. In Example 7, the consigner had bought the comic book in the late 1960s for $100. (Source: Heritage Auction Galleries.) Assume that the year of purchase was 1969 and that the value V of the comic book has since grown exponentially, as given by dV = kV, dt where t is the number of years since 1969. a) Use the data points (0, $100) and (41, $1,075,000) to find the function that satisfies the equation. b) Estimate the value of the comic book in 2020, and compare your answer to that of Example 7. c) What is the doubling time for the value of the comic book? d) In what year will the value of the comic book be $30 million? Compare your answer to that of Example 7.

Rounding to the nearest thousandth

The desired function is V1t2 = 0.10e 0.228t, where V is in dollars and t is the number of years since 1939. b) To estimate the value of the comic book in 2020, which is 2020 - 1939 = 81 years after 1939, we substitute 81 for t in the equation: V1t2 = 0.10e 0.228t V1812 = 0.10e 0.2281812 L $10,484,567. Not a bad resale value for a 10¢ comic book, presuming someone will pay the price! c) The doubling time T is given by T =

ln 2 ln 2 = L 3.04 yr. k 0.228

d) We substitute $30,000,000 for V1t2 and solve for t: V1t2 = 0.10e 0.228t 30,000,000 = 0.10e 0.228t 30,000,000 0.10 300,000,000 ln 300,000,000 ln 300,000,000 ln 300,000,000 0.228 19.519293 0.228 86

= e 0.228t = e 0.228t = ln e 0.228t = 0.228t = t L t

Skip this step when using a calculator.

L t.

Rounding to the nearest year

We add 86 to 1939 to get 2025 as the year in which the value of the comic book will reach $30 million. Note that in part (a) of this example, we find ln 10,750,000 and divide by 71, obtaining approximately 0.228. We then use that value for k in part (b). Answers are found this way in the exercises. You may note some variation in the last one or two decimal places of your answers if you round as you go.

O Quick Check 6

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Models of Limited Growth The growth model P1t2 = P0e kt has many applications to unlimited population growth, as we have seen in this section. However, there are often factors that prevent a population from exceeding some limiting value L—perhaps a limitation on food, living space, or other natural resources. One model of such growth is P1t2 =

L , for k 7 0, 1 + be -kt

P(t) L

which is called the logistic equation, or logistic function.

L , 1 + be−kt for k > 0

P(t) =

P0

t

■ EXAMPLE 8 Business: Satellite Radio Subscribers. Satellite radio companies provide subscribers with clear signals of hundreds of radio stations, including music, talk, and sports. The provider XM started up in 2001, followed by Sirius in 2002. Both companies did well, experiencing what seemed like exponential growth, but the slowing of this growth led Sirius to buy out XM in 2008, forming Sirius XM. The combined number of subscribers N, in millions, after time t, in years since 2000, with t = 1 corresponding to 2001, can be modeled by the logistic equation N1t2 =

19.362 . 1 + 295.393e -1.11t

(Source: Sirius XM Radio, Inc.) a) b) c) d)

Find the combined number of subscribers after 1 yr (in 2001), 3 yr, 5 yr, and 8 yr. Find the rate at which the number of subscribers was growing after 8 yr. Graph the equation. Explain why an uninhibited growth model is inappropriate but a logistic equation is appropriate to model this growth.

Solution a) We use a calculator to find the function values: N112 N132 N152 N182

= = = =

0.197 million, 1.673 million, 9.013 million, 18.598 million.

After 1 yr, there were about 197,000 subscribers. After 3 yr, there were about 1,673,000 subscribers. After 5 yr, there were about 9,013,000 subscribers. After 8 yr, there were about 18,598,000 subscribers. b) We find the rate of change using the Quotient Rule: N1t2 = N¿1t2 = =

19.362 , 1 + 295.393e -1.11t

11 + 295.393e -1.11t2 # 0 - 19.3621295.393e -1.11t21- 1.112 11 + 295.393e -1.11t22

6348.533e -1.11t . 11 + 295.393e -1.11t22

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345

Next, we use a calculator to evaluate the derivative at t = 8: N¿182 = 0.815.

Life Science: Spread of an Epidemic. In a town whose population is 3500, an epidemic of a disease occurs. The number of people N infected t days after the disease first appears is given by N1t2 =

After 8 yr, the number of subscribers was growing at a rate of 0.815 million, or 815,000, per year. c) The graph follows. N(t)

Number of subscribers (in millions)

Quick Check 7

3500 . 1 + 19.9e -0.6t

a) How many people are initially infected with the disease 1t = 02? b) Find the number infected after 2 days, 5 days, 8 days, 12 days, and 18 days. c) Graph the equation. d) Find the rate at which the disease is spreading after 16 days. e) Using this model, determine whether all 3500 residents will ever be infected.

20

N(t) =

10

19.362 1 + 295.393e –1.11 t

10

t

Number of years since 2000

d) An uninhibited growth model is inappropriate because as more and more subscribers are added, the population contains fewer who have not subscribed, perhaps because of the cost or a lack of awareness or interest, or simply because the population is finite. The logistic equation, graphed in part (c), displays the rapid rise in the number of subscribers over the early years of the business as well as the slower growth in later years. It would appear that the limiting value of subscribers is between 19 and 20 million.

O Quick Check 7 Another model of limited growth is provided by the function P1t2 = L11 - e -kt2, for k 7 0,

which is shown graphed below. This function also increases over the entire interval 30, q 2, but increases most rapidly at the beginning, unlike the logistic equation. P(t)

L P(t) = L(1 – e –kt), for k  0

0

t

Business Application: An Alternative Derivation of e and P1t2  P0ekt The number e can also be found using the compound-interest formula (which was developed in Chapter R), A = Pa1 +

i nt b , n

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where A is the amount that an initial investment P will be worth after t years at interest rate i, expressed as a decimal, compounded n times per year. Suppose that $1 is invested at 100% interest 1i = 100% = 12 for 1 yr (though obviously no bank would pay this). The formula becomes A = a1 +

1 n b . n

Suppose that the number of compounding periods, n, increases indefinitely. Let’s investigate the behavior of the function. We obtain the following table of values and graph. 1 n A = a1 + b n

n 1 2 3 4 12 52 365 8,760 525,600

A e ≈ 2.71828

$2.00000 $2.25000 $2.37037 $2.44141 $2.61304 $2.69260 $2.71457 $2.71813 $2.71828

3.0 2.5 2.0

⎛ 1⎛n A = ⎝1 + –n ⎝

1.5 1.0 0.5 10 20 30 40 50

n

1 n If interest is compounded continuously, we have A = lim a1 + b , or n n: q equivalently, A = lim a1 + h:0

1 1>h b , or A = lim 11 + h21>h. h:0 1>h

Recall from Section 3.1 that lim 11 + h21>h = e. Thus, h:0

lim a1 +

n: q

1 n b = e. n

This result is confirmed by the graph and table above. For $1, invested at an interest rate of 100% with increasingly frequent compounding periods, the greatest value it could grow to in 1 yr is about $2.7183. To develop the formula P1t2 = P0e kt, we again start with the compound-interest formula, A = Pa1 +

i nt b , n

and assume that interest will be compounded continuously. Let P = P0 and i = k to obtain P1t2 = P0 a1 +

k nt b . n

We are interested in what happens as n approaches q . To find this limit, we first let k 1 = , so that qk = n. n q

Exercise Set 3.3

TECHNOLOGY CONNECTION

347

Note that since k is a positive constant, as n gets large, so must q. Thus,

Exploratory

P1t2 = lim cP0 a1 + n: q

Graph 1 x y = a1 + b x using the viewing window 30, 5000, 0, 54, with Xscl = 1000 and Yscl = 1. Trace along the graph. Why does the graph appear to be horizontal? As you trace to the right, note the value of the y-coordinate. Is it approaching a constant? What seems to be its limiting value?

k nt b d n

1 qkt = P0 lim c a1 + b d q q: q = P0 c lim a1 + q: q

1 q kt b d q

Letting the number of compounding periods become infinite The limit of a constant times a function is the constant times the limit. We also substitute 1>q for k>n and qk for n. Also, q : q because n : q. The limit of a power is the power of the limit: a form of Limit Property L2 in Section 1.2.

= P03e4kt.

Section Summary • Uninhibited growth can be modeled by a differential dP equation of the type = kP, whose solutions are dt kt P1t2 = P0e .

• Certain kinds of limited growth can be modeled by equaL tions such as P1t2 = and P1t2 = L11 - e -kt2, 1 + be -kt for k 7 0.

• The rate of exponential growth k and the doubling time T ln 2 ln 2 are related by the equation T = , or k = . k T

EXERCISE SET

3.3 1. Find the general form of f if f ¿1x2 = 4f 1x2. 2. Find the general form of g if g¿1x2 = 6g1x2. 3. Find the general form of the function that satisfies dA>dt = - 9A. 4. Find the general form of the function that satisfies dP>dt = - 3P1t2. 5. Find the general form of the function that satisfies dQ>dt = kQ. 6. Find the general form of the function that satisfies dR>dt = kR.

APPLICATIONS Business and Economics 7. U.S. patents. The number of applications for patents, N, grew dramatically in recent years, with growth averaging about 4.6% per year. That is, N¿1t2 = 0.046N1t2. (Source: New York Times, 11/13/05, p. C1.)

a) Find the function that satisfies this equation. Assume that t = 0 corresponds to 1980, when approximately 112,000 patent applications were received. b) Estimate the number of patent applications in 2020. c) Estimate the doubling time for N(t). 8. Franchise expansion. Pete Zah’s, Inc., is selling franchises for pizza shops throughout the country. The marketing manager estimates that the number of franchises, N, will increase at the rate of 10% per year, that is, dN = 0.10N. dt a) Find the function that satisfies this equation. Assume that the number of franchises at t = 0 is 50. b) How many franchises will there be in 20 yr? c) In what period of time will the initial number of 50 franchises double? 9. Compound interest. Suppose that P0 is invested in the Mandelbrot Bond Fund for which interest is compounded continuously at 5.9% per year. That is, the balance P grows at the rate given by dP = 0.059P. dt

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a) Find the function that satisfies the equation. Write it in terms of P0 and 0.059. b) Suppose that $1000 is invested. What is the balance after 1 yr? After 2 yr? c) When will an investment of $1000 double itself? 10. Compound interest. Suppose that P0 is invested in a savings account for which interest is compounded continuously at 4.3% per year. That is, the balance P grows at the rate given by dP = 0.043P. dt a) Find the function that satisfies the equation. Write it in terms of P0 and 0.043. b) Suppose that $20,000 is invested. What is the balance after 1 yr? After 2 yr? c) When will an investment of $20,000 double itself? 11. Bottled water sales. Since 2000, sales of bottled water have increased at the rate of approximately 9.3% per year. That is, the volume of bottled water sold, G, in billions of gallons, t years after 2000 is growing at the rate given by dG = 0.093G. dt (Source: The Beverage Marketing Corporation.)

a) Find the function that satisfies the equation, given that approximately 4.7 billion gallons of bottled water were sold in 2000. b) Predict the number of gallons of water sold in 2025. c) What is the doubling time for G(t)? 12. Annual net sales. Green Mountain Coffee Roasters produces many varieties of flavored coffees, teas, and K-cups. The net sales S of the company have grown exponentially at the rate of 36.1% per year, and the growth can be approximated by dS = 0.361S, dt where t is the number of years since 2004. (Source: Green Mountain Coffee Roasters financial statements.) a) Find the function that satisfies the equation, given that net sales in 2004 1t = 02 were approximately $120,400. b) Estimate net sales in 2006, 2008, and 2015. c) What is the doubling time for S(t)? 13. Annual interest rate. Euler Bank advertises that it compounds interest continuously and that it will double your money in 15 yr. What is its annual interest rate?

14. Annual interest rate. Hardy Bank advertises that it compounds interest continuously and that it will double your money in 12 yr. What is its annual interest rate? 15. Oil demand. The growth rate of the demand for oil in the United States is 10% per year. When will the demand be double that of 2006? 16. Coal demand. The growth rate of the demand for coal in the world is 4% per year. When will the demand be double that of 2006? Interest compounded continuously. For Exercises 17–20, complete the following. Initial Investment at t  0, P0 17. $75,000

Interest Rate, k

Doubling Time, T (in years)

6.2%

18. $5,000 19. 20.

Amount after 5 yr

$7,130.90 8.4%

$11,414.71 11

$17,539.32

21. Art masterpieces. In 2004, a collector paid $104,168,000 for Pablo Picasso’s “Garcon à la Pipe.” The same painting sold for $30,000 in 1950. (Source: BBC News, 5/6/04.)

a) Find the exponential growth rate k, to three decimal places, and determine the exponential growth function V, for which V(t) is the painting’s value, in dollars, t years after 1950. b) Predict the value of the painting in 2015. c) What is the doubling time for the value of the painting? d) How long after 1950 will the value of the painting be $1 billion? 22. Per capita income. In 2000, U.S. per capita personal income I was $29,849. In 2008, it was $39,742. (Source: U.S. Bureau of Economic Analysis.) Assume that the growth of U.S. per capita personal income follows an exponential model. a) Letting t = 0 be 2000, write the function. b) Predict what U.S. per capita income will be in 2020. c) In what year will U.S. per capita income be double that of 2000?

Exercise Set 3.3

23. Federal receipts. In 1990, U.S. federal receipts (money taken in), E, were $1.031 billion. In 2009, federal receipts were $2.523 billion. (Source: National Center for Education Statistics.) Assume that the growth of federal receipts follows an exponential model and use 1990 as the base year 1t = 02. a) Find the value of k to six decimal places, and write the function, with E1t2 in billions of dollars. b) Estimate federal receipts in 2015. c) When will federal receipts be $10 billion? 24. Consumer price index. The consumer price index compares the costs, c, of goods and services over various years, where 1983 is used as a base 1t = 02. The same goods and services that cost $100 in 1983 cost $216 in 2010. Assuming an exponential model: a) Write the function, rounding k to five decimal places. b) Estimate what the goods and services costing $100 in 1983 will cost in 2020. c) In what year did the same goods and services cost twice the 1983 price?

349

d) What is the doubling time for sales of shredders? e) Compare your answers to parts (a)–(d) with those from Exercise 25. Decide which exponential function seems better to you, and explain why. 27. Value of Manhattan Island. Peter Minuit of the Dutch West India Company purchased Manhattan Island from the natives living there in 1626 for $24 worth of merchandise. Assuming an exponential rate of inflation of 5%, how much will Manhattan be worth in 2020? 28. Total revenue. Intel, a computer chip manufacturer, reported $1265 million in total revenue in 1986. In 2005, the total revenue was $38.8 billion. (Source: U.S. Securities and Exchange Commission.) Assuming an exponential model, find the growth rate k, to four decimal places, and write the revenue function R, with R(t) in billions of dollars. Then predict the company’s total revenue for 2012.

Sales of paper shredders. Data in the following bar graph show paper shredder sales in recent years. Use these data for Exercises 25 and 26. SHREDDER BOOM: ESTIMATED PAPER SHREDDER SALES

Sales (in millions of dollars)

y $406 million

425 400 375 $324 $309 million 325 $294 million 300 $280 million million 275 350

2000

2001

2002 2003 Year

$350 million

2004

2005

x

(Source: www.sfgate.com.)

25. a) Use REGRESSION to fit an exponential function y = a # b x to the data. Let y be in millions of dollars. Then convert that formula to an exponential function, base e, where x is the number of years after 1990, and determine the exponential growth rate. (See the Technology Connection on p. 341.) b) Estimate the total sales of paper shredders in 2007 and in 2012. c) After what amount of time will sales be $500 million? d) What is the doubling time for sales of shredders? 26. a) To find an exponential function, base e, that fits the data, find k using the points (10, 280) and (15, 406). Then write the function. (Assume that x is the number of years after 1990.) b) Estimate the total sales of paper shredders in 2007 and in 2012. c) After what amount of time will total sales be $500 million?

29. The U.S. Forever Stamp. On May 12, 2008, the U.S. Postal Service reissued the Forever Stamp (which features an image of the Liberty Bell). The Forever Stamp is always valid as first-class postage on standard envelopes weighing 1 ounce or less, regardless of any subsequent increases in the first-class rate. (Source: U.S. Postal Service.) a) The cost of first-class postage stamp was 4¢ in 1962 and 44¢ in 2010. This increase represents exponential growth. Write the function S for the cost of a stamp t years after 1962 1t = 02. b) What was the growth rate in the cost? c) Predict the cost of a first-class postage stamp in 2013, 2016, and 2019. d) An advertising firm spent $4400 on 10,000 first-class postage stamps in 2009. Knowing it will need 10,000 first-class stamps in each of the years 2010–2020, it decides at the beginning of 2010 to try to save money by spending $4400 on 10,000 Forever Stamps, but also buying enough of the stamps to cover the years 2011 through 2020. Assuming there is a postage increase in each of the years 2013, 2016, and 2019, to the cost predicted in part (c), how much money will the firm save by buying Forever Stamps? e) Discuss the pros and cons of the purchase decision described in part (d). 30. Average salary of Major League baseball players. In 1970, the average salary of Major League baseball players was $29,303. In 2005, the average salary was $2,632,655. (Source: Baseball Almanac.)Assuming exponential growth occurred, what was the growth rate to the nearest hundredth of a percent? What will the average salary be in 2015? In 2020?

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31. Effect of advertising. Suppose that SpryBorg Inc. introduces a new computer game in Houston using television advertisements. Surveys show that P% of the target audience buy the game after x ads are broadcast, satisfying P1x2 = a) b)

c) d)

Life and Physical Sciences Population growth. For Exercises 36–40, complete the following.

100

. 1 + 49e -0.13x What percentage buy the game without seeing a TV ad 1x = 02? What percentage buy the game after the ad is run 5 times? 10 times? 20 times? 30 times? 50 times? 60 times? Find the rate of change, P¿1x2. Sketch a graph of the function.

32. Cost of a Hershey bar. The cost of a Hershey bar was $0.05 in 1962 and $0.75 in 2010 (in a supermarket, not in a movie theater). a) Find an exponential function that fits the data. b) Predict the cost of a Hershey bar in 2015 and 2025. 33. Superman comic book. Three days before the sale of the Batman comic book (Example 7) in 2010, a 1938 comic book with the first appearance of Superman sold at auction in Dallas for a record $1.0 million. The comic book originally cost 10¢ ($0.10). (Source: Heritage Auction Galleries.) Using two representative data points, (0, $0.10) and (72, $1,000,000), we can approximate the data with an exponential function. The modeling assumption is that the value V of the comic book has grown exponentially, as given by dV = kV. dt (In the summer of 2010, a family in the southern United States was facing foreclosure on their mortgage and loss of their home. Then, as they were packing, an amazing twist of fate occurred; they came across some old comic books in the basement and one of them was this first superman comic. They sold it and saved their house.) a) Find the function that satisfies this equation. Assume that V0 = $0.10. b) Estimate the value of the comic book in 2020. c) What is the doubling time for the value of the comic book? d) After what time will the value of the comic book be $30 million, assuming there is no change in the growth rate? 34. Batman comic book. Refer to Example 7. In what year will the value of the comic book be $5 million? 35. Batman comic book. Refer to Example 7. In what year will the value of the comic book be $10 million?

Population

Exponential Growth Rate, k

36. Mexico

Doubling Time, T (in years)

3.5%>yr

37. Europe

69.31

38. Oil reserves

6.931

39. Coal reserves

17.3

40. Alaska

2.794%>yr

41. Yellowstone grizzly bears. In 1972, the population of grizzly bears in Yellowstone National Park had shrunk to approximately 190. In 2005, the number of Yellowstone grizzlies had grown to about 610. (Source: New York Times, 9/26/05.) Find an exponential function that fits the data, and then predict Yellowstone’s grizzly bear population in 2016. Round k to three decimal places.

42. Bicentennial growth of the United States. The population of the United States in 1776 was about 2,508,000. In the country’s bicentennial year, the population was about 216,000,000. a) Assuming an exponential model, what was the growth rate of the United States through its bicentennial year? b) Is exponential growth a reasonable assumption? Explain. 43. Limited population growth. A ship carrying 1000 passengers has the misfortune to be wrecked on a small island from which the passengers are never rescued. The natural resources of the island restrict the growth of the population to a limiting value of 5780, to which the population gets closer and closer but which it never reaches. The population of the island after time t, in years, is approximated by the logistic equation P1t2 =

5780 1 + 4.78e -0.4t

.

Exercise Set 3.3

a) Find the population after 0 yr, 1 yr, 2 yr, 5 yr, 10 yr, and 20 yr. b) Find the rate of change, P¿1t2. c) Sketch a graph of the function.

351

months. Suppose that the percentage P of physicians prescribing a new cancer medication after t months is approximated by P1t2 = 10011 - e -0.4t2.

a) What percentage of doctors are prescribing the medication after 0 months? 1 month? 2 months? 3 months? 5 months? 12 months? 16 months? b) Find P¿172, and interpret its meaning. c) Sketch a graph of the function. 48. Spread of infection. Spread by skin-to-skin contact or via shared towels or clothing, methicillin-resistant Staphylococcus aureus (MRSA) can easily infect growing numbers of students at a university. Left unchecked, the number of cases of MRSA on a university campus t weeks after the first 9 cases occur can be modeled by N1t2 = This island may have achieved its limited population growth.

44. Limited population growth. A lake is stocked with 400 rainbow trout. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 2500. (See Exercise 43.) The population of trout in the lake after time t, in months, is approximated by P1t2 =

2500 1 + 5.25e -0.32t

.

a) Find the population after 0 months, 1 month, 5 months, 10 months, 15 months, and 20 months. b) Find the rate of change, P¿1t2. c) Sketch a graph of the function.

Social Sciences 45. Women college graduates. The number of women graduating from 4-yr colleges in the United States grew from 1930, when 48,869 women earned a bachelor’s degree, to 2005, when approximately 832,000 women received such a degree. (Source: National Center for Education Statistics.) Find an exponential function that fits the data, and the exponential growth rate, rounded to the nearest hundredth of a percent. 46. Hullian learning model. The Hullian learning model asserts that the probability p of mastering a task after t learning trials is approximated by p1t2 = 1 - e -kt, where k is a constant that depends on the task to be learned. Suppose that a new dance is taught to an aerobics class. For this particular dance, the constant k = 0.28. a) What is the probability of mastering the dance’s steps in 1 trial? 2 trials? 5 trials? 11 trials? 16 trials? 20 trials? b) Find the rate of change, p¿1t2. c) Sketch a graph of the function. 47. Diffusion of information. Pharmaceutical firms invest significantly in testing new medications. After a drug is approved by the Federal Drug Administration, it still takes time for physicians to fully accept and start prescribing the medication. The acceptance by physicians approaches a limiting value of 100%, or 1, after time t, in

568.803 1 + 62.200e -0.092t

.

(Source: Vermont Department of Health, Epidemiology Division.) a) Find the number of infected students beyond the first 9 cases after 3 weeks, 40 weeks, and 80 weeks. b) Find the rate at which the disease is spreading after 20 weeks. c) Explain why an unrestricted growth model is inappropriate but a logistic equation is appropriate for this situation. Then use a calculator to graph the equation. 49. Spread of a rumor. The rumor “People who study math all get scholarships” spreads across a college campus. Data in the following table show the number of students N who have heard the rumor after time t, in days. a) Use REGRESSION to fit a logistic equation, N1t2 =

c 1 + ae -bt

,

to the data. b) Estimate the limiting value of the function. At most, how many students will hear the rumor? Time, t (in days)

Number, N, Who Have Heard the Rumor

1 2 3 4 5 6 7 8 9 10 11 12

1 2 4 7 12 18 24 26 28 28 29 30

c) Graph the function. d) Find the rate of change, N¿1t2. e) Find lim N¿1t2, and explain its meaning. t: q

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•

Exponential and Logarithmic Functions

SYNTHESIS We have now studied models for linear, quadratic, exponential, and logistic growth. In the real world, understanding which is the most appropriate type of model for a given situation is an important skill. For each situation in Exercises 50–56, identify the most appropriate type of model and explain why you chose that model. List any restrictions you would place on the domain of the function. 50. The growth in value of a U.S. savings bond 51. The growth in the length of Zachary’s hair following a haircut 52. The growth in sales of cellphones 53. The drop and rise of a lake’s water level during and after a drought 54. The rapidly growing sales of organic foods 55. The number of manufacturing jobs that have left the United States since 1995 56. The life expectancy of the average American 57. Find an expression relating the exponential growth rate k and the quadrupling time T4. 58. Find an expression relating the exponential growth rate k and the tripling time T3. 59. A quantity Q1 grows exponentially with a doubling time of 1 yr. A quantity Q2 grows exponentially with a doubling time of 2 yr. If the initial amounts of Q1 and Q2 are the same, how long will it take for Q1 to be twice the size of Q2? 60. To what exponential growth rate per hour does a growth rate of 100% per day correspond? Business: effective annual yield. Suppose that $100 is invested at 7%, compounded continuously, for 1 yr. We know from Example 4 that the ending balance will be $107.25. This would also be the ending balance if $100 were invested at 7.25%, compounded once a year (simple interest). The rate of 7.25% is called the effective annual yield. In general, if P0 is invested at interest rate k, compounded continuously, then the effective annual yield is that number i satisfying P011  i2  P0e k. Then, 1  i  e k, or Effective annual yield  i  e k  1. 61. An amount is invested at 7.3% per year compounded continuously. What is the effective annual yield? 62. An amount is invested at 8% per year compounded continuously. What is the effective annual yield? 63. The effective annual yield on an investment compounded continuously is 9.42%. At what rate was it invested? 64. The effective annual yield on an investment compounded continuously is 6.61%. At what rate was it invested? 65. To show that the exponential growth rate can be determined using any two points, let y1 = Ce kt1, and y2 = Ce kt2. Solve this system of equations for k to show that k can be calculated directly, using 1t1, y12 and 1t2, y22.

66. Complete the table below, which relates growth rate k and doubling time T. Growth Rate, k (per year)

1%

2%

Doubling Time, T (in years)

14%

15

10

Graph T = 1ln 22>k. Is this a linear relationship? Explain. 67. Describe the differences in the graphs of an exponential function and a logistic function. 68. Explain how the Rule of 70 could be useful to someone studying inflation. 69. Business: total revenue. The revenue of Red Rocks, Inc., in millions of dollars, is given by the function R1t2 =

4000 1 + 1999e -0.5t

,

where t is measured in years. a) What is R(0), and what does it represent? b) Find lim R1t2. Call this value Rmax, and explain what t: q

it means. c) Find the value of t (to the nearest integer) for which R1t2 = 0.99Rmax.

Answers to Quick Checks

1. f ¿1x2 = 20e 4x; f ¿1x2 = 4f 1x2 2. N1t2 = ce kt, where c is an arbitrary constant 3. (a) P1t2 = P0e 0.04t; (b) $104.08; (c) 17.3 yr 4. 0.69% per day 5. (a) P1t2 = 1.314e 0.006t; (b) 1.429 billion; (c) 115.5 yr 6. (a) V1t2 = 100e 0.226t, which is almost the same as the growth rate found in Example 7; (b) $10,131,604, which is quite close to the estimate found in Example 7; (c) 3.07 yr; (d) after 55.8 yr, or in 2025, again about the same as was found in Example 7 7. (a) 167; (b) 500, 1758, 3007, 3449, 3499 (c) N(t) 3000 2000

N(t) =

3500 1 + 19.9e −0.6 t

1000 5

10

15

t

(d) After 16 days, the number of people infected is growing at the rate of about 2.8 people per day. (e) According to the model, lim N1t2 = 3500, so virtually all of the town’s t: q

residents will be affected.

3.4

3.4 OBJECTIVES • Find a function that satisfies dP>dt = - kP. • Convert between decay rate and half-life. • Solve applied problems involving exponential decay.

•

353

Applications: Decay

Applications: Decay In the equation of population growth, dP>dt = kP, the constant k is actually given by k = 1Birth rate2 - 1Death rate2.

Thus, a population “grows” only when the birth rate is greater than the death rate. When the birth rate is less than the death rate, k will be negative, and the population will be decreasing, or “decaying,” at a rate proportional to its size. For convenience in our computations, we will express such a negative value as - k, where k 7 0. The equation dP = - kP, where k 7 0, dt shows P to be decreasing as a function of time, and the solution P1t2 = P0e -kt shows it to be decreasing exponentially. This is called exponential decay. The amount present initially at t = 0 is again P0. y

y

TECHNOLOGY CONNECTION

Exploratory Using the same set of axes, graph y1 = e 2x and y2 = e -2x. Compare the graphs. Then, using the same set of axes, graph y1 = 100e -0.06x and y2 = 100e 0.06x. Compare each pair of graphs.

Decay – P = P0e kt, for k  0

Growth P = P0ekt, for k  0 P0

P0 t

t

Radioactive Decay Radioactive elements decay exponentially; that is, they disintegrate at a rate that is proportional to the amount present.

■ EXAMPLE 1 Life Science: Decay. Strontium-90 has a decay rate of 2.8% per year. The rate of change of an amount N of this radioactive isotope is given by dN = - 0.028N. dt a) Find the function that satisfies the equation. Let N0 represent the amount present at t = 0. b) Suppose that 1000 grams (g) of strontium-90 is present at t = 0. How much will remain after 70 yr? c) After how long will half of the 1000 g remain?

Solution a) N1t2 = N0e -0.028t b) N1702 = = L L

1000e -0.0281702 1000e -1.96 100010.14085842092 140.8584209.

After 70 yr, about 140.9 g of the strontium-90 remains.

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Exponential and Logarithmic Functions

Quick Check 1

c) We are asking, “At what time T will N1T2 be half of N0, or 12 # 1000?” The number T is called the half-life. To find T, we solve the equation 500 = 1000e -0.028T 1 -0.028T 2 = e

O

Life Science: Decay. Xenon133 has a decay rate of 14% per day. The rate of change of an amount N of this radioactive isotope is given by dN = - 0.14N. dt a) Find the function that satisfies the equation. Let N0 represent the amount present at t = 0. b) Suppose 1000 g of xenon133 is present at t = 0. How much will remain after 10 days? c) After how long will half of the 1000 g remain?

How can scientists determine that the remains of an animal or plant have lost 30% of the carbon-14? The assumption is that the percentage of carbon-14 in the atmosphere and in living plants and animals is the same. When a plant or animal dies, the amount of carbon-14 decays exponentially. The scientist burns the remains and uses a Geiger counter to determine the percentage of carbon-14 in the smoke. The amount by which this varies from the percentage in the atmosphere indicates how much carbon-14 has been lost through decay. The process of carbon-14 dating was developed by the American chemist Willard F. Libby in 1952. It is known that the radioactivity of a living plant measures 16 disintegrations per gram per minute. Since the half-life of carbon-14 is 5730 years, a dead plant with an activity of 8 disintegrations per gram per minute is 5730 years old, one with an activity of 4 disintegrations per gram per minute is about 11,500 years old, and so on. Carbon-14 dating can be used to determine the age of organic objects from 30,000 to 40,000 years old. Beyond such an age, it is too difficult to measure the radioactivity, and other methods are used.

1 2

ln ln 1 - ln 2 0 - ln 2 - ln 2 - 0.028 ln 2 0.028

-0.028T

We use 500 because 500 =

1 2

# 1000.

Dividing both sides by 1000

= ln e = - 0.028T = - 0.028T

Taking the natural logarithm of both sides

= T

Dividing both sides by - 0.028

Using the properties of logarithms

= T

0.693147 L T 0.028 25 L T. Thus, the half-life of strontium-90 is about 25 yr.

O Quick Check 1 We can find a general expression relating the decay rate k and the half-life T by solving the equation 1 2

P0 = P0e -kT

1 2 ln 12

ln 1 - ln 2 0 - ln 2 - ln 2 ln 2

= e -kT = = = = =

ln e -kT - kT - kT - kT kT.

Again, we have the following.

THEOREM 10 The decay rate, k, and the half-life, T, are related by kT = ln 2 = 0.693147, or k =

ln 2 T

and

T =

ln 2 . k

Thus, the half-life, T, depends only on the decay rate, k. In particular, it is independent of the initial population size.

3.4

•

Applications: Decay

355

The effect of half-life is shown in the radioactive decay curve below. Note that the exponential function gets close to, but never reaches, 0 as t gets larger. Thus, in theory, a radioactive substance never completely decays. Radioactive Decay Curve

Population

P0

1 – 2 P0

Exponential decay

1 – 4 P0 1 – 8 P0

T

2T

3T

Time

■ EXAMPLE 2 Life Science: Half-life. Plutonium-239, a common product of a functioning nuclear reactor, can be deadly to people exposed to it. Its decay rate is about 0.0028% per year. What is its half-life?

Solution We have Quick Check 2

O

Life Science: Half-life. a) The decay rate of cesium137 is 2.3% per year. What is its half-life? b) The half-life of barium-140 is 13 days. What is its decay rate?

ln 2 k ln 2 = 0.000028 L 24,755.

T =

Converting the percentage to decimal notation

Thus, the half-life of plutonium-239 is about 24,755 yr.

O Quick Check 2 ■ EXAMPLE 3 Life Science: Carbon Dating. The radioactive element carbon-14 has a half-life of 5730 yr. The percentage of carbon-14 present in the remains of plants and animals can be used to determine age. Archaeologists found that the linen wrapping from one of the Dead Sea Scrolls had lost 22.3% of its carbon-14. How old was the linen wrapping?

Solution Our plan is to find the exponential equation of the form N1t2 = N0e -kt,

replace N1t2 with 11 - 0.2232N0, and solve for t. First, however, we must find the decay rate, k: k = In 1947, a Bedouin youth looking for a stray goat climbed into a cave at Kirbet Qumran on the shores of the Dead Sea near Jericho and came upon earthenware jars containing an incalculable treasure of ancient manuscripts, which concern the Jewish books of the Bible. Shown here are fragments of those so-called Dead Sea Scrolls, a portion of some 600 or so texts found so far. Officials date them before A.D. 70, making them the oldest biblical manuscripts by 1000 years.

ln 2 0.693147 = L 0.00012097, or T 5730

0.012097% per year.

Thus, the amount N1t2 that remains from an initial amount N0 after t years is given by: N1t2 = N0e -0.00012097t.

Remember: k is positive, so - k is negative.

(Note: This equation can be used for all subsequent carbon-dating problems.)

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If the linen wrapping of a Dead Sea Scroll lost 22.3% of its carbon-14 from an initial amount P0, then 77.7% # P0 remains. To find the age t of the wrapping, we solve the following equation for t: 77.7% N0 0.777 ln 0.777 ln 0.777 ln 0.777 - 0.00012097 2086

Quick Check 3 Life Science: Carbon Dating. How old is a skeleton found at an archaeological site if tests show that it has lost 60% of its carbon-14?

= = = =

N0e -0.00012097t e -0.00012097t ln e -0.00012097t - 0.00012097t

= t L t.

Thus, the linen wrapping of the Dead Sea Scroll is about 2086 yr old.

O Quick Check 3 A Business Application: Present Value A representative of a financial institution is often asked to solve a problem like the following.

■ EXAMPLE 4 Business: Present Value. Following the birth of their granddaughter, two grandparents want to make an initial investment of P0 that will grow to $10,000 by the child’s 20th birthday. Interest is compounded continuously at 6%. What should the initial investment be?

Solution Using the equation P = P0e kt, we find P0 such that 0.06 # 20 10,000 = P0e

,

or 10,000 = P0e 1.2. Now 10,000 e 1.2

= P0,

or 10,000e -1.2 = P0, and, using a calculator, we have P0 = 10,000e -1.2 L $3011.94 Thus, the grandparents must deposit $3011.94, which will grow to $10,000 by the child’s 20th birthday. Economists call $3011.94 the present value of $10,000 due 20 yr from now at 6%, compounded continuously. The process of computing present value is called discounting. Another way to pose this problem is to ask “What must I invest now, at 6%, compounded continuously, in order to have $10,000 in 20 years?” The answer is $3011.94, and it is the present value of $10,000.

O

3.4

•

Applications: Decay

357

Computing present value can be interpreted as exponential decay from the future back to the present. P

Future value $10,000

Present value $3011.94

Quick Check 4 Business: Present Value. Repeat Example 4 for an interest rate of 4%.

20

t

O Quick Check 4 In general, the present value P0 of an amount P due t years later is found by solving the following equation for P0: P0e kt = P P P0 = kt = Pe -kt. e

THEOREM 11 The present value P0 of an amount P due t years later, at interest rate k, compounded continuously, is given by P0 = Pe -kt.

Newton’s Law of Cooling Consider the following situation. A hot cup of soup, at a temperature of 200°, is placed in a 70° room.* The temperature of the soup decreases over time t, in minutes, according to the model known as Newton’s Law of Cooling.

Newton’s Law of Cooling The temperature T of a cooling object drops at a rate that is proportional to the difference T - C, where C is the constant temperature of the surrounding medium. Thus, dT = - k1T - C2. dt

(1)

The function that satisfies equation (1) is T = T1t2 = ae -kt + C.

(2)

To check that T1t2 = ae -kt + C is the solution, find dT>dt and substitute dT>dt and T(t) into equation (1). This check is left to the student.

*Assume throughout this section that all temperatures are in degrees Fahrenheit unless noted otherwise.

O

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■ EXAMPLE 5 Life Science: Scalding Coffee. McDivett’s Pie Shoppes, a national restaurant firm, finds that the temperature of its freshly brewed coffee is 130°. The company fears that if customers spill hot coffee on themselves, lawsuits might result. Room temperature in the restaurants is generally 72°. The temperature of the coffee cools to 120° after 4.3 min. The company determines that it is safer to serve the coffee at a temperature of 105°. How long does it take a cup of coffee to cool to 105°?

Solution Note that C, the surrounding air temperature, is 72°. To find the value of a

in equation (2) for Newton’s Law of Cooling, we observe that at t = 0, we have T102 = 130°. We solve for a as follows:

#

130 = ae -k 0 + 72 130 = a + 72 58 = a. Note that a = 130 - 72, the difference between the original temperatures. Next, we find k using the fact that T14.32 = 120: 120 48 48 58 48 ln 58 - 0.1892420 k

#

= 58e -k 14.32 + 72 = 58e -4.3k = e -4.3k = ln e -4.3k

L - 4.3k L 0.044.

We now have T1t2 = 58e -0.044t. To see how long it will take the coffee to cool to 105°, we set T1t2 = 105 and solve for t:

Quick Check 5

O

Life Science: Scalding Coffee. Repeat Example 5, but assume that the coffee is sold in an ice cream shop, where the room temperature is 70°, and it cools to a temperature of 120° in 4 min.

105 33 33 58 33 ln 58 - 0.5639354

= e -0.044t = ln e -0.044t L - 0.044t

t L 12.8 min. Thus, to cool to 105°, the coffee should be allowed to cool for about 13 min.

O Quick Check 5

T a +C T(t) = ae –kt + C C

0

= 58e -0.044t + 72 = 58e -0.044t

t

The graph of T1t2 = ae -kt + C shows that lim T1t2 = C. The t: q temperature of the object decreases toward the temperature of the surrounding medium. Mathematically, this model tells us that the object’s temperature never quite reaches C. In practice, the temperature of the cooling object will get so close to that of the surrounding medium that no device could detect a difference. Let’s now see how Newton’s Law of Cooling can be used in solving a crime.

3.4

•

Applications: Decay

359

■ EXAMPLE 6 Forensics: When Was the Murder Committed? Found stabbed by a collection of number 2 pencils, Prof. Cal Kulice’s body was slumped over a stack of exams with plenty of red marks on them. A coroner arrives at noon, immediately takes the temperature of the body, and finds it to be 94.6°. She waits 1 hr, takes the temperature again, and finds it to be 93.4°. She also notes that the temperature of the room is 70°. When was the murder committed?

Solution Note that C, the surrounding air temperature, is 70°. To find a in T1t2 = ae -kt + C, we assume that the temperature of the body was normal when the murder occurred. Thus, T = 98.6° at t = 0: #

98.6 = ae -k 0 + 70, a = 28.6. This gives T1t2 = 28.6e -kt + 70. To find the number of hours N since the murder was committed, we must first determine k. From the two temperature readings the coroner made, we have 94.6 = 28.6e -kN + 70, or 24.6 = 28.6e -kN; 93.4 = 28.6e -k1N + 12 + 70, or 23.4 = 28.6e -k1N + 12.

(3) (4)

Dividing equation (3) by equation (4), we get 24.6 28.6e -kN = 23.4 28.6e -k1N + 12 = e -kN + k1N + 12 = e -kN + kN + k = e k. We solve this equation for k: ln

24.6 = ln e k 23.4 0.05 L k.

Taking the natural logarithm on both sides

Next, we substitute back into equation (3) and solve for N: Quick Check 6 Forensics. Repeat Example 6, assuming that the coroner arrives at 2 A.M., immediately takes the temperature of the body, and finds it to be 92.8°. She waits 1 hr, takes the temperature again, and finds it to be 90.6°. She also notes that the temperature of the room is 72°. When was the murder committed?

24.6 = 28.6e -0.05N 24.6 = e -0.05N 28.6 24.6 = ln e -0.05N ln 28.6 - 0.150660 L - 0.05N 3 L N. Since the coroner arrived at noon, or 12 o’clock, the murder occurred at about 9:00 A.M.

O Quick Check 6

O

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Section Summary • Several types of functions are additional candidates for curve fitting and applications: y

y

y

x a

a x

Exponential: f(x) = ab x, or ae kx a, b, k > 0, b ≠ 1

x Exponential: f(x) = ab –x, or ae –kx a, b, k > 0, b ≠ 1

P(t)

P(t)

L

L

Logarithmic: f(x) = a + b ln x, b > 0

P0

t Logistic Growth: L , for k > 0 1 + be−kt

P(t) =

t Inhibited Growth (not Logistic): P(t) = L(1 − e−kt), for k > 0

When we analyze a set of data, we can consider these models, as well as linear, quadratic, polynomial, and rational functions.

EXERCISE SET

3.4 APPLICATIONS Life and Physical Sciences 1. Radioactive decay. Iodine-131 has a decay rate of 9.6% per day. The rate of change of an amount N of iodine-131 is given by dN = - 0.096 N, dt where t is the number of days since the decay began. a) Let N0 represent the amount of iodine-131 present at t = 0. Find the exponential function that models the situation. b) Suppose that 500 g of iodine-131 is present at t = 0. How much will remain after 4 days? c) After how many days will half of the 500 g of iodine-131 remain?

2. Radioactive decay. Carbon-14 has a decay rate of 0.012097% per year. The rate of change of an amount N of carbon-14 is given by dN = - 0.00012097N, dt where t is the number of years since the decay began. a) Let N0 represent the amount of carbon-14 present at t = 0. Find the exponential function that models the situation. b) Suppose 200 g of carbon-14 is present at t = 0. How much will remain after 800 yr? c) After how many years will half of the 200 g of carbon14 remain?

361

Exercise Set 3.4

3. Chemistry. Substance A decomposes at a rate proportional to the amount of A present. a) Write an equation relating A to the amount left of an initial amount A0 after time t. b) It is found that 10 lb of A will reduce to 5 lb in 3.3 hr. After how long will there be only 1 lb left? 4. Chemistry. Substance A decomposes at a rate proportional to the amount of A present. a) Write an equation relating A to the amount left of an initial amount A0 after time t. b) It is found that 8 g of A will reduce to 4 g in 3 hr. After how long will there be only 1 g left? Radioactive decay. For Exercises 5–8, complete the following. Radioactive Substance

Decay Rate, k

5. Polonium-218

Business and Economics 16. Present value. Following the birth of a child, a parent wants to make an initial investment P0 that will grow to $30,000 by the child’s 20th birthday. Interest is compounded continuously at 6%. What should the initial investment be? P

Future value $30,000

Present value ? 20

Half-life, T

t

3 min

6. Radium-226

1600 yr

7. Lead-210

3.15%>yr

8. Strontium-90

2.77%>yr

9. Half-life. Of an initial amount of 1000 g of lead-210, how much will remain after 100 yr? See Exercise 7 for the value of k. 10. Half-life. Of an initial amount of 1000 g of polonium218, how much will remain after 20 min? See Exercise 5 for the value of k. 11. Carbon dating. How old is an ivory tusk that has lost 40% of its carbon-14? 12. Carbon dating. How old is a piece of wood that has lost 90% of its carbon-14? 13. Cancer treatment. Iodine-125 is often used to treat cancer and has a half-life of 60.1 days. In a sample, the amount of iodine-125 decreased by 25% while in storage. How long was the sample sitting on the shelf? 14. Carbon dating. How old is a Chinese artifact that has lost 60% of its carbon-14? 15. Carbon dating. Recently, while digging in Chaco Canyon, New Mexico, archaeologists found corn pollen that had lost 38.1% of its carbon-14. The age of this corn pollen was evidence that Indians had been cultivating crops in the Southwest centuries earlier than scientists had thought. (Source: American Anthropologist.) What was the age of the pollen?

17. Present value. Following the birth of a child, a parent wants to make an initial investment P0 that will grow to $40,000 by the child’s 20th birthday. Interest is compounded continuously at 5.3%. What should the initial investment be? 18. Present value. A homeowner wants to have $15,000 available in 5 yr to pay for new siding. Interest is 4.3%, compounded continuously. How much money should be invested? 19. Sports salaries. An athlete signs a contract that guarantees a $9-million salary 6 yr from now. Assuming that money can be invested at 5.7%, with interest compounded continuously, what is the present value of that year’s salary? 20. Actors’ salaries. An actor signs a film contract that will pay $12 million when the film is completed 3 yr from now. Assuming that money can be invested at 6.2%, with interest compounded continuously, what is the present value of that payment? 21. Estate planning. A person has a trust fund that will yield $80,000 in 13 yr. A CPA is preparing a financial statement for this client and wants to take into account the present value of the trust fund in computing the client’s net worth. Interest is compounded continuously at 4.8%. What is the present value of the trust fund? 22. Supply and demand. The supply and demand for stereos produced by a sound company are given by S1x2 = ln x and D1x2 = ln

163,000 , x

where S1x2 is the number of stereos that the company is willing to sell at price x and D1x2 is the quantity that the public is willing to buy at price x. Find the equilibrium point. (See Section R.5.) 23. Salvage value. A business estimates that the salvage value V1t2, in dollars, of a piece of machinery after t years is given by V1t2 = 40,000e -t. Chaco Canyon, New Mexico

a) What did the machinery cost initially? b) What is the salvage value after 2 yr?

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•

Exponential and Logarithmic Functions

Salvage value (in dollars)

c) Find the rate of change of the salvage value, and explain its meaning. V(t) 40,000 35,000 30,000

– V(t) = 40,000e t

25,000 20,000 15,000 10,000 5,000

2

4 Years

6

t

24. Salvage value. A company tracks the value of a particular photocopier over a period of years. The data in the table below show the value of the copier at time t, in years, after the date of purchase. Time, t (in years)

Salvage Value

0 1 2 3 4 5 6

$34,000 22,791 15,277 10,241 6,865 4,600 3,084

(Source: International Data Corporation.)

a) Use REGRESSION to fit an exponential function y = a # b x to the data. Then convert that formula to V1t2 = V0e -kt, where V0 is the value when the copier is purchased and t is the time, in years, from the date of purchase. (See the Technology Connection on p. 341.) b) Estimate the salvage value of the copier after 7 yr; 10 yr. c) After what amount of time will the salvage value be $1000? d) After how long will the copier be worth half of its original value? e) Find the rate of change of the salvage value, and interpret its meaning. 25. Actuarial science. An actuary works for an insurance company and calculates insurance premiums. Given an actual mortality rate (probability of death) for a given age, actuaries sometimes need to project future expected mortality rates of people of that age. An example of a formula that is used to project future mortality rates is Q1t2 = 1Q0 - 0.000552e 0.163t + 0.00055,

where t is the number of years into the future and Q0 is the mortality rate when t = 0. a) Suppose the initial actual mortality rate of a group of females aged 25 is 0.014 (14 deaths per 1000). What is the future expected mortality rate of this group of females 3, 5, and 10 yr in the future? b) Sketch the graph of the mortality function Q(t) for the group in part (a) for 0 … t … 10.

26. Actuarial science. Use the formula from Exercise 25. a) Suppose the initial actual mortality rate of a group of males aged 25 is 0.023 (23 deaths per 1000). What is the future expected mortality rate of this group of males 3, 5, and 10 yr in the future? b) Sketch the graph of the mortality function Q(t) for the group in part (a) for 0 … t … 10. c) What is the ratio of the mortality rate for 25-year-old males 10 yr in the future to that for 25-year-old females 10 yr in the future (Exercise 25a)? 27. U.S. farms. The number N of farms in the United States has declined continually since 1950. In 1950, there were 5,650,000 farms, and in 2005, that number had decreased to 2,100,990. (Sources: U.S. Department of Agriculture; National Agricultural Statistics Service.) Number of farms (in millions)

362

y 6 5,650,000 5 4 3 2,100,990

2 1 10

20 30 40 50 60 Number of years since 1950

x

Assuming the number of farms decreased according to the exponential decay model: a) Find the value of k, and write an exponential function that describes the number of farms after time t, where t is the number of years since 1950. b) Estimate the number of farms in 2009 and in 2015. c) At this decay rate, when will only 1,000,000 farms remain?

Social Sciences 28. Forgetting. In an art history class, students took a final exam. They were subsequently retested with an equivalent test at monthly intervals. Their average retest scores t months later are given in the following table. Time, t (in months)

Score, y

1 2 3 4 5 6

84.9% 84.6% 84.4% 84.2% 84.1% 83.9%

a) Use REGRESSION to fit a logarithmic function y = a + b ln x to the data. b) Use the function to predict the average test score after 8 months, 10 months, 24 months, and 36 months. c) After how long will the test scores fall below 82%? d) Find the rate of change of the scores, and interpret its meaning.

Exercise Set 3.4

29. Decline in beef consumption. The annual consumption of beef per person was about 64.6 lb in 2000 and about 61.2 lb in 2008. Assuming that B1t2, the annual beef consumption t years after 2000, is decreasing according to the exponential decay model: a) Find the value of k, and write the equation. b) Estimate the consumption of beef in 2015. c) In what year (theoretically) will the consumption of beef be 20 lb per person? 30. Population decrease of Russia. The population of Russia dropped from 150 million in 1995 to 140 million in 2009. (Source: CIA–The World Factbook.) Assume that P1t2, the population, in millions, t years after 1995, is decreasing according to the exponential decay model. a) Find the value of k, and write the equation. b) Estimate the population of Russia in 2016. c) When will the population of Russia be 100 million? 31. Population decrease of Ukraine. The population of Ukraine dropped from 51.9 million in 1995 to 45.7 million in 2009. (Source: CIA–The World Factbook.) Assume that P1t2, the population, in millions, t years after 1995, is decreasing according to the exponential decay model. a) Find the value of k, and write the equation. b) Estimate the population of Ukraine in 2015. c) After how many years will the population of Ukraine be 1 million, according to this model?

Life and Natural Sciences 32. Cooling. After warming the water in a hot tub to 100°, the heating element fails. The surrounding air temperature is 40°, and in 5 min the water temperature drops to 95°. a) Find the value of the constant a in Newton’s Law of Cooling. b) Find the value of the constant k. Round to five decimal places. c) What is the water temperature after 10 min? d) How long does it take the water to cool to 41°? e) Find the rate of change of the water temperature, and interpret its meaning. 33. Cooling. The temperature in a whirlpool bath is 102°, and the room temperature is 75°. The water cools to 90° in 10 min. a) Find the value of the constant a in Newton’s Law of Cooling. b) Find the value of the constant k. Round to five decimal places. c) What is the water temperature after 20 min? d) How long does it take the water to cool to 80°? e) Find the rate of change of the water temperature, and interpret its meaning. 34. Forensics. A coroner arrives at a murder scene at 2 A.M. He takes the temperature of the body and finds it to be 61.6°. He waits 1 hr, takes the temperature again, and

363

finds it to be 57.2°. The body is in a meat freezer, where the temperature is 10°. When was the murder committed? 35. Forensics. A coroner arrives at a murder scene at 11 P.M. She finds the temperature of the body to be 85.9°. She waits 1 hr, takes the temperature again, and finds it to be 83.4°. She notes that the room temperature is 60°. When was the murder committed? 36. Prisoner-of-war protest. The initial weight of a prisoner of war is 140 lb. To protest the conditions of her imprisonment, she begins a fast. Her weight t days after her last meal is approximated by W = 140e -0.009t. a) How much does the prisoner weigh after 25 days? b) At what rate is the prisoner’s weight changing after 25 days? 37. Political protest. A monk weighing 170 lb begins a fast to protest a war. His weight after t days is given by W = 170e -0.008t. a) When the war ends 20 days later, how much does the monk weigh? b) At what rate is the monk losing weight after 20 days (before any food is consumed)? 38. Atmospheric pressure. Atmospheric pressure P at altitude a is given by P = P0e -0.00005a, where P0 is the pressure at sea level. Assume that P0 = 14.7 lb>in2 (pounds per square inch). a) Find the pressure at an altitude of 1000 ft. b) Find the pressure at an altitude of 20,000 ft. c) At what altitude is the pressure 14.7 lb>in2? d) Find the rate of change of the pressure, and interpret its meaning. 39. Satellite power. The power supply of a satellite is a radioisotope (radioactive substance). The power output P, in watts (W), decreases at a rate proportional to the amount present; P is given by P = 50e -0.004t, where t is the time, in days. a) How much power will be available after 375 days? b) What is the half-life of the power supply? c) The satellite’s equipment cannot operate on fewer than 10 W of power. How long can the satellite stay in operation? d) How much power did the satellite have to begin with? e) Find the rate of change of the power output, and interpret its meaning. 40. Cases of tuberculosis. The number of cases N of tuberculosis in the United States has decreased continually since 1956, as shown in the following graph. In 1956 1t = 02, there were 69,895 cases. By 2006 1t = 502, this number had decreased by over 80%, to 13,767 cases.

364

•

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Exponential and Logarithmic Functions

CASES OF TUBERCULOSIS IN THE UNITED STATES

y Population (in thousands)

Number of cases (in thousands)

80

45.

69,895

60 40 13,767

20

8 6 4 2 0

1

2

3

4

5

x

6

Year

1956

1966

1976 1986 Year

1996

2006

y

46. Number of available apartments

0

(Source: Centers for Disease Control and Prevention.)

a) Find the value of k, and write an exponential function that describes the number of tuberculosis cases after time t, where t is the number of years since 1956. b) Estimate the number of cases in 2012 and in 2020. c) At this decay rate, in what year will there be 5000 cases?

Number of fruit flies (in hundreds)

30 0

1

2

3

4

5

x

6

47.

48. Sales (in millions)

Week

y 4 2 0 0

4

2

6 x

Year

y 1 0.5 0

49.

50. Sales (in millions)

41.

y

60

Sales (in millions)

For each of the scatterplots in Exercises 41–50, determine which, if any, of these functions might be used as a model for the data: a) Quadratic: f 1x2  ax 2  bx  c b) Polynomial, not quadratic c) Exponential: f 1x2  ae kx, k>0 d) Exponential: f 1x2  ae kx, k>0 e) Logarithmic: f 1x2  a  b ln x a f) Logistic: f 1x2 = 1  be kx

90

Sales (in millions)

Modeling

120

y 8 4 0

0

2

4 6 Year

8 x

50 Year

0

100 x

y 20 10 0 0

2

4

6x

Year

15

SYNTHESIS

10

51. Economics: supply and demand. The demand, D1x2, and supply, S1x2, functions for a certain type of multipurpose printer are as follows:

5

0

1

2

3

4

5

x

D1x2 = q = 480e -0.003x

Day

and 42.

Number of mosquitoes (in thousands)

y

S1x2 = q = 150e 0.004x. Find the equilibrium point. Assume that x is the price in dollars.

15 10

The Beer–Lambert Law. A beam of light enters a medium such as water or smoky air with initial intensity I0. Its intensity is decreased depending on the thickness (or concentration) of the medium. The intensity I at a depth (or concentration) of x units is given by

5

0

3

6

9

x

12

Hour

I  I0e Mx.

44. y

3.4 3.2 3.0 0

2

4 Year

6x

Sales (in millions)

Sales (in millions)

43.

The constant m (“mu”), called the coefficient of absorption, varies with the medium. Use this law for Exercises 52 and 53.

y 10 5 0 0

2

4 Year

6x

52. Light through smog. Concentrations of particulates in the air due to pollution reduce sunlight. In a smoggy area, m = 0.01 and x is the concentration of particulates measured in micrograms per cubic meter 1mcg>m32.

Exercise Set 3.4

What change is more significant—dropping pollution levels from 100 mcg>m3 to 90 mcg>m3 or dropping them from 60 mcg>m3 to 50 mcg>m3? Why? 53. Light through sea water. Sea water has m = 1.4 and x is measured in meters. What would increase cloudiness more—dropping x from 2 m to 5 m or dropping x from 7 m to 10 m? Explain.

365

a) What was the temperature of the water when you began? b) At what temperature does there seem to be a leveling off of the graph? c) What is the difference between your answers to parts (a) and (b)? d) How does the water temperature in part (b) compare with the room temperature? e) Find an equation that fits the data. Use this equation to check values of other data points. How do they compare? f) Is it ever “theoretically” possible for the temperature of the water to be the same as the room temperature? Explain. g) Find the rate of change of the temperature, and interpret its meaning. 55. An interest rate decreases from 8% to 7.2%. Explain why this increases the present value of an amount due 10 yr later.

Answers to Quick Checks 54. Newton’s Law of Cooling. Consider the following exploratory situation. Fill a glass with hot tap water. Place a thermometer in the glass and measure the temperature. Check the temperature every 30 min thereafter. Plot your data on this graph, and connect the points with a smooth curve.

160 150

Temperature (°F)

140 130 120 110 100 90 80 70 60 50 0

0

(Start)

30 60 90 120 150 180 210

Time (in minutes)

1. (a) N1t2 = N0e -0.14t; (b) 246.6 g; (c) 4.95 days 2. (a) 30.1 yr; (b) 5.3% per day 3. 7574 yr 4. $4493.29 5. 11.7 min 6. N = 2.2 hr, so the murder was committed 2 hr and 12 min before the coroner arrived, at about 11:48 P.M. on the previous day.

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Exponential and Logarithmic Functions

3.5 OBJECTIVES • Differentiate functions involving ax. • Differentiate functions involving loga x.

The Derivatives of ax and loga x The Derivative of ax To find the derivative of a x, for any base a, we first express a x as a power of e. To do this, we recall that log b x is the power to which b is raised in order to get x. Thus, blogb x = x. In particular, it follows that e loge A = A, or e ln A = A. If we replace A with a x, we have x

x

e ln a = a x, or a x = e ln a .

(1)

x

To find the derivative of a , we differentiate both sides: d x d ln ax a = e dx dx d x ln a = e dx d 1ln a2x = e dx = e 1ln a2x # ln a x = e ln a # ln a = a x # ln a.

Using a property of logarithms

Differentiating e kx with respect to x Using a property of logarithms Using equation (1)

Thus, we have the following theorem.

THEOREM 12 d x a = 1ln a2a x dx

■ EXAMPLE 1 Differentiate: a) y = 2 x; b) y = 11.42x; c) f1x2 = 32x. Solution d x Using Theorem 12 2 = 1ln 222 x dx Note that ln 2 L 0.7, so this equation verifies our earlier approximation of the derivative of 2 x as 10.722 x in Section 3.1. d b) 11.42x = 1ln 1.4211.42x dx c) Since f1x2 = 32x is of the form f1x2 = 3g1x2, the Chain Rule applies: a)

Quick Check 1 Differentiate: a) y = 5x; b) f1x2 = 4 x; c) y = 14.32x.

f ¿1x2 = 1ln 3232x #

d 12x2 dx = ln 3 # 32x # 2 = 2 ln 3 # 32x.

O Quick Check 1

O

3.5

•

The Derivatives of ax and loga x

367

Compare these formulas: d x a = 1ln a2a x dx

and

d x e = e x. dx

The simplicity of the latter formula is a reason for the use of base e in calculus. The many applications of e in natural phenomena provide additional reasons. One other result also follows from what we have done. If f1x2 = a x, we now know that

f ¿1x2 = a x 1ln a2.

Alternatively, in Section 3.1, we showed that if f1x2 = a x, then f ¿1x2 = a x # lim

ah - 1 . h:0 h

Thus, a x1ln a2 = a x # lim

ah - 1 . h:0 h

Dividing both sides by a x, we have the following.

THEOREM 13 ah - 1 h:0 h

ln a = lim

The Derivative of loga x Just as the derivative of a x is expressed in terms of ln a, so too is the derivative of log a x. To find this derivative, we first express log a x in terms of ln a using the change-of-base formula (P7 of Theorem 3) from Section 3.2: d log e x d log a x = a b dx dx log e a d ln x a b dx ln a 1 # d = 1ln x2 ln a dx

Using the change-of-base formula

=

=

1 is a constant. ln a

1 #1 . ln a x

THEOREM 14 d 1 #1 log a x = dx ln a x

Comparing this equation with d 1 ln x = , x dx we see another reason for the use of base e in calculus: we avoid obtaining the constant 1>1ln a2 when taking the derivative.

368

CHAPTER 3

•

Exponential and Logarithmic Functions

■ EXAMPLE 2 Differentiate: a) y = log 8 x; b) y = log x; c) f1x2 = log 3 1x 2 + 12; d) f1x2 = x 3 log 5 x.

Solution a) b)

d 1 #1 log 8 x = dx ln 8 x d log x = log 10 x dx 1 #1 = ln 10 x

Using Theorem 14

log x means log 10 x.

c) Note that f1x2 = log 3 1x 2 + 12 is of the form f1x2 = log 3 1g1x22, so the Chain Rule is required: f ¿1x2 = TECHNOLOGY CONNECTION

Exploratory Using the nDeriv feature, check the results of Examples 1 and 2 graphically. Then differentiate y = log 2 x, and check the result with your calculator.

=

1 # 1 # 2x ln 3 x 2 + 1

=

2x . 1ln 321x 2 + 12

Using the Chain Rule

d) Since f1x2 = x 3 log 5 x is of the form f1x2 = g1x2 # h1x2, the Product Rule is applied: f ¿1x2 = x 3 #

d log 5 x + log 5 x # 3x 2 Using the Product Rule dx 1 #1 = x3 # + log 5 x # 3x 2 ln 5 x x2 1 = + 3x 2 log 5 x, or x 2 a + 3 log 5 xb. ln 5 ln 5

Quick Check 2

O

Differentiate: a) y = log 2 x; b) f1x2 = - 7 log x; c) g1x2 = x 6 log x; d) y = log 8 1x 3 - 72.

1 # 1 # d 2 1x + 12 ln 3 x 2 + 1 dx

O Quick Check 2

Section Summary • The following rules apply when we differentiate exponential and logarithmic functions whose bases are positive but not the number e: d x a = 1ln a2a x, dx

and

d 1 #1 log a x = . dx ln a x

EXERCISE SET

3.5 Differentiate. 1. y = 7x

3. f 1x2 = 8x

5. g1x2 = x 315.42x

2. y = 6x

4. f 1x2 = 15x

6. g1x2 = x 513.72x

4

7. y = 7x

+2

8. y = 4 x

2

+5

2

9. y = e 8x

11. f 1x2 = 3x

4

10. y = e x

+1

12. f 1x2 = 12 7x - 4

369

Exercise Set 3.5

14. y = log 8 x

15. y = log 17 x

16. y = log 23 x

17. g1x2 = log 6 15x + 12

18. g1x2 = log 32 19x - 22

19. F1x2 = log 16x - 72

20. G1x2 = log 15x + 42

21. y = log 8 1x + x2 3

25. y = 6x # log 7 x

27. G1x2 = 1log 12 x25

#

log 16x + 52

32. y = log 17x + 32 # 4 2x

3

26. y = 5x # log 2 x

4

7000 6000 5000 4000 3000 2000

33. F1x2 = 7x # 1log 4 x29

34. G1x2 = log 9 x # 14 x26

35. f 1x2 = 13x 5 + x25 log 3 x

36. g1x2 = 2x 3 - x 1log 5 x2

APPLICATIONS Business and Economics 37. Double declining balance depreciation. An office machine is purchased for $5200. Under certain assumptions, its salvage value, V, in dollars, is depreciated according to a method called double declining balance, by basically 80% each year, and is given by V1t2 = 520010.802t, where t is the time, in years, after purchase. a) Find V¿1t2. b) Interpret the meaning of V¿1t2. 38. Recycling aluminum cans. It is known that 45% of all aluminum cans distributed will be recycled each year. A beverage company uses 250,000 lb of aluminum cans. After recycling, the amount of aluminum, in pounds, still in use after t years is given by N1t2 = 250,00010.452t. (Source: The Container Recycling Institute.) a) Find N¿1t2. b) Interpret the meaning of N¿1t2. 39. Household liability. The total financial liability, in billions of dollars, of U.S. households can be modeled by the function L1t2 = 154711.0832t, where t is the number of years after 1980. The graph of this function follows.

5

0

6x 30. f 1x2 = 5x - 1

+8

L(t) = 1547(1.083)t

1000 t

10 15 20 Years since 1980

a) Using this model, predict the total financial liability of U.S. households in 2012. b) Find L¿1252. c) Interpret the meaning of L¿1252. 40. Small business. The number of nonfarm proprietorships, in thousands, in the United States can be modeled by the function N1t2 = 8400 ln t - 10,500. where t is the number of years after 1970. The graph of this function is given below. Number of nonfarm proprietorships (in thousands)

-1

24. g1x2 = - log 6 A 2x + 5 B 28. F1x2 = 1log 9 x27

7x 29. g1x2 = 4x + 1 3

22. y = log 9 1x - x2 4

23. f 1x2 = 4 log 7 A 2x - 2 B

31. y = 52x

L(t) 8000 Liability amount (in billions of dollars)

13. y = log 4 x

N(t) 18,000 16,000 14,000 12,000 10,000 8000 6000 4000 2000 0

N(t) = 8400 ln t – 10,500

5

10

15 20 25 Years since 1970

30

t

a) Using this model, predict the number of nonfarm proprietorships in the United States in 2014. b) Find N¿1452. c) Interpret the meaning of N¿1452.

Life and Physical Sciences 41. Agriculture. Farmers wishing to avoid the use of genetically modified (GMO) seeds are increasingly concerned about inadvertently growing GMO plants as a result of pollen drifting from nearby farms. Assuming that these farmers raise their own seeds, the fractional portion of their crop that remains free of GMO plants t years later can be approximated by P1t2 = 10.982t.

370

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•

Exponential and Logarithmic Functions

a) Using this model, predict the fractional portion of the crop that will be GMO-free 10 yr after a neighboring farm begins to use GMO seeds. b) Find P¿1152. c) Interpret the meaning of P¿1152. Earthquake magnitude. The magnitude R (measured on the Richter scale) of an earthquake of intensity I is defined as I R  log , I0 where I0 is a minimum intensity used for comparison. When one earthquake is 10 times as intense as another, its magnitude on the Richter scale is 1 higher. If one earthquake is 100 times as intense as another, its magnitude on the Richter scale is 2 higher, and so on. Thus, an earthquake whose magnitude is 6 on the Richter scale is 10 times as intense as an earthquake whose magnitude is 5. Earthquake intensities can be interpreted as multiples of the minimum intensity I0. Use this information for Exercises 42 and 43. 42. On January 12, 2010, a devastating earthquake struck the Caribbean nation of Haiti. It had an intensity of I0107. What was its magnitude on the Richter scale?

44. Earthquake intensity. The intensity of an earthquake is given by I = I010R, where R is the magnitude on the Richter scale and I0 is the minimum intensity, at which R = 0, used for comparison. a) Find I, in terms of I0, for an earthquake of magnitude 7 on the Richter scale. b) Find I, in terms of I0, for an earthquake of magnitude 8 on the Richter scale. c) Compare your answers to parts (a) and (b). d) Find the rate of change dI>dR. e) Interpret the meaning of dI>dR. 45. Intensity of sound. The intensity of a sound is given by I = I0100.1L, where L is the loudness of the sound as measured in decibels and I0 is the minimum intensity detectable by the human ear. a) Find I, in terms of I0, for the loudness of a power mower, which is 100 decibels. b) Find I, in terms of I0, for the loudness of a just audible sound, which is 10 decibels. c) Compare your answers to parts (a) and (b). d) Find the rate of change dI>dL. e) Interpret the meaning of dI>dL. 46. Earthquake magnitude. The magnitude R (measured on the Richter scale) of an earthquake of intensity I is defined as I R = log , I0

This photograph shows part of the damage in Haiti due to the earthquake of January 2010.

43. On February 27, 2010, an earthquake more intense than the one in Haiti struck the nation of Chile in South America. It had an intensity of I0108.8. What was its magnitude on the Richter scale?

Chile South America

Santiago Chile Santiago

The location of the February 2010 earthquake in Chile.

where I0 is the minimum intensity (used for comparison). (The exponential form of this definition is given in Exercise 44.) a) Find the rate of change dR>dI. b) Interpret the meaning of dR>dI. 47. Loudness of sound. The loudness L of a sound of intensity I is defined as I L = 10 log , I0 where I0 is the minimum intensity detectable by the human ear and L is the loudness measured in decibels. (The exponential form of this definition is given in Exercise 45.) a) Find the rate of change dL>dI. b) Interpret the meaning of dL>dI. 48. Response to drug dosage. The response y to a dosage x of a drug is given by y = m log x + b, where m and b are constants. The response may be hard to measure with a number. The patient might perspire more, have an increase in temperature, or faint. a) Find the rate of change dy>dx. b) Interpret the meaning of dy>dx.

3.6

SYNTHESIS 3h - 1 . (Hint: See p. 367.) h:0 h

Use the Chain Rule, implicit differentiation, and other techniques to differentiate each function given in Exercises 50–57. x

4

54. f 1x2 = x e , for x 7 0 x

55. y = a f 1x2

56. y = log a f 1x2, for f 1x2 positive

57. y = 3 f 1x24g1x2, for f 1x2 positive

OBJECTIVES • Find the elasticity of a demand function. • Find the maximum of a total-revenue function. • Characterize demand in terms of elasticity.

59. In your own words, derive the formula for finding the derivative of f 1x2 = log a x.

1. (a) 1ln 525x; (b) 1ln 424 x; (c) 1ln 4.3214.32x 1 #1 -7 # 1 2. (a) ; (b) ; ln 2 x ln 10 x 3x 2 1 (c) x 5 a + 6 log xb; (d) ln 10 1ln 821x 3 - 72

53. y = log 3 1log x2

3.6

371

Answers to Quick Checks

51. y = 2 x

52. y = x x, for x 7 0

An Economics Application: Elasticity of Demand

58. In your own words, derive the formula for finding the derivative of f 1x2 = a x.

49. Find lim

50. f 1x2 = 312 2

•

An Economics Application: Elasticity of Demand Retailers and manufacturers often need to know how a small change in price will affect the demand for a product. If a small increase in price produces no change in demand, a price increase may make sense; if a small increase in price creates a large drop in demand, the increase is probably ill advised. To measure the sensitivity of demand to a small percent increase in price, economists calculate the elasticity of demand. Suppose that q represents a quantity of goods purchased and x is the price per unit of the goods. Recall that q and x are related by the demand function q = D1x2. Suppose that there is a change ¢x in the price per unit. The percent change in price is given by ¢x # 100 ¢x ¢x # 100 = = %. x x 100 x

A change in the price produces a change ¢q in the quantity sold. The percent change in quantity is given by ¢q ¢q # 100 = %. q q

The ratio of the percent change in quantity to the percent change in price is ¢q>q ¢x>x

,

which can be expressed as x # ¢q . q ¢x Note that for differentiable functions, lim

¢x : 0

¢q dq = , ¢x dx

so the limit as ¢x approaches 0 of the expression in equation (1) becomes lim

¢x : 0

x # ¢q x dq x x # = # = # D¿1x2 = D¿1x2. q ¢x q dx q D1x2

This result is the basis of the following definition.

(1)

372

CHAPTER 3

•

Exponential and Logarithmic Functions

DEFINITION The elasticity of demand E is given as a function of price x by E1x2 = -

x # D¿1x2 D1x2

.

To understand the purpose of the negative sign in the preceding definition, note that the price, x, and the demand, D1x2, are both nonnegative. Since D1x2 is normally decreasing, D¿1x2 is usually negative. By inserting a negative sign in the definition, economists make E1x2 nonnegative and easier to work with.

■ EXAMPLE 1 Economics: Demand for DVD Rentals. Klix Video has found that demand for rentals of its DVDs is given by q = D1x2 = 120 - 20x, where q is the number of DVDs rented per day at x dollars per rental. Find each of the following. a) The quantity demanded when the price is $2 per rental b) The elasticity as a function of x c) The elasticity at x = 2 and at x = 4. Interpret the meaning of these values of the elasticity. d) The value of x for which E1x2 = 1. Interpret the meaning of this price. e) The total-revenue function, R1x2 = x # D1x2 f) The price x at which total revenue is a maximum

Solution a) For x = 2, we have D122 = 120 - 20122 = 80. Thus, 80 DVDs per day will be rented at a price of $2 per rental. b) To find the elasticity, we first find the derivative D¿1x2: D¿1x2 = - 20. Then we substitute - 20 for D¿1x2 and 120 - 20x for D1x2 in the expression for elasticity: E1x2 = c) E122 =

x # D¿1x2 D1x2

= -

x # 1- 202

120 - 20x

=

20x x = . 120 - 20x 6 - x

2 1 = 6 - 2 2

At x = 2, the elasticity is 12, which is less than 1. Thus, the ratio of the percent change in quantity to the percent change in price is less than 1. A small percentage increase in price will cause an even smaller percentage decrease in the quantity sold. E142 =

4 = 2 6 - 4

At x = 4, the elasticity is 2, which is greater than 1. Thus, the ratio of the percent change in quantity to the percent change in price is greater than 1. A small percentage increase in price will cause a larger percentage decrease in the quantity sold.

3.6

TECHNOLOGY CONNECTION

x 6 - x x 2x x

A company determines that the demand function for in-ear radios is

where q is the number sold per day when the price is x dollars per radio.

EXERCISES 1. Find the elasticity E and the

An Economics Application: Elasticity of Demand

373

d) We set E1x2 = 1 and solve for p:

Economics: Demand for In-Ear Radios

q = D1x2 = 300 - x,

•

= 1 = 6 - x = 6 = 3.

We multiply both sides by 6 - x, assuming that x Z 6.

Thus, when the price is $3 per rental, the ratio of the percent change in quantity to the percent change in price is 1. e) Recall that the total revenue R(x) is given by x # D1x2. Then R1x2 = x # D1x2 = x1120 - 20x2 = 120x - 20x 2.

f) To find the price x that maximizes total revenue, we find R¿1x2:

total revenue R.

R¿1x2 = 120 - 40x.

We see that R¿1x2 exists for all x in the interval 30, q 2. Thus, we solve:

2. Using only the first quadrant,

graph the demand, elasticity, and total-revenue functions on the same set of axes.

R¿1x2 = 120 - 40x = 0 - 40x = - 120 x = 3.

3. Find the price x for which the

total revenue is a maximum. Use the method of Example 1. Check the answer graphically.

Since there is only one critical value, we can try to use the second derivative to see if we have a maximum: R–1x2 = - 40 6 0. Thus, R–132 is negative, so R(3) is a maximum. That is, total revenue is a maximum at $3 per rental. Note in parts (d) and (f ) of Example 1 that the value of x for which E1x2 = 1 is the same as the value of x for which total revenue is a maximum. The following theorem states that this is always the case.

THEOREM 15 Total revenue is increasing at those x-values for which E1x2 6 1. Total revenue is decreasing at those x-values for which E1x2 7 1. Total revenue is maximized at the value(s) of x for which E1x2 = 1.

Proof. so

We know that

R1x2 = x # D1x2, R¿1x2 = x # D¿1x2 + D1x2 # 1 = D1x2c

x # D¿1x2 D1x2

+ 1d

Using the Product Rule Check this by multiplying.

= D1x23- E1x2 + 14 = D1x231 - E1x24. (continued)

CHAPTER 3

•

Exponential and Logarithmic Functions

Since we can assume that D1x2 7 0, it follows that R¿1x2 is positive for E1x2 6 1, is negative for E1x2 7 1, and is 0 when E1x2 = 1. Thus, total revenue is increasing for E1x2 6 1, is decreasing for E1x2 7 1, and is maximized when E1x2 = 1. ■

Elasticity and Revenue For a particular value of the price x: 1. The demand is inelastic if E1x2 6 1. An increase in price will bring an increase in revenue. If demand is inelastic, then revenue is increasing. 2. The demand has unit elasticity if E1x2 = 1. The demand has unit elasticity when revenue is at a maximum. 3. The demand is elastic if E1x2 7 1. An increase in price will bring a decrease in revenue. If demand is elastic, then revenue is decreasing. q Inelastic

Demand function, q = D(x)

Unit elasticity

Elastic

E1 x1 Price

x

R(x) Inelastic demand

E=1

Elastic demand

Total revenue

374

Increasing revenue

Decreasing revenue

x1 Price

x

In summary, suppose that Klix Video in Example 1 raises the price per rental and that the total revenue increases. Then we say the demand is inelastic. If the total revenue decreases, we say the demand is elastic. Some price elasticities in the U.S. economy are listed in the following table.

3.6

•

An Economics Application: Elasticity of Demand

375

Price Elasticities in the U.S. Economy Industry Elastic Demands Metals Electrical engineering products Mechanical engineering products Furniture Motor vehicles Instrument engineering products Professional services Transportation services Inelastic Demands Gas, electricity, and water Oil Chemicals Beverages (all types) Tobacco Food Banking and insurance services Housing services Clothing Agricultural and fish products Books, magazines, and newspapers Coal

Quick Check 1

O

Economics: Demand for DVD Rentals. Internet rentals affect Klix Video in such a way that the demand for rentals of its DVDs changes to q = D1x2 = 30 - 5x. a) Find the quantity demanded when the price is $2 per rental, $3 per rental, and $5 per rental. b) Find the elasticity of demand as a function of x. c) Find the elasticity at x = 2, x = 3, and x = 5. Interpret the meaning of these values. d) Find the value of x for which E1x2 = 1. Interpret the meaning of this price. e) Find the total-revenue function, R1x2 = x # D1x2. f) Find the price x at which total revenue is a maximum.

Elasticity

1.52 1.39 1.30 1.26 1.14 1.10 1.09 1.03 0.92 0.91 0.89 0.78 0.61 0.58 0.56 0.55 0.49 0.42 0.34 0.32

(Source: Ahsan Mansur and John Whalley, “Numerical specification of applied general equilibrium models: Estimation, calibration, and data.” In H. E. Scarf and J. B. Shoven (eds.), Applied General Equilibrium Analysis. (New York: Cambridge University Press, 1984), p. 109.)

O Quick Check 1

Section Summary • The elasticity of demand E is given as a function of price x by E1x2 = -

x # D¿1x2 D1x2

.

Elasticity provides a means of evaluating the change in revenue that results from an increase in price.

376

CHAPTER 3

•

Exponential and Logarithmic Functions

EXERCISE SET

3.6 For the demand function given in each of Exercises 1–12, find the following. a) The elasticity b) The elasticity at the given price, stating whether the demand is elastic or inelastic c) The value(s) of x for which total revenue is a maximum (assume that x is in dollars) 1. q = D1x2 = 400 - x; x = 125 2. q = D1x2 = 500 - x; x = 38 3. q = D1x2 = 200 - 4x; x = 46 4. q = D1x2 = 500 - 2x; x = 57 5. q = D1x2 =

400 ; x = 50 x

6. q = D1x2 =

3000 ; x = 60 x

7. q = D1x2 = 2600 - x ; x = 100 8. q = D1x2 = 2300 - x ; x = 250

where q is measured in millions of barrels of oil per day at a price of x dollars per barrel. a) Find the elasticity. b) Find the elasticity at a price of $10 per barrel, stating whether the demand is elastic or inelastic at that price. c) Find the elasticity at a price of $20 per barrel, stating whether the demand is elastic or inelastic at that price. d) Find the elasticity at a price of $30 per barrel, stating whether the demand is elastic or inelastic at that price. e) At what price is the revenue a maximum? f) What quantity of oil will be sold at the price that maximizes revenue? Compare the current world price to your answer. g) At a price of $30 per barrel, will a small increase in price cause the total revenue to increase or decrease? 15. Demand for computer games. High Wire Electronics determines the following demand function for a new game: q = D1x2 = 2200 - x 3, where q is the number of games sold per day when the price is x dollars per game.

9. q = D1x2 = 100e -0.25x; x = 10

11. q = D1x2 = 12. q = D1x2 =

100

1x + 322 500

; x = 1

12x + 122

; x = 8 2

Number of games sold per day

10. q = D1x2 = 200e -0.05x; x = 80

y 15

D(x) = √200 − x3 1 6 x

1

Price (in dollars)

Business and Economics 13. Demand for chocolate chip cookies. Good Times Bakers works out a demand function for its chocolate chip cookies and finds it to be q = D1x2 = 967 - 25x, where q is the quantity of cookies sold when the price per cookie, in cents, is x. a) Find the elasticity. b) At what price is the elasticity of demand equal to 1? c) At what prices is the elasticity of demand elastic? d) At what prices is the elasticity of demand inelastic? e) At what price is the revenue a maximum? f) At a price of 20¢ per cookie, will a small increase in price cause the total revenue to increase or decrease? 14. Demand for oil. Suppose that you have been hired as an economic consultant concerning the world demand for oil. The demand function is q = D1x2 = 63,000 + 50x - 25x 2, 0 … x … 50,

a) Find the elasticity. b) Find the elasticity when x = 3. c) At x = 3, will a small increase in price cause the total revenue to increase or decrease? 16. Demand for tomato plants. Sunshine Gardens determines the following demand function during early summer for tomato plants: q = D1x2 =

2x + 300 , 10x + 11

where q is the number of plants sold per day when the price is x dollars per plant. Number of plants sold per day

APPLICATIONS

q 30

2x + 300 D(x) = ———— 10x + 11 5 20 x

1 Price (in dollars)

Exercise Set 3.6

a) Find the elasticity. b) Find the elasticity when x = 3. c) At $3 per plant, will a small increase in price cause the total revenue to increase or decrease?

b) Is the value of the elasticity dependent on the price per unit? c) Does the total revenue have a maximum? At what value of x? 19. Let

SYNTHESIS 17. Economics: constant elasticity curve. a) Find the elasticity of the demand function q = D1x2 =

377

k , xn

where k is a positive constant and n is an integer greater than 0. b) Is the value of the elasticity dependent on the price per unit? c) Does the total revenue have a maximum? When? 18. Economics: exponential demand curve. a) Find the elasticity of the demand function q = D1x2 = Ae -kx, where A and k are positive constants.

L1x2 = ln D1x2. Describe the elasticity in terms of L¿1x2. 20. Explain in your own words the concept of elasticity and its usefulness to economists. Do some library or online research or consult an economist in order to determine when and how this concept was first developed. 21. Explain how the elasticity of demand for a product can be affected by the availability of substitutes for the product.

Answers to Quick Check x ; 6 - x (c) 0.5, 1, 5 (see Example 1 for the interpretations); (d) $3; (e) R1x2 = 30x - 5x 2; (f) $3 1. (a) 20, 15, 5; (b) E1x2 =

CHAPTER 3 SUMMARY KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 3.1 An exponential function f is a function of the form f 1x2 = a x, where x is any real number and a, the base, is any positive number other than 1.

y

y

10 9 8 7 6 5 4 3 2 1

10 9 8 7 6 5 4 3 2 1

54321

1 2 3 4 5

x

f(x) = (−12 )

x

54321

1 2 3 4 5

x

g(x) = 3x

To differentiate a function like f 1x2 = 15x 2 + 32e 7x, both the Product Rule and the Chain Rule are needed:

The natural base e is such that d x e = e x, dx

where e L 2.718 and e = lim 11 + h2

1>h

d 315x 2 + 32e 7x4 = 15x 2 + 32e 7x # 7 + e 7x # 10x. dx

h:0

d f1x2 and e = f ¿1x2e x. dx

SECTION 3.2 A logarithmic function g is any function of the form g1x2 = log a x, where a, the base, is a positive number other than 1:

y 5 4 3 2 1

y = log a x means x = a y.

1 1 2 3 4 5

Logarithmic functions are inverses of exponential functions.

1 2 3 4 5 6 7 8 9

x

y = log 3 x

Remember that log b c is the exponent to which b is raised to get c. The notation for common logarithms and natural logarithms is

log 1000 = log 10 103 = 3 and ln e 15 = log e e 15 = 25

log c = log 10 c and ln c = log e c. The following important properties of logarithms allow us to manipulate expressions. We assume that M, N, and a are positive, with a Z 1, and k is a real number: log a 1MN2 = log a M + log a N M = log a M - log a N log a N log a 1M k2 = k log a M log a a = 1 log a a k = k log a 1 = 0 log a M log b M = log a b

378

CHAPTER 3

•

Exponential and Logarithmic Functions

log 2 5x = log 2 5 + log 2 x 2 = log 2 2 - log 2 Q = 1 - log 2 Q log 2 Q ln e 7 = 7 ln t log a t = ln a log 10 1 = 0

Chapter 3 Summary

KEY TERMS AND CONCEPTS

EXAMPLES

Logarithms are used to solve certain exponential equations.

Solve: 5e 2t = 80. We have 5e 2t e 2t ln e 2t 2t

= = = =

80 16 ln 16 ln 16 ln 16 t = 2 t L 1.386

The derivative of the natural logarithm function of x, where x is any positive number, is the reciprocal of x: d 1 ln x = , x dx

379

Dividing both sides by 5 Taking the natural log of both sides Using a property of logarithms

1 # d 3ln 15x2 # 1x 3 - 7x24 = ln 15x2 # 13x 2 - 72 + 1x 3 - 7x2 # 5 dx 5x 1 # = ln 15x2 # 13x 2 - 72 + x # 1x 2 - 72 # 5 5x = ln 15x2 # 13x 2 - 72 + 1x 2 - 72

and f ¿1x2 d . ln f 1x2 = dx f 1x2 Differentiating a natural logarithmic function may require applying the Product, Quotient, and Chain Rules.

SECTION 3.3 Because exponential functions are the only functions for which the derivative (rate of change) is directly proportional to the function value at any point in time, they can be used to model many real-world situations involving uninhibited growth. dP = kP, with k 7 0, then P1t2 = P0e kt, dt where P0 is the initial population at t = 0. If

Business. The balance P in an account with Turing Mutual Funds grows at a rate given by dP = 0.04P, dt where t is time, in years. Find the function that satisfies the equation 1let P102 = P02. After what period of time will an initial investment, P0, double itself? The function is P1t2 = P0e 0.04t. Check: d P e 0.04t = P0e 0.04t # 0.04 dt 0 = 0.04P0e 0.04t = 0.04P1t2 To find the time for the amount to double, we set P1t2 = 2P0 and solve for t: 2P0 2 ln 2 ln 2 ln 2 0.04

= = = =

P0e 0.04t e 0.04t ln 1e 0.04t2 0.04t

Dividing both sides by P0 Taking the natural log of both sides

= t, or t L 17.3 yr. (continued)

380

•

CHAPTER 3

Exponential and Logarithmic Functions

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 3.3 (continued ) The exponential growth rate k and the doubling time T are related by ln 2 k = T

ln 2 and T = . k

The Rule of 70 expresses the doubling time for a sum of money in terms of the interest rate k expressed as a decimal: T L

70 . 100k

If the exponential growth rate is 4% per year, then the doubling time is T =

or using the Rule of 70, T L

k =

and

L 1 + be

70 L 17.5 yr. 10010.042

The doubling time for the number of downloads per day from iTunes is 614 days. The exponential growth rate is

Two models for limited or inhibited growth are P1t2 =

0.693147 ln 2 L L 17.3 yr, 4% 0.04

P(t)

0.693147 ln 2 L L 0.0011289 = 0.11% per day. T 614

P(t)

Logistic Growth

Inhibited Growth (not Logistic)

L

L

, -kt

P1t2 = L11 - e -kt2,

P0

where k 7 0 and L is the limiting value.

P(t) =

L , for k > 0 1 + be−kt t

P(t) = L(1 − e−kt), for k > 0 t

SECTION 3.4 P

Exponential growth is modeled by P1t2 = P0e kt, k 7 0, and exponential decay is modeled by P1t2 = P0e -kt, k 7 0.

P

Growth P = P0 e kt, for k > 0

Decay P = P0 e − kt, for k > 0 P0

P0 t

Exponential decay characterizes many real-world phenomena. One of the most common is radioactive decay.

t

Physical Science. Lead-210 has a decay rate of 3.15% per year. The rate of change of an amount N of lead-210 is given by dN = - 0.0315N. dt Find the function that satisfies the equation. How much of an 80-g sample of lead-210 will remain after 20 yr? The function is N1t2 = N0e -0.0315t. Check: d 1N0e -0.0315t2 = N0e -0.0315t1- 0.03152 dt = - 0.0315 # N1t2. Then the amount remaining after 20 yr is, N1202 = 80e -0.03151202 L 42.6 g.

Chapter 3 Summary

KEY TERMS AND CONCEPTS

EXAMPLES

Half-life, T, and decay rate, k, are related by

The half-life of a radioactive isotope is 38 days. The decay rate is

k =

ln 2 T

and

T =

ln 2 . k

T =

381

ln 2 0.693147 = L 0.0182 = 1.82% per day. 38 38

The decay equation, P0 = Pe -kt, can also be used to calculate present value. The present value P0 of an amount P due t years later, at interest rate k, compounded continuously, is given by P0 = Pe

The present value of $200,000 due 8 yr from now, at 4.6% interest, compounded continuously, is given by P0 = Pe -kt = 200,000 e -0.046

-kt

#8

= $138,423.44.

.

SECTION 3.5 d 0.3x = ln 7 # 70.3x # 0.3 7 dx d 1 #1 log 6 x = dx ln 6 x

The following formulas can be used to differentiate exponential and logarithmic functions for any base a, other than e: d 1 #1 d x a = 1ln a2a x and log a x = . dx dx ln a x

SECTION 3.6 The elasticity of demand E is a function of x, the price: E1x2 = -

x # D¿1x2 D1x2

.

When E1x2 7 1, total revenue is decreasing; when E1x2 6 1, total revenue is increasing; and when E1x2 = 1, total revenue is maximized.

Business. The Leslie Davis Band finds that demand for its CD at performances is given by q = D1x2 = 50 - 2x, where x is the price, in dollars, of each CD sold and q is the number of CDs sold at a performance. Find the elasticity when the price is $10 per CD, and interpret the result. Then, find the price at which revenue is maximized. Elasticity at x is given by E1x2 = = -

x # D¿1x2 D1x2 x1- 22

50 - 2x x - 2x = . = 50 - 2x 25 - x Thus, E1102 =

2 10 = . 25 - 10 3

Since E(10) is less than 1, the demand for the CD is inelastic, and an increase in price will increase revenue. Revenue is maximized when E1x2 = 1: x 25 - x x 2x x

= 1 = 25 - x = 25 = 12.5.

At a price of $12.50 per CD, revenue will be maximized.

CHAPTER 3 REVIEW EXERCISES These review exercises are for test preparation. They can also be used as a practice test. Answers are at the back of the book. The blue bracketed section references tell you what part(s) of the chapter to restudy if your answer is incorrect.

6. P1t2 = 50 + ln t

CONCEPT REINFORCEMENT

f)

P(t)

50

t

Match each equation in column A with the most appropriate graph in column B. [3.1–3.4] Column A 1. P1t2 = 50e 0.03t

Column B a)

Classify each statement as either true or false.

P(t)

7. The base a in the exponential function given by f 1x2 = a x must be greater than 1. [3.1]

50

8. The base a in the logarithmic function given by g1x2 = log a x must be greater than 0. [3.2] t

2. P1t2 =

50

b)

1 + 2e -0.02t

9. If f ¿1x2 = c # f 1x2 for c Z 0 and f 1x2 Z 0, then f must be an exponential function. [3.3] 10. With exponential growth, the doubling time depends on the size of the original population. [3.3]

P(t)

11. A radioactive isotope’s half-life determines the value of its decay constant. [3.4] 12. A radioactive isotope’s half-life depends on how much of the substance is initially present. [3.4]

50

t

3. P1t2 = 50e -0.20t

c)

P(t)

14. For any logarithmic function of the form g1x2 = log a x, 1 1 it follows that g¿1x2 = # . [3.5] a x

1

4. P1t2 = ln t

d)

13. For any exponential function of the form f 1x2 = a x, it follows that f ¿1x2 = ln a # a x. [3.5]

t

P(t)

15. Revenue is maximized when the elasticity of demand is 1. [3.6]

REVIEW EXERCISES

50

Differentiate each function. 16. y = ln x [3.2]

18. y = ln 1x 4 + 52 [3.2]

1

5. P1t2 = 5011 - e -0.04t2

e)

t

20. f 1x2 = ln2x [3.2] 22. f 1x2 =

P(t) 50

ln x x3

[3.2]

23. f 1x2 = e x # ln 4x [3.1, 3.2] 2

24. f 1x2 = e 4x - ln t

382

CHAPTER 3

•

Exponential and Logarithmic Functions

x [3.1, 3.2] 4

25. g1x2 = x 8 - 8 ln x [3.2]

17. y = e x [3.1] 19. y = e 22x [3.1]

21. f 1x2 = x 4e 3x [3.1]

Chapter 3 Review Exercises

26. y =

ln e x [3.1, 3.2] ex

45. Life science: decay rate. A certain radioactive isotope has a decay rate of 7% per day, that is, dA = - 0.07A, dt

27. F1x2 = 9x [3.5] 28. g1x2 = log 2 x [3.5]

29. y = 3x # log 4 12x + 12 [3.5] Graph each function. [3.1] 30. f 1x2 = 4 x

383

31. g1x2 =

A 13 B x

Given log a 2  1.8301 and log a 7  5.0999, find each logarithm. [3.2] 32. log a 14

33. log a 27

34. log a 28

35. log a 3.5

36. log a 27

37. log a 14

38. Find the function that satisfies dQ>dt = 7Q, given that Q102 = 25. [3.3] 39. Life science: population growth. The population of Boomtown doubled in 16 yr. What was the growth rate of the city? Round to the nearest tenth of a percent. [3.3] 40. Business: interest compounded continuously. Suppose that $8300 is invested in Noether Bond Fund, where the interest rate is 6.8%, compounded continuously. How long will it take for the $8300 to double itself? Round to the nearest tenth of a year. [3.3] 41. Business: cost of a prime-rib dinner. The average cost C of a prime-rib dinner was $15.81 in 1986. In 2010, it was $27.95. Assuming that the exponential growth model applies: [3.3] a) Find the exponential growth rate to three decimal places, and write the function that models the situation. b) What will the cost of such a dinner be in 2012? In 2020? 42. Business: franchise growth. A clothing firm is selling franchises throughout the United States and Canada. It is estimated that the number of franchises N will increase at the rate of 12% per year, that is, dN = 0.12N, dt where t is the time, in years. [3.3] a) Find the function that satisfies the equation, assuming that the number of franchises in 2007 1t = 02 is 60. b) How many franchises will there be in 2013? c) After how long will the number of franchises be 120? Round to the nearest tenth of a year. 43. Life science: decay rate. The decay rate of a certain radioactive substance is 13% per year. What is its half-life? Round to the nearest tenth of a year. [3.4] 44. Life science: half-life. The half-life of radon-222 is 3.8 days. What is its decay rate? Round to the nearest tenth of a percent. [3.4]

where A is the amount of the isotope present at time t, in days. [3.4] a) Find a function that satisfies the equation if the amount of the isotope present at t = 0 is 800 g. b) After 20 days, how much of the 800 g will remain? Round to the nearest gram. c) After how long will half of the original amount remain? 46. Social science: Hullian learning model. The probability p of mastering a certain assembly-line task after t learning trials is given by p1t2 = 1 - e -0.7t. [3.3] a) What is the probability of learning the task after 1 trial? 2 trials? 5 trials? 10 trials? 14 trials? b) Find the rate of change, p¿1t2. c) Interpret the meaning of p¿1t2. d) Sketch a graph of the function. 47. Business: present value. Find the present value of $1,000,000 due 40 yr later at 4.2%, compounded continuously. [3.4] 48. Economics: elasticity of demand. Consider the demand function 600

q = D1x2 =

1x + 422

. [3.6]

a) Find the elasticity. b) Find the elasticity at x = $1, stating whether the demand is elastic or inelastic. c) Find the elasticity at x = $12, stating whether the demand is elastic or inelastic. d) At a price of $12, will a small increase in price cause the total revenue to increase or decrease? e) Find the value of x for which the total revenue is a maximum.

SYNTHESIS 49. Differentiate: y =

e 2x + e -2x e 2x - e -2x

. [3.1]

50. Find the minimum value of f 1x2 = x 4 ln 14x2. [3.2]

TECHNOLOGY CONNECTION 51. Graph: f 1x2 = 52. Find lim

x:0

e 1>x

11 + e 1>x22

e 1>x

11 + e 1>x22

. [3.1]

. [3.1]

53. Business: shopping on the Internet. Online sales of all types of consumer products increased at an exponential rate in the last decade or so. Data in the following table show online retail sales, in billions of dollars. [3.3]

384

CHAPTER 3

•

Exponential and Logarithmic Functions

Years, t, after 1998

U.S. Online Retail Sales (in billions)

0 1 2 3 4 5 6 7

$ 4.9 14.7 28.0 34.3 44.7 55.7 69.2 86.3

a) Use REGRESSION to fit an exponential function y = a # b x to the data. Then convert that formula to an exponential function, base e, where t is the number of years after 1998, and determine the exponential growth rate. b) Estimate online sales in 2010; in 2020. c) After what amount of time will online sales be $400 billion? d) What is the doubling time of online sales?

(Source: U.S. Census Bureau.)

CHAPTER 3 TEST Differentiate. 1. y = 2e

2. y = 1ln x2

3x

4

3. f 1x2 = e -x

4. f 1x2 = ln

5. f 1x2 = e x - 5x 3

6. f 1x2 = 3e x ln x

7. y = 7 + 3

8. y = log 14 x

2

x

x

x 7

Given log b 2  0.2560 and log b 9  0.8114, find each of the following. 9. log b 18

10. log b 4.5

11. log b 3

12. Find the function that satisfies dM>dt = 6M, with M102 = 2. 13. The doubling time for a certain bacteria population is 3 hr. What is the growth rate? Round to the nearest tenth of a percent.

APPLICATIONS 14. Business: interest compounded continuously. An investment is made at 6.931% per year, compounded continuously. What is the doubling time? Round to the nearest tenth of a year. 15. Business: cost of milk. The cost C of a gallon of milk was $3.22 in 2006. In 2010, it was $3.50. (Source: U.S. Department of Labor, Bureau of Labor Statistics.) Assuming that the exponential growth model applies: a) Find the exponential growth rate to the nearest tenth of a percent, and write the equation. b) Find the cost of a gallon of milk in 2012 and 2018.

16. Life science: drug dosage. A dose of a drug is injected into the body of a patient. The drug amount in the body decreases at the rate of 10% per hour, that is, dA = - 0.1A, dt where A is the amount in the body and t is the time, in hours. a) A dose of 3 cubic centimeters (cc) is administered. Assuming A0 = 3, find the function that satisfies the equation. b) How much of the initial dose of 3 cc will remain after 10 hr? c) After how long does half of the original dose remain? 17. Life science: decay rate. The decay rate of radium-226 is 4.209% per century. What is its half-life? 18. Life science: half-life. The half-life of bohrium-267 is 17 sec. What is its decay rate? Express the rate as a percentage rounded to four decimal places. 19. Business: effect of advertising. Twin City Roasters introduced a new coffee in a trial run. The firm advertised the coffee on television and found that the percentage P of people who bought the coffee after t ads had been run satisfied the function 100 . 1 + 24e -0.28t What percentage of people bought the coffee before seeing the ad 1t = 02? What percentage bought the coffee after the ad had been run 1 time? 5 times? 10 times? 15 times? 20 times? 30 times? 35 times? Find the rate of change, P¿1t2. Interpret the meaning of P¿1t2. Sketch a graph of the function.

P1t2 = a) b)

c) d) e)

Extended Technology Application

20. In 2010, a professional athlete signed a contract paying him $13 million in 2016. Find the present value of that amount in 2010, assuming 4.3% interest, compounded continuously. 21. Economics: elasticity of demand. Consider the demand function q = D1x2 = 400e -0.2x. a) Find the elasticity. b) Find the elasticity at x = 3, and state whether the demand is elastic or inelastic. c) Find the elasticity at x = 18, and state whether the demand is elastic or inelastic. d) At a price of $3, will a small increase in price cause the total revenue to increase or decrease? e) Find the price for which the total revenue is a maximum.

SYNTHESIS

22. Differentiate: y = x 1ln x22 - 2x ln x + 2x.

23. Find the maximum and minimum values of f 1x2 = x 4e -x over 30, 104.

TECHNOLOGY CONNECTION 24. Graph: f 1x2 =

e x - e -x . e x + e -x

e x - e -x x -x . x:0 e + e

25. Find lim

26. Business: average price of a television commercial. The cost of a 30-sec television commercial that runs during the Super Bowl was increasing exponentially

from 1991 to 2006. Data in the table below show costs for those years. Years, t, after 1990

Cost of commercial

1 3 5 8 13 16

$ 800,000 850,000 1,000,000 1,300,000 2,100,000 2,600,000

(Source: National Football League.)

a) Use REGRESSION to fit an exponential function y = a # b x to the data. Then convert that formula to an exponential function, base e, where t is the number of years after 1990. b) Estimate the cost of a commercial run during the Super Bowl in 2012 and 2015. c) After what amount of time will the cost be $1 billion? d) What is the doubling time of the cost of a commercial run during the Super Bowl? e) The cost of a Super Bowl commercial in 2009 turned out to be $3 million, and in 2010 it dropped to about $2.8 million, possibly due to the decline in the world economy. Expand the table of costs, and make a scatterplot of the data. Does the cost still seem to follow an exponential function? Explain. What kind of curve seems to fit the data best? Fit that curve using REGRESSION, and predict the cost of a Super Bowl commercial in 2012 and in 2015. Compare your answers to those of part (b).

Extended Technology Application The Business of Motion Picture Revenue and DVD Release There has been increasing pressure by motion picture executives to narrow the gap between the theater release of a movie and the release to DVD. The executives want to reduce marketing expenses, adapt to audiences’ increasing consumption of on-demand movies, and boost decreasing DVD sales. Theater owners, on the other hand, want to protect declining ticket sales;

385

the number of people attending movies in the United States and Canada decreased from 1.57 billion in 2002 to 1.42 billion in 2009. The owners are fearful that if movie executives shorten the time between theater release and DVD release, the number of ticket buyers will drop even more, as many people will be willing to wait for the DVD.

386

CHAPTER 3

•

Exponential and Logarithmic Functions

The following table presents the number of days between theater release and DVD release for 10 movies. Note that average gap in time is about 4 months.

MOVIE

Avatar The Blind Side The Twilight Saga: New Moon Slumdog Millionaire Sherlock Holmes UP Julie & Julia Precious Juno Iron Man

RELEASE DATES

NUMBER OF DAYS SEPARATING RELEASE DATES

Theater: Dec. 18, 2009 126 DVD: Apr. 22, 2010 Theater: Nov. 27, 2009 119 DVD: Mar. 26, 2010 Theater: Nov. 20, 2009 121 DVD: Mar. 22, 2010 Theater: Jan. 23, 2009 67 DVD: Mar. 31, 2009 Theater: Dec. 25, 2009 94 DVD: Mar. 30, 2010 Theater: May 29, 2009 164 DVD: Nov. 10, 2009 Theater: Aug. 7, 2009 122 DVD: Dec. 8, 2009 Theater: Nov. 6, 2009 121 DVD: Mar. 9, 2010 Theater: Dec. 25, 2007 112 DVD: Apr. 15, 2008 Theater: May 2, 2008 151 DVD: Sept. 30, 2008 Average = 120 days, or about 4 months

Extended Technology Application

Let’s examine the data for The Blind Side, for which Sandra Bullock won the 2010 Oscar for best actress. The movie was based on a true story about a family that takes in a destitute young man and nurtures him into adulthood, when he becomes a player for the Baltimore Ravens in the National Football League. The following table presents weekly estimates of gross revenue, G, for the movie. The total revenue, R, is approximated by adding each week’s gross revenue to the preceding week’s total revenue. (Occasionally, other revenues, such as from permission fees, are added to box office revenue, so total revenue might be more than this sum.)

Revenue for The Blind Side WEEK IN RELEASE, t (week 1 = Nov. 20 to Nov. 26, 2009)

GROSS REVENUE, G (current week estimates, in millions)

TOTAL REVENUE, R (or cumulative box office revenue, in millions)

1, Nov. 27, 2009

$60.12

$ 60.12

2, Dec. 4

$48.71

$108.83

3, Dec. 11

$25.94

$134.77

4, Dec. 18

$19.94

$154.71

5, Dec. 25

$17.95

$172.66

6, Jan. 1, 2010

$23.75

$196.41

7, Jan. 8

$15.04

$211.45

8, Jan. 15

$ 9.76

$221.21

9, Jan. 22

$ 8.30

$229.51

10, Jan. 29

$ 5.38

$234.89

11, Feb. 5

?

?

12, Feb. 12

?

?

13, Feb. 19

?

?

14, Feb. 26

?

?

15, Mar. 5

?

?

16, Mar. 12

?

?

17, Mar. 19

?

?

18, Mar. 26, week of DVD release

?

?

387

388

CHAPTER 3

•

Exponential and Logarithmic Functions

Revenue for Avatar WEEK IN RELEASE, t

GROSS REVENUE, G (current week estimates, in millions)

TOTAL REVENUE, R (or cumulative box office revenue, in millions)

1, Dec. 25

$137.27

$137.27

2, Jan. 1, 2010

$146.54

$283.81

3, Jan. 8

$ 96.73

$380.54

4, Jan. 15

$ 69.93

$450.47

5, Jan. 22

$ 66.33

$516.80

6, Jan. 29

$ 47.67

$564.47

7, Feb. 5

$ 42.02

$606.49

8, Feb. 12

$ 31.11

$637.60

9, Feb. 19

$ 34.12

$671.72

10, Feb. 26

$ 21.18

$692.90

11, Mar. 5

?

?

12, Mar. 12

?

?

13, Mar. 19

?

?

14, Mar. 26

?

?

15, Apr. 2

?

?

16, Apr. 9, week of DVD release

?

?

(week 1 = Dec. 18 to Dec. 24, 2009)

EXERCISES 1. Assume that you are a movie executive and you want to select a function that seems to fit the data best. First, make a scatterplot of the data points (t, G). Then use REGRESSION to fit linear, quadratic, cubic, and exponential functions to the data, and graph each equation with the scatterplot. Then decide which function seems to fit best and give your reasons. 2. Presuming you have selected the exponential function, use it to predict gross revenue, G, for week 11 through week 18, the week in which the DVD is released. 3. Compute the values of total revenue, R, by successively adding the values of G for weeks 11 through 18 to each week’s total revenue, R.

Discuss why you think the time selected for DVD release is appropriate. 4. Use REGRESSION to fit a logistic function of the form c R1t2 = 1 + ae -bt to the data, and graph it with the scatterplot of Exercise 1. Based on these results, what dollar amount would seem to be a limiting value for the total revenue from The Blind Side? 5. Find the rate of change R¿1t2, and explain its meaning. Find lim R¿1t2, and explain its meaning. t: q

6. Now, consider the data for the movie Avatar. Using the procedures in Exercises 1–5, what dollar amount would seem to be a limiting value on the gross revenue for Avatar?

4

Integration Chapter Snapshot What You’ll Learn

Why It’s Important

4.1 4.2 4.3 4.4 4.5 4.6 4.7

Is it possible to determine the distance a vehicle has traveled if we know its velocity function? Can we determine a company’s total profit if we know its marginal-profit function? We can, using a process called integration, which is one of the two main branches of calculus, the other being differentiation.We will see that we can use integration to find the area under a curve, which has many practical applications in science, business, and statistics.

Antidifferentiation Antiderivatives as Areas Area and Definite Integrals Properties of Definite Integrals Integration Techniques: Substitution Integration Techniques: Integration by Parts Integration Techniques: Tables

Where It’s Used

y

CHANGING RATE OF CREDIT M ARKET DEBT

For the years 2005 through 2009, the annual rate of change in the national credit market debt, in billions of dollars per year, could be modeled by the function 2

D ¿ (t) = -810.3t + 1730.3t + 3648, where t is the number of years since 2005. (Source: Federal Reserve System.) Find the national credit market debt in 2009, given that D(0) = 41,267. This problem appears as Exercise 60 in Exercise Set 4.1.

Credit market debt

NATIONAL CREDIT MARKET DEBT

2005

D⬘(t) = −810.3t 2 + 17

30.3t + 3648

2006

2007

Year

2008

2009

t

389

390

CHAPTER 4

• Integration

4.1 OBJECTIVES • Find an antiderivative of a function. • Evaluate indefinite integrals using basic rules of antidifferentiation. • Use initial conditions to determine an antiderivative.

Antidifferentiation Suppose we do the reverse of differentiation: given a function, we find another function whose derivative is the given function. This is called antidifferentiation, and it is a part of the larger process of integration. Integration, the main topic of this chapter, is the second main branch of calculus, the first being differentiation. We will see that integration can be used to find the area under a curve over a closed interval, which has many important applications. Antidifferentiation is the process of differentiation in reverse. Given a function f1x2, we determine another function F1x2 such that the derivative of F1x2 is f1x2; that d is, F1x2 = f1x2. dx For example, let f1x2 = 2x. The function F1x2 = x 2 is an antiderivative of f1x2 d since x 2 = 2x. However, other functions also have a derivative of 2x. For example, dx y = x 2 + 1, y = x 2 - 10, and y = x 2 + 250 also differentiate to 2x; the x 2 term differentiates to 2x, and the constant term differentiates to zero. Therefore, an antiderivative of f1x2 = 2x is any function that can be written in the form F1x2 = x 2 + C, where C is a constant. This leads us to the following theorem.

THEOREM 1 The antiderivative of f1x2 is the set of functions F1x2 + C such that d 3F1x2 + C4 = f1x2. dx The constant C is called the constant of integration. Theorem 1 can be restated as follows: if two functions F1x2 and G1x2 have the same derivative f1x2, then F1x2 and G1x2 differ by at most a constant: F1x2 = G1x2 + C. If F1x2 is an antiderivative of a function f1x2, we write f1x2 dx = F1x2 + C. L This equation is read as “the antiderivative of f1x2, with respect to x, is the set of functions F1x2 + C.” The expression on the left side is called an indefinite integral. The symbol 1 is the integral sign and is a command for antidifferentiation. The function f1x2 is called the integrand, and the meaning of dx will be made clear when we develop the geometry of integration in Section 4.2.

■ EXAMPLE 1 Determine these indefinite integrals. That is, find the antiderivative of each integrand: a) 1 8 dx;

b) 1 3x 2 dx;

c) 1 e x dx;

d)

1 dx. Lx

Solution You have seen these integrands before as derivatives of other functions. a) 1 8 dx = 8x + C

Check:

d 18x + C2 = 8. dx

b) 1 3x 2 dx = x 3 + C

Check:

d 3 1x + C2 = 3x 2 . dx

c) 1 e x dx = e x + C

Check:

d x 1e + C2 = e x . dx

Check:

d 1 1ln x + C2 = . x dx

d)

1 dx = ln x + C Lx

Always check each antiderivative you determine by differentiating it.

4.1

• Antidifferentiation

391

The results of Example 1 suggest several useful rules of antidifferentiation, which are summarized in Theorem 2.

THEOREM 2

Rules of Antidifferentiation A1. Constant Rule: L

k dx = kx + C.

A2. Power Rule (where n Z - 1): L

x n dx =

1 x n + 1 + C, n + 1

n Z - 1.

A3. Natural Logarithm Rule: 1 dx = ln x + C, Lx

x 7 0.

A4. Exponential Rule (base e): L

e ax dx =

1 ax e + C, a

a Z 0.

Let’s use these rules in the following examples.

■ EXAMPLE 2 Find the antiderivative of f1x2 = x 4. That is, determine 1 x 4 dx. Solution We know that the derivative of a power function has an exponent decreased by 1, so we might guess that F1x2 = x 5 + C is an antiderivative of f1x2 = x 4. However, d 5 x = 5x 4, so our guess is not correct. It is close, however: including a coefficient of 15 dx gives us the desired antiderivative: L

x 4 dx =

1 5 x + C. 5

Check:

d 1 5 1 a x + Cb = 15x 42 = x 4. dx 5 5

Note that 15 times 5 gives the coefficient 1. Using the Power Rule of Antidifferentiation can be viewed as a two-step process:

L

1 x n dx =

1. Raise the power by 1. 2. Divide the term by the new power.

1 xn+1 + C n + 1 2

■ EXAMPLE 3 Use the Power Rule of Antidifferentiation to determine these indefinite integrals: a) 1 x 7 dx;

b) 1 x 99 dx;

c) 1 1x dx;

d)

1 dx. 3 Lx

Be sure to check each answer by differentiation.

Solution a)

L

x 7 dx =

x7+1 1 + C = x8 + C 7 + 1 8

Check:

d 1 8 1 a x + Cb = 18x 72 = x 7. dx 8 8

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CHAPTER 4

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b)

L

x 99 + 1 + C 99 + 1 1 100 = x + C 100

x 99dx =

Check:

1 100 d 1 a x + Cb = 1100x 992 = x 99. dx 100 100

c) We note that 2x = x 1>2. Therefore, L

2x dx = =

Quick Check 1

O

Determine these indefinite integrals: a) 1 x 10 dx; b) 1 x 200 dx; 6 c) 1 2x dx; d)

1 dx. 4 Lx

d) We note that

L

x 1>2 dx =

2 3>2 x + C. 3

x 11>22 + 1

A 12 B + 1

+ C =

Check:

x 3>2 3 2

+ C

d 2 3>2 2 3 a x + Cb = a x 1>2 b = x 1>2 = 2x. dx 3 3 2

1 = x -3. Therefore, x3

1 x -3 + 1 1 dx = x -3 dx = + C = - x -2 + C 3 3 + 1 2 x L L 1 d 1 1 1 = - 2 + C. Check: ¢ - x -2 + C ≤ = - 1- 2x -32 = x -3 = 3 . dx 2 2 x 2x

O Quick Check 1 The Power Rule of Antidifferentiation is valid for all real numbers n, except for n = - 1. Attempting to use the Power Rule when n = - 1 will result in a 0 in the denominator of the coefficient. However, as we saw in Example 1(d), if n = - 1, we have 1 x -1 = , which is the derivative of the natural logarithm function, y = ln x. Therefore, x 1 dx = ln x + C, for x 7 0. Lx 1 Caution! Note the key difference between the indefinite integrals dx and 3 Lx 1 dx. Although they look similar, the first of these integrals is determined by the Lx Power Rule, while the second is determined by the Natural Logarithm Rule. d x The exponential function f1x2 = e x has the property that e = e x; therefore, we dx can conclude that 1 e x dx = e x + C. In Example 4, we explore the case of f1x2 = e ax.

■ EXAMPLE 4 Determine the indefinite integral 1 e 4x dx. Solution Since we know that L

d x e = e x, it is reasonable to make this initial guess: dx

e 4x dx = e 4x + C.

d 4x 1e + C2 = 4e 4x, with the coefficient 4 in the derivadx tive resulting from application of the Chain Rule. We modify our guess by inserting 14 to obtain the correct antiderivative: But this is (slightly) wrong, since

Quick Check 2 Find each antiderivative: a) 1 e -3x dx; b) 1 e 11>22x dx.

L

e 4x dx =

This checks:

1 4x e + C. 4

d 1 4x 1 1 a e + Cb = 14e 4x2 = e 4x; multiplying and 4 gives 1. dx 4 4 4

O Quick Check 2

O

• Antidifferentiation

4.1

393

Two useful properties of antidifferentiation are presented in Theorem 3.

THEOREM 3

Properties of Antidifferentiation P1. A constant factor can be moved to the front of an indefinite integral: L

3c # f1x24 dx = c #

L

f1x2 dx.

P2. The antiderivative of a sum or a difference is the sum or the difference of the antiderivatives: L

3f1x2 ; g1x24 dx =

L

f1x2 dx ;

L

g1x2 dx.

In Example 5, we use the rules of antidifferentiation in conjunction with the properties of antidifferentiation. In part (b), we algebraically simplify the integrand before performing the antidifferentiation steps.

■ EXAMPLE 5 Determine these indefinite integrals. Assume x 7 0. a) 1 13x 5 + 7x 2 + 82 dx;

b)

4 + 3x + 2x 4 dx. x L

Solution a) We antidifferentiate each term separately: L

13x 5 + 7x 2 + 82 dx =

L

3x 5 dx +

L

7x 2 dx +

L

8 dx

By Antidifferentiation Property P2 By Antidifferentiation Property P1 and Rules A1 and A2

= 3 A 16 x 6 B + 7 A 13 x 3 B + 8x

= 12 x 6 + 73 x 3 + 8x + C.

Note the simplification of coefficients and the inclusion of just one constant of integration. b) We algebraically simplify the integrand by noting that x is a common denominator and then reducing each ratio as much as possible: Quick Check 3 Determine these indefinite integrals: a) 1 12x 4 + 3x 3 - 7x 2 + x - 52 dx; b) 1 1x - 522 dx; x 2 - 7x + 2 c) dx. x2 L

4 + 3x + 2x 4 4 3x 2x 4 4 = + + = + 3 + 2x 3 + C. x x x x x Therefore, 4 + 3x + 2x 4 4 dx = ¢ + 3 + 2x 3 ≤ dx x L L x = 4 ln x + 3x + 12 x 4 + C.

By Antidifferentiation Properties P1 and P2 and Rules A1, A2, and A3

O Quick Check 3 Initial Conditions The constant of integration C may be of interest in some applications. In such cases, we may specify a point that is a solution of the antiderivative, thereby allowing us to solve for C. This point is called an initial condition.

O

394

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• Integration

■ EXAMPLE 6 Find a function f such that f ¿1x2 = 2x + 3 and f112 = - 2. Solution The antiderivative of f ¿1x2 = 2x + 3 is f1x2 =

L

12x + 32 dx = x 2 + 3x + C.

Since f112 = - 2, we let x = 1 and f112 = - 2, and solve for C: - 2 = 1122 + 3112 + C.

Simplifying, we have - 2 = 4 + C, which gives C = - 6. Therefore, the specific antiderivative of f ¿1x2 = 2x + 3 that satisfies the initial condition is f1x2 = x 2 + 3x - 6. y 10

5

⫺3

Quick Check 4 Find an antiderivative of g1x2 = e 2x such that the point 10, 32 is a solution of the antiderivative.

⫺2

⫺1

⫺2 ⫺5

1

x

f(x) = x 2 + 3x − 2 f(x) = x 2 + 3x − 4 f(x) = x 2 + 3x − 6 f(x) = x 2 + 3x − 8 f(x) = x 2 + 3x − 10

(1, ⫺2)

⫺10

O Quick Check 4 The antiderivative of a function has many applications. For example, in Section 1.8, we saw that velocity is the derivative of a distance function, and, therefore, distance is the antiderivative of a velocity function. If information about the distance of an object at some time t is known, it provides us with an initial condition.

■ EXAMPLE 7 Physical Sciences: Height of a Thrown Object. A rock is thrown

directly upward with an initial velocity of 50 ft> sec from an initial height of 10 ft. The velocity of this rock is modeled by the function v1t2 = - 32t + 50, where t is in seconds, v is in feet per second, and t = 0 represents the moment the rock is released. a) Determine a distance function h as a function of t (in this case, “distance” is the same as “height”). Be sure to consider the fact that at t = 0, the rock is 10 ft above the ground. b) Determine the height and the velocity of the rock after 3 sec.

Solution a) Since distance (height) is the antiderivative of velocity, we have the following: 1- 32t + 502 dt L = - 16t 2 + 50t + C. The constant of integration C can be determined since we know the initial height of the rock, which gives us the ordered pair (0, 10) as an initial condition. We substitute 0 for t and 10 for h(t), and solve for C: h1t2 =

10 = - 161022 + 50102 + C 10 = C. Therefore, the distance function is h1t2 = - 16t 2 + 50t + 10. b) To determine the height of the rock after 3 sec, we substitute 3 for t in our distance function: h132 = - 161322 + 50132 + 10 = 16 ft.

O

4.1

• Antidifferentiation

395

The velocity after 3 sec is v132 = - 32132 + 50 = - 46 ft>sec. Thus, the rock is 16 ft above the ground, but the negative velocity indicates that it is moving downward.

■ EXAMPLE 8 Life Sciences: Change in Population. The rate of change of the population of Phoenix, Arizona, is modeled by the exponential function P¿1t2 = 11.7e0.026t , where t is the number of years since 1960 and P¿1t2 is in thousands of people per year. In 1980, Phoenix had a population of 790,000. (Source: U.S. Census Bureau.) a) Find the population model P(t). b) Estimate the population of Phoenix in 2012.

Solution a) We antidifferentiate the rate-of-change model: P1t2 =

Quick Check 5

O

A town’s rate of population change is modeled by P¿1t2 = 34t + 16, where t is the number of years since 1990 and P¿1t2 is in people per year. a) Find the population model for this town if it is known that in 2000, the town had a population of 2500. b) Forecast the town’s population in 2015.

11.7e 0.026t dt

L 11.7 0.026t e + C By Antidifferentiation Property P1 and Rule A4 = 0.026 = 450e 0.026t + C The population in 1980 is treated as the initial condition: (20, 790). We make the substitutions and solve for C: 790 = 450e 0.0261202 + C 790 = 756.9 + C C = 33.1. Therefore, the population model is P1t2 = 450e 0.026t + 33.1. b) The year 2012 corresponds to t = 52, so we make the substitution: P1522 = 450e 0.0261522 + 33.1 = 1772. According to this model, the population of Phoenix in 2012 should be about 1,772,000. Given that Phoenix had a population of 1,567,000 in 2008, this prediction is reasonable.

O Quick Check 5

TECHNOLOGY CONNECTION

Antiderivatives and Area A graphing calculator can calculate the area under the graph of a function. In the Y= window, enter the function f 1x2 = 2x, for x Ú 0, and graph it in 30, 10, 0, 204. Press F and m and then select 1 f 1x2 dx from the list. For “Lower Limit,” type in 0, and press [. For “Upper Limit,” let x = 1, and press [. The calculator will shade in the region and report the area in the lower-left corner. Do this for a series of x-values, and put the information in a table like that to the right.

Base, x

Height, f 1x2

Area of region, A1x2

1

2

1

2

4

4

3 4 5 6 7 (continued)

396

CHAPTER 4

• Integration

Antiderivatives and Area (continued) 2. Repeat parts (a) through (e) of Exercise 1 for f 1x2 = 3,

EXERCISES 1. a) Fill in the entire table. b) If x = 20, what is the area under the graph of f ? c) What is the relationship between the value of x in the

first column and the area in the third column? d) Form your observation from part (c) into an area function A1x2. e) What is the relationship between the area function A from part (d) and the given function f ?

and look for a pattern in an relationship between the area function A and the given function f. 3. Repeat parts (a) through (e) of Exercise 1 for f 1x2 = 3x 2 ,

and look for a pattern in the relationship between the area function A and the given function f.

Section Summary • The antiderivative of a function f1x2 is a set of functions F1x2 + C such that d 3F1x2 + C4 = f1x2, dx

x n dx =

where the constant C is called the constant of integration. • An antiderivative is denoted by an indefinite integral using the integral sign, 1 . If F1x2 is an antiderivative of f1x2, we write L

• The Constant Rule of Antidifferentiation is 1 k dx = kx + C. • The Power Rule of Antidifferentiation is L • The Natural Logarithm Rule of Antidifferentiation is 1 dx = ln x + C, for x 7 0. Lx • The Exponential Rule ( base e) of Antidifferentiation is

f1x2 dx = F1x2 + C.

e ax dx =

We check the correctness of an antiderivative we have found by differentiating it.

1 x n + 1 + C, for n Z - 1. n + 1

1 ax e + C, for a Z 0. a

L • An initial condition is an ordered pair that is a solution of a particular antiderivative of an integrand.

EXERCISE SET

4.1 Determine these indefinite integrals. 1.

L

x 6 dx

2.

17. L

x 7 dx 19.

3. 5. 7. 9. 11. 13. 15.

L L L L

4.

2 dx x

1>4

dx

1x 2 + x - 12 dx

8.

12t 2 + 5t - 32 dt

10.

1

dx

12.

2x dx

14.

2x 5 dx

16.

Lx

3

3

L L

6.

L L L L

4 dx x

dx 23.

13t 2 - 4t + 72 dt

25.

1

dx

27.

2x dx

29.

5

3

L

21.

1x 2 - x + 22 dx

Lx L

1>3

2x 2 dx

31.

dx Lx 1 Lx

18.

4

20.

dx

3 5 a + 2 b dx x L x -7 L 2x 3

L L L L

2

dx

22.

24.

2e 2x dx

26.

e 3x dx

28.

e 7x dx

30.

5e 3x dx

32.

dx L x2 2 Lx

dx

4 7 a + b dx x L x3 5 L 2x 3 4

L L L L

dx

4e 4x dx e 5x dx e 6x dx 2e 5x dx

Exercise Set 4.1

33. 35. 37.

39. 40. 41. 42.

43. 44. 45.

46.

L L L L L L

6e 8x dx

34.

2 -9x dx 3e

36.

15x 2 - 2e 7x2 dx

38.

a x2 ax 4 +

L L L

where t is the number of years since 2005. (Source: Federal Reserve System.) Use the preceding information for Exercises 59 and 60.

12e 3x dx 4 -10x dx 5e

12x 5 - 4e 3x2 dx

3 2x + x -4>3 b dx 2 1 82x

13x + 222 dx

-

59. Find the national credit market debt, D1t2, during the years 2005 through 2009 given that D102 = 41,267.

4 -2>5 x b dx 5

60. What was the national credit market debt in 2009, given that D102 = 41,267?

(Hint: Expand first.)

61. Total cost from marginal cost. A company determines that the marginal cost, C¿, of producing the xth unit of a product is given by

1x + 42 dx 2

L

C¿1x2 = x 3 - 2x.

3 - 5e 2x + 2x 7 b dx L x a

L

a 2e 6x -

3 3 + 2x 4 b dx x

a

-

2 5x 8 e - b dx x 3

+

3 6x 7 e - b dx x 4

7

L 2x a

4

L 2x 5

Find the total-cost function, C, assuming that C1x2 is in dollars and that fixed costs are $7000. 62. Total cost from marginal cost. A company determines that the marginal cost, C¿, of producing the xth unit of a product is given by C¿1x2 = x 3 - x. Find the total-cost function, C, assuming that C1x2 is in dollars and that fixed costs are $6500. 63. Total revenue from marginal revenue. A company determines that the marginal revenue, R¿, in dollars, from selling the xth unit of a product is given by

Find f such that:

47. f ¿1x2 = x - 3, f 122 = 9 48. f ¿1x2 = x - 5, f 112 = 6

R¿1x2 = x 2 - 3.

49. f ¿1x2 = x 2 - 4, f 102 = 7

a) Find the total-revenue function, R, assuming that R102 = 0. b) Why is R102 = 0 a reasonable assumption?

50. f ¿1x2 = x 2 + 1, f 102 = 8

51. f ¿1x2 = 5x 2 + 3x - 7, f 102 = 9

64. Total revenue from marginal revenue. A company determines that the marginal revenue, R¿, in dollars, from selling the xth unit of a product is given by

52. f ¿1x2 = 8x 2 + 4x - 2, f 102 = 6 53. f ¿1x2 = 3x 2 - 5x + 1, f 112 =

R¿1x2 = x 2 - 1.

7 2

54. f ¿1x2 = 6x 2 - 4x + 2, f 112 = 9 55. f ¿1x2 = 5e 2x , f 102 = 57. f ¿1x2 = 58. f ¿1x2 =

397

4 2x 2 3

2x

1 2

56. f ¿1x2 = 3e 4x , f 102 =

, f 112 = - 5 , f 112 = 1

APPLICATIONS Business and Economics Credit market debt. From 2005 to 2009, the annual rate of change in the national credit market debt, in billions of dollars per year, could be modeled by the function D œ 1t2  810.3t 2  1730.3t  3648,

7 4

a) Find the total-revenue function, R, assuming that R102 = 0. b) Why is R102 = 0 a reasonable assumption? 65. Demand from marginal demand. A company finds that the rate at which the quantity of a product that consumers demand changes with respect to price is given by the marginal-demand function D¿1x2 = -

4000 x2

,

where x is the price per unit, in dollars. Find the demand function if it is known that 1003 units of the product are demanded by consumers when the price is $4 per unit. 66. Supply from marginal supply. A company finds that the rate at which a seller’s quantity supplied changes with respect to price is given by the marginal-supply function S¿1x2 = 0.24x 2 + 4x + 10,

398

CHAPTER 4

• Integration

where x is the price per unit, in dollars. Find the supply function if it is known that the seller will sell 121 units of the product when the price is $5 per unit.

70. Memory. In a memory experiment, the rate at which students memorize Spanish vocabulary is found to be given by M¿1t2 = 0.2t - 0.003t 2,

67. Efficiency of a machine operator. The rate at which a machine operator’s efficiency, E (expressed as a percentage), changes with respect to time t is given by

where M1t2 is the number of words memorized in t minutes. a) Find M1t2 if it is known that M102 = 0. b) How many words are memorized in 8 min?

dE = 30 - 10t, dt where t is the number of hours the operator has been at work.

Physical Sciences 71. Physics: height of a thrown baseball. A baseball is thrown directly upward with an initial velocity of 75 ft/sec from an initial height of 30 ft. The velocity of the baseball is given by the function v1t2 = - 32t + 75, where t is the number of seconds since the ball was released and v is in feet per second. a) Find the function h that gives the height (in feet) of the baseball after t seconds, using the fact that at t = 0, the ball is 30 ft above the ground. b) What are the height and the velocity of the baseball after 2 sec of flight? c) After how many seconds does the ball reach its highest point? (Hint: The ball “stops” for a moment before starting its downward fall.) d) How high does the ball get at its highest point? e) After how many seconds will the ball hit the ground? f) What is the ball’s velocity at the moment it hits the ground?

A machine operator’s efficiency changes with respect to time.

a) Find E1t2, given that the operator’s efficiency after working 2 hr is 72%; that is, E122 = 72. b) Use the answer to part (a) to find the operator’s efficiency after 3 hr; after 5 hr. 68. Efficiency of a machine operator. The rate at which a machine operator’s efficiency, E (expressed as a percentage), changes with respect to time t is given by

General Interest 72. Population growth. The rates of change in population for two cities are as follows:

dE = 40 - 10t, dt

Alphaville: P¿1t2 = 45, Betaburgh: Q¿1t2 = 105e 0.03t,

where t is the number of hours the operator has been at work. a) Find E1t2, given that the operator’s efficiency after working 2 hr is 72%; that is, E122 = 72. b) Use the answer to part (a) to find the operator’s efficiency after 4 hr; after 8 hr.

where t is the number of years since 1990, and both P¿ and Q¿ are measured in people per year. In 1990, Alphaville had a population of 5000, and Betaburgh had a population of 3500. a) Determine the population models for both cities. b) What were the populations of Alphaville and Betaburgh, to the nearest hundred, in 2000? c) Sketch the graph of each city’s population model and estimate the year in which the two cities have the same population.

Social and Life Sciences 69. Spread of an influenza. During 18 weeks from November 2009 to February 2010, the rate at which the number of cases of swine flu changed could be approximated by I¿1t2 = - 6.34t + 141.6, where I is the total number of people who have contracted swine flu and t is time measured in weeks. (Source: Centers for Disease Control and Prevention.) a) Estimate I1t2, the total number who have contracted influenza by time t. Assume that I102 = 1408. b) Approximately how many people contracted influenza during the first 8 weeks? c) Approximately how many people contracted influenza during the whole 18 weeks? d) Approximately how many people per 100,000 contracted influenza during the last 7 of the 18 weeks?

SYNTHESIS Find f. 73. f ¿1t2 = 2t +

1 2t

, f 142 = 0

74. f ¿1t2 = t 23, f 102 = 8 Evaluate. Each of the following can be determined using the rules developed in this section, but some algebra may be required beforehand. 75.

L

15t + 422 t 4 dt

76.

L

1x - 122 x 3 dx

4.2

77.

79. 81. 83.

85.

L

11 - t22t dt x 4 - 6x 2 - 7

L

x3

78.

dx

1 dx L In 10 x L

80. 82.

13x - 5212x + 122 dx

x2 - 1 dx Lx + 1

84.

86.

1t + 322

dt 2t SupersetV

L L L

399

Antiderivatives as Areas

Answers to Quick Checks 1 11 1 201 x + C; (b) x + C; 11 201 6 6 1 (c) 2x 7 + C; (d) - 3 + C 7 3x 1 2. (a) - e -3x + C; (b) 2e 11>22 x + C 3 3 7 1 2 5 3. (a) x + x 4 - x 3 + x 2 - 5x + C; 5 4 3 2 1 2 (b) x 3 - 5x 2 + 25x + C; (c) x - 7 ln x + C x 3 1 5 4. G1x2 = e 2x + 5. (a) P1t2 = 17t 2 + 16t + 640; 2 2 (b) P1252 = 11,665 people 1. (a)

1t + 123 dt be ax dx 3

L

•

264x 4 dx

t3 + 8 dt Lt + 2

87. On a test, a student makes this statement: “The function f 1x2 = x 2 has a unique antiderivative.” Is this a true statement? Why or why not? 88. Describe the graphical interpretation of an antiderivative.

• Find the area under a graph to solve real-world problems. • Use rectangles to approximate the area under a graph.

■ EXAMPLE 1 Physical Sciences: Distance as Area. A vehicle travels at 50 mi/hr for 2 hr. How far has the vehicle traveled?

Solution The answer is 100 mi. We treat the vehicle’s veloc-

y

ity as a function, v1x2 = 50. We graph this function, sketch a vertical line at x = 2, and obtain a rectangle. This rectangle measures 2 units horizontally and 50 units vertically. Its area is the distance the vehicle has traveled: 2 hr #

50 mi = 100 mi. 1 hr

Velocity (in miles/hour)

OBJECTIVES

Integral calculus is primarily concerned with the area below the graph of a function (specifically, the area between the graph of a function and the x-axis). There are many situations where the area can be interpreted in a meaningful way. In this section, we assume that all functions are nonnegative; that is, f1x2 Ú 0. Consider the following examples.

2 hr = 2 1 hr

v(x) = 50

50

Area = 2 · 50 = 100

Note that the units of hours cancel.

0

1

2

x

Time (in hours)

■ EXAMPLE 2 Business: Total Cost as Area. Green Leaf Skateboards determines that for the first 50 skateboards produced, its cost is $40 per skateboard. What is the total cost to produce 50 skateboards?

Solution The marginal-cost function is C¿1x2 = 40,

0 … x … 50. Its graph is a horizontal line. If we mark off 50 units along the x-axis, we get a rectangle, as in Example 1. The area of this rectangle is 40 # 50 = 2000. Therefore, the total cost to produce 50 skateboards is $2000: dollars a50 skateboards # 40 = 2000 dollarsb. skateboard

y Cost (in dollars)

4.2

Antiderivatives as Areas

C⬘(x) = 40

40

Area = 40 · 50 = 2000

0

50

Number of skateboards

x

400

CHAPTER 4

•

Integration

Geometry and Areas For the time being, we will deal with linear functions. In these cases, we can use geometry to find the area formed by the graph of a function. The following two formulas, where b = base and h = height, will be useful. Area of a rectangle: A = bh

Area of a triangle: A = 12 bh

h

h b

b

Examples 3 and 4 continue to explore the themes of Example 1 (distance as area) and Example 2 (total cost as area), respectively. In these cases, the graph of the function is linear with a nonzero slope.

■ EXAMPLE 3 Physical Sciences: Distance as Area. The velocity of a moving object

is given by the function v1x2 = 3x, where x is in hours and v is in miles per hour. Use geometry to find the area under the graph, which is the distance the object has traveled: a) during the first 3 hr 10 … x … 32; b) between the third hour and the fifth hour 13 … x … 52.

Solution y Velocity (in miles/hour)

a) The graph of the velocity function is shown at the right. We see the region corresponding to the time interval 0 … x … 3 is a triangle with base 3 and height 9 (since v132 = 9). Therefore, the area of this region is A = 12 132192 = 27 2 = 13.5. The object traveled 13.5 mi during the first 3 hr.

15

v(x) = 3x

10

5

0

1

2

3

4

x

5

Time (in hours)

An object moves with a velocity of v1t2 = 12 t, where t is in minutes and v is in feet per minute. a) How far does the object travel during the first 30 min? b) How far does the object travel between the first hour and the second hour?

b) The region corresponding to the time interval 3 … x … 5 is a trapezoid. It can be decomposed into a rectangle and a triangle as indicated in the figure to the right. The rectangle has base 2 and height 9, and thus an area A = 122192 = 18; the triangle has base 2 and height 6, for an area A = 12 122162 = 6. Summing these, we get 24. Therefore, the object traveled 24 mi between the third hour and the fifth hour.

y Velocity (in miles/hour)

Quick Check 1

15

v(x) = 3x

2 9

5

0

O Quick Check 1

6

10

1

2

3

4

5

Time (in hours)

x

O

•

4.2

401

Antiderivatives as Areas

■ EXAMPLE 4 Business: Total Profit as Area. Cousland, Inc., has a marginal-

profit function modeled by the linear function P¿1x2 = 0.15x, where x is in months and P¿ is in thousands of dollars per month. Sketch this graph and use it to determine the total profit earned by Cousland, Inc., in a year 10 … x … 122.

Solution The graph of P¿ is shown below: Profit (in thousands of dollars per month)

y 1.8 1.4

P⬘(x) = 0.15x

1 0.6 0

0

2

4

6

8

10

12

x

Time (in months)

Quick Check 2 Calculate the total profit of Cousland, Inc., between the fifth and the twelfth month.

For the 12-month period, the area is calculated by using the formula for a triangle: thousands of dollars A = 12 112 months2 a1.8 b = 10.8 thousand dollars. Cousland, Inc., month earned a total profit of $10,800 in a year.

O Quick Check 2 In each of the Examples 1 through 4, the function was a rate function; its output units formed a rate (miles per hour in Examples 1 and 3, dollars per skateboard in Example 2, thousands of dollars per month in Example 4). The units of the area were derived by multiplying input units by output units.

Riemann Summation How does an antiderivative of a function translate into the area below that function’s graph? The Technology Connection on p. 395 and Examples 1 through 4 in this section suggest a pattern: • If f1x2 = k, where k is a constant, its graph is a horizontal line of height k. The region under this graph over the interval 30, x4 is a rectangle, and its area is A = k # x (height times base). • If f1x2 = mx, its graph is a line of slope m, passing through the origin. The region under this graph over an interval 30, x4 is a triangle, and its area is A = 12 1x21mx2 = 12 mx 2.

y

f(x) = k

(x, k)

k

0

A(x) = kx

k

A(x)

x

x

x

y (x, mx)

mx f(x) = mx

0

x

1 A(x) = − mx 2 2

mx

A(x) x

x

In these two cases, the area function is an antiderivative of the function that generated the graph. Is this always true? Is the formula for the area under the graph of any function that function’s antiderivative? How do we handle curved graphs for which area formulas may not be known? We investigate these questions using geometry, in a procedure called Riemann summation (pronounced “Ree-mahn”) in honor of the great German mathematician G. F. Bernhard Riemann (1826–1866). Before we consider areas under curves, let’s revisit Green Leaf Skateboards.

O

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■ EXAMPLE 5 Business: Total Cost. Green Leaf Skateboards has the following marginal-cost function for producing skateboards: For up to 50 skateboards, the cost is $40 per skateboard. For quantities from 51 through 125 skateboards, the cost drops to $30 per skateboard. After 125 skateboards, it drops to $25 per skateboard. If x represents the number of skateboards produced, the marginal-cost function C¿ is 40, for 0 … x … 50, C¿1x2 = c 30, for 50 6 x … 125, 25, for 125 6 x … 150. Find the total cost to produce 150 skateboards.

Cost per skateboard (in dollars)

Solution We are extending Example 2. We calculate the areas of the rectangles formed by the horizontal lines of the graph of the marginal-cost function: 40, for 0 ≤ x ≤ 50, 30, for 50 < x ≤ 125, 25, for 125 < x ≤ 150

40

C⬘(x) =

35 30 25 20 15 10 5 0

25

50

75

100

125

150 x

Number of skateboards produced

The total cost to produce 150 skateboards is found by summing those areas: Total cost = 14021502 + 13021752 + 12521252 = $4875.

Example 5 illustrates the first steps of Riemann summation, a method that allows us to determine the area under curved graphs. We use rectangles to approximate the area under a curve given by y = f1x2, a continuous function, over an interval 3a, b4. Riemann summation is accomplished with the use of summation notation, introduced below. In the following figure, 3a, b4 is divided into four subintervals, each having width ¢x = 1b - a2>4. y f(x 2)

f(x 1)

f(x 2)

f

f(x 3) f(x 4)

a = x1

x2

x3

x4

b

x

Δx

The area under a curve can be approximated by a sum of rectangular areas.

The heights of the rectangles shown are f1x 12,

f1x 22,

f1x 32,

and

f1x 42.

The area of the region under the curve is approximately the sum of the areas of the four rectangles: f1x 12 ¢x + f1x 22 ¢x + f1x 32 ¢x + f1x 42 ¢x.

4.2

•

403

Antiderivatives as Areas

We can denote this sum with summation notation, or sigma notation, which uses the Greek capital letter sigma, © : 4

a f1x i2 ¢x.

i=1

This is read “the sum of the product f1x i2 ¢x from i = 1 to i = 4.” To recover the original expression, we substitute the numbers 1 through 4 successively for i in f1x i2 ¢x and write plus signs between the results. Before we continue, let’s consider some examples involving summation notation.

■ EXAMPLE 6 Write summation notation for 2 + 4 + 6 + 8 + 10. Quick Check 3

O

Write summation notation for each expression: a) 5 + 10 + 15 + 20 + 25; b) 33 + 44 + 55 + 66.

Solution Note that we are adding consecutive multiples of 2: 5

2 + 4 + 6 + 8 + 10 = a 2i. i=1

O Quick Check 3 ■ EXAMPLE 7 Write summation notation for

g1x 12 ¢x + g1x 22 ¢x + Á + g1x 192 ¢x.

Solution 19

g1x 12 ¢x + g1x 22 ¢x + Á + g1x 192 ¢x = a g1x i2 ¢x i=1

4

■ EXAMPLE 8 Express a 3i without using summation notation. i=1 Solution Quick Check 4 6

Express a 1i + i2 without 2

i=1

using summation notation.

4

i 1 2 3 4 a 3 = 3 + 3 + 3 + 3 , or 120

i=1

O Quick Check 4 30

■ EXAMPLE 9 Express a h1xi2 ¢x without using summation notation. i=1 Solution

30

Á + h1x 302 ¢x a h1x i2 ¢x = h1x 12 ¢x + h1x 22 ¢x +

i=1

Approximation of area by rectangles becomes more accurate as we use smaller subintervals and hence more rectangles, as shown in the following figures. y

y f

f(x 3)

a = x1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 b 8 subintervals

x

f

a = x1

… 24 subintervals

x 24

b

x

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Integration

y f

In general, suppose that the interval 3a, b4 is divided into n equally sized subintervals, each of width ¢x = 1b - a2>n. We construct rectangles with heights f1x 12, f1x 22, Á , f1x n2.

The width of each rectangle is ¢x, so the first rectangle has an area of f1x 12¢x, the second rectangle has an area of f1x 22¢x, and so on. The area of the region under the curve is approximated by the sum of the areas of the rectangles:

…

n

a = x1 x2 x3 x4 x5

…

xn

b

x

a f1x i 2¢x.

i=1

■ EXAMPLE 10 Consider the graph of f1x2 = 600x - x 2

over the interval 30, 6004. a) Approximate the area by dividing the interval into 6 subintervals. b) Approximate the area by dividing the interval into 12 subintervals.

Solution y 90,000

a) We divide 30, 6004 into 6 subintervals of size

Total area = 35,000,000

¢x =

600 - 0 = 100, 6

with x i ranging from x 1 = 0 to x 6 = 500. Thus, the area under the curve is approximately

# # # a f1x i2 ¢x = f102 100 + f11002 100 + f12002 100 i=1 + f13002 # 100 + f14002 # 100 + f15002 # 100 6

10,000 0

y 90,000

100 200 300 400 500 600

= 0 # 100 + 50,000 # 100 + 80,000 # 100 + 90,000 # 100 + 80,000 # 100 + 50,000 # 100 = 35,000,000.

x

b) We divide 30, 6004 into 12 subintervals of size ¢x = 1600 - 02> 12 = 50, with x i ranging from x 1 = 0 to x 12 = 550. Thus, we have another approximation of the area under the curve:

Total area = 35,750,000

# # # # a f1x i2 ¢x = f102 50 + f1502 50 + f11002 50 + f11502 50 + f12002 # 50 + f12502 # 50 + f13002 # 50 + f13502 # 50 + f14002 # 50 + f14502 # 50 + f15002 # 50 + f15502 # 50 = 0 # 50 + 27,500 # 50 + 50,000 # 50 + 67,500 # 50 + 80,000 # 50 + 87,500 # 50 + 90,000 # 50 + 87,500 # 50 + 80,000 # 50 + 67,500 # 50 + 50,000 # 50 + 27,500 # 50 = 35,750,000. 12

i=1

10,000 0

100 200 300 400 500 600

x

Note that in Example 10 the approximation using n = 12 is closer to the exact value than the one using n = 6. The sums used in Example 10 to approximate the area under a curve are called Riemann sums. Riemann sums can be calculated using any x-value within each subinterval. For simplicity, in this text we will always use the left endpoint of each subinterval.

■ EXAMPLE 11 Use 5 subintervals to approximate the area under the graph of f1x2 = 0.1x 3 - 2.3x 2 + 12x + 25 over the interval 31, 164.

•

4.2

405

Antiderivatives as Areas

Solution We divide 31, 164 into 5 subintervals of size ¢x = 116 - 12>5 = 3, with x i ranging from x 1 = 1 to x 5 = 13. Although a drawing is not required, we can make one to help visualize the area. y 60 50

f (x) = 0.1x3 − 2.3x2 + 12x + 25

40 30 20 10 0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

x

The area under the curve from 1 to 16 is approximately Quick Check 5 Use 6 subintervals to approximate the area under the graph of the function in Example 11 over the interval 30, 124.

# # # # # a f1x i2 ¢x = f112 3 + f142 3 + f172 3 + f1102 3 + f1132 3 i=1 = 34.8 # 3 + 42.6 # 3 + 30.6 # 3 + 15 # 3 + 12 # 3 = 405. 5

O Quick Check 5 Steps for the Process of Riemann Summation 1. Draw the graph of f1x2. 2. Subdivide the interval 3a, b4 into n subintervals of equal width. Calculate b - a . the width of each rectangle by using the formula ¢x = n 3. Construct rectangles above the subintervals such that the top left corner of each rectangle touches the graph. 4. Determine the area of each rectangle. 5. Sum these areas to arrive at an approximation for the total area under the curve.

Definite Integrals The key concept being developed in this section is that the more subintervals we use, the more accurate the approximation of area becomes. As the number of subdivisions n increases, the width of each rectangle ¢x decreases. If n is allowed to approach infinity, then ¢x approaches 0; these are limits, and the approximations of area become more and more exact to the true area under the graph. The exact area underneath the graph of a continuous function y = f1x2 over an interval 3a, b4 is, by definition, given by a definite integral.

DEFINITION

Let y = f1x2 be continuous and nonnegative, f1x2 Ú 0, over an interval 3a, b4. A definite integral is the limit as n : q (equivalently, ¢x : 0) of the Riemann sum of the areas of rectangles under the graph of the function y = f1x2 over the interval 3a, b4. Exact area = lim a f1x i2 # ¢x = ¢x:0

b

n

i=1

La

f1x2 dx.

O

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Notice that the summation symbol becomes an integral sign (the elongated “s” is Leibniz notation representing “sum”) and ¢x becomes dx. The interval endpoints a and b are placed at the bottom right and top right, respectively, of the integral sign. If f1x2 Ú 0 over an interval 3a, b4, the definite integral represents area. The definite integral is also defined for f1x2 6 0. We will discuss its interpretation in Section 4.3. We can use geometry to determine the value of some definite integrals, as the following example suggests. 2

13x + 22 dx. L0 y Solution This definite integral is a command to calculate the exact area underneath the graph of the 8 function f1x2 = 3x + 2 over the interval 30, 24. We sketch the graph and note that the region is a trape6 zoid. Thus, we can use geometry to determine this area (a Riemann sum is not needed here).

■ EXAMPLE 12 Determine the value of

(2, 8) f(x) = 3x + 2

4

Quick Check 6

(0, 2)

O

Use geometry to determine the values of these definite integrals:

0

L0

1x + 12 dx;

2

L0

7

b)

L4

115 - 2x2 dx.

2 x

Using a method similar to that in Example 3(b), we find that the area is 10. Therefore,

3

a)

1

13x + 22 dx = 10.

O Quick Check 6

Section Summary • The area under a curve can often be interpreted in a meaningful way. • The units of the area are found by multiplying the units of the input variable by the units of the output variable. It is crucial that the units are consistent. • Geometry can be used to find areas of regions formed by graphs of linear functions.

• A Riemann sum uses rectangles to approximate the area under a curve. The more rectangles, the better the approximation. b • The definite integral, 1a f1x2 dx, is a representation of the exact area under the graph of a continuous function y = f1x2, where f1x2 Ú 0, over an interval 3a, b4.

Exercise Set 4.2

407

EXERCISE SET

4.2 APPLICATIONS Business and Economics In Exercises 1–8, calculate total cost, disregarding any fixed costs. 1. Total cost from marginal cost. Redline Roasting has found that the cost, in dollars per pound, of the coffee it roasts is C¿1x2 = - 0.012x + 6.50, for x … 300, where x is the number of pounds of coffee roasted. Find the total cost of roasting 200 lb of coffee. 2. Total cost from marginal cost. Sylvie’s Old World Cheeses has found that the cost, in dollars per kilogram, of the cheese it produces is C¿1x2 = - 0.003x + 4.25, for x … 500, where x is the number of kilograms of cheese produced. Find the total cost of producing 400 kg of cheese. 3. Total cost from marginal cost. Photos from Nature has found that the cost per card of producing x note cards is given by C¿1x2 = - 0.04x + 85, for x … 1000, where C¿1x2 is the cost, in cents, per card. Find the total cost of producing 650 cards. 4. Total cost from marginal cost. Cleo’s Custom Fabrics has found that the cost per yard of producing x yards of a particular fabric is given by C¿1x2 = - 0.007x + 12, for x … 350, where C¿1x2 is the cost in dollars. Find the total cost of producing 200 yd of this material. 5. Total revenue from marginal profit. A concert promoter sells x tickets and has a marginal-profit function given by P¿1x2 = 2x - 1150, where P¿1x2 is in dollars per ticket. This means that the rate of change of total profit with respect to the number of tickets sold, x, is P¿1x2. Find the total profit from the sale of the first 300 tickets. 6. Total profit from marginal profit. Poyse Inc. has a marginal-profit function given by P¿1x2 = - 2x + 80, where P¿1x2 is in dollars per unit. This means that the rate of change of total profit with respect to the number of units produced, x, is P¿1x2. Find the total profit from the production and sale of the first 40 units. 7. Total cost from marginal costs. Raggs, Ltd., determines that its marginal cost, in dollars per dress, is given by 2 x + 50, for x … 450. C¿1x2 = 25 Find the total cost of producing the first 200 dresses.

8. Total cost from marginal cost. Using the information and answer from Exercise 7, find the cost of producing the 201st dress through the 400th dress. 9. Total cost from marginal cost. Ship Shape Woodworkers has found that the marginal cost of producing x feet of custom molding is given by C¿1x2 = - 0.00002x 2 - 0.04x + 45, for x … 800, where C¿1x2 is in cents. Approximate the total cost of manufacturing 800 ft of molding, using 5 subintervals over 30, 8004 and the left endpoint of each subinterval. 10. Total cost from marginal cost. Soulful Scents has found that the marginal cost of producing x ounces of a new fragrance is given by C¿1x2 = 0.0005x 2 - 0.1x + 30, for x … 125, where C¿1x2 is in dollars. Use 5 subintervals over [0, 100] and the left endpoint of each subinterval to approximate the total cost of producing 100 oz of the fragrance. 11. Total cost from marginal cost. Shelly’s Roadside Fruit has found that the marginal cost of producing x pints of fresh-squeezed orange juice is given by C¿1x2 = 0.000008x 2 - 0.004x + 2, for x … 350, where C¿1x2 is in dollars. Approximate the total cost of producing 270 pt of juice, using 3 subintervals over 30, 2704 and the left endpoint of each subinterval. 12. Total cost from marginal cost. Mangianello Paving, Inc., has found that the marginal cost, in dollars, of paving a road surface with asphalt is given by 1 2 x - 20x + 1800, for x … 80, 6

C¿1x2 =

where x is measured in hundreds of feet. Use 4 subintervals over 30, 404 and the left endpoint of each subinterval to approximate the total cost of paving 4000 ft of road surface. In Exercises 13–18, write summation notation for each expression. 13. 3 + 6 + 9 + 12 + 15 + 18 14. 5 + 10 + 15 + 20 + 25 + 30 + 35 15. f 1x 12 + f 1x 22 + f 1x 32 + f 1x 42

16. g1x 12 + g1x 22 + g1x 32 + g1x 42 + g1x 52 17. G1x 12 + G1x 22 + Á + G1x 152 18. F 1x 12 + F 1x 22 + Á + F 1x 172 4

19. Express a 2 i without using summation notation. i=1

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CHAPTER 4

•

Integration

the area of each rectangle and then adding. Compare your answer to that for part (a).

5

20. Express a 1- 22i without using summation notation. i=0

y 5

21. Express a f 1x i2 without using summation notation. i=1

f(4)

4

22. Express a g1x i2 without using summation notation. i=1

23. a) Approximate the area under the following graph of 1 f 1x2 = 2 over the interval 31, 74 by computing x the area of each rectangle to four decimal places and then adding.

25. Total profit from marginal profit. Holcomb Hill Fitness has found that the marginal profit, P¿1x2, in cents, is given by

y f(1)

P¿1x2 = - 0.0006x 3 + 0.28x 2 + 55.6x, for x … 500,

1

2

3

4

5

6

7

where x is the number of members currently enrolled at the health club.

x

b) Approximate the area under the graph of f 1x2 =

1

x2 over the interval 31, 74 by computing the area of each rectangle to four decimal places and then adding. Compare your answer to that for part (a).

Marginal profit (in cents)

0

x

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

y 30,000

P′

24,000 18,000 12,000 6000 0

y

100

200

300

400

500 x

Number of members

f(1)

Approximate the total profit when 300 members are enrolled by computing the sum 6

0

1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7

x

a P¿1x i2 ¢x,

i=1

24. a) Approximate the area under the graph of f 1x2 = x 2 + 1 over the interval 30, 54 by computing the area of each rectangle and then adding.

with ¢x = 50. 26. Total cost from marginal cost. Raggs, Ltd., has found that the marginal cost, in dollars, for the xth jacket produced is given by C¿1x2 = 0.0003x 2 - 0.2x + 50.

y

f(4)

0

1

2

3

4

5

x

b) Approximate the area under the graph of f 1x2 = x 2 + 1 over the interval 30, 54 by computing

Total cost (in dollars)

y 50 40 30 C'(x)

20 10 0

100

200

300

Number of jackets

400

x

409

Exercise Set 4.2

Approximate the total cost of producing 400 jackets by computing the sum 4

a C¿1x i2 ¢x,

The Trapezoidal Rule. Another way to approximate an integral is to replace each rectangle in the sum with a trapezoid (see Figs. 1 and 2). The area of a trapezoid is h1c1  c22/2, where c1 and c2 are the lengths of the parallel sides. Thus, in Fig. 2,

i=1

Area under f over 7a, b8 f 1a2  f 1m2 f 1m2  f 1b2 « ≤x  ¢x 2 2 f 1a2 f 1b2 « ≤x c  f 1m2  d. 2 2

with ¢x = 100. 27. Approximate the area under the graph of f 1x2 = 0.01x 4 - 1.44x 2 + 60

over the interval 32, 104 by dividing the interval into 4 subintervals.

f(b)

28. Approximate the area under the graph of

f(b) f(m)

f(m)

g1x2 = - 0.02x 4 + 0.28x 3 - 0.3x 2 + 20

over the interval 33, 124 by dividing the interval into 4 subintervals. 29. Approximate the area under the graph of F 1x2 = 0.2x 3 + 2x 2 - 0.2x - 2

over the interval 3- 8, - 34 using 5 subintervals. 30. Approximate the area under the graph of G1x2 = 0.1x 3 + 1.2x 2 - 0.4x - 4.8

over the interval 3- 10, - 44 using 6 subintervals. In Exercises 31–39, use geometry to evaluate each definite integral. 2

31.

5

32.

2 dx

L0

L0

6

33.

34.

L-1

3

35.

L0

36.

L0

10

37. 39.

L2

4x dx 5

1 x dx L0 2 4

4 dx

5

x dx

38.

L0

12x + 52 dx

110 - 2x2 dx

SYNTHESIS 40. Show that, for any function f defined for all x i’s, and any constant k, we have 4

4

i=1

i=1

a

m

b

a

m

b

FIGURE 2

FIGURE 1

For an interval 7a, b8 subdivided into n equal subintervals of length ¢x  1b  a2>n, we get the approximation Area under f over 7a, b8 f 1a2  f 1x 22  f 1x 32  « ≤xc 2

Á

 f 1x n2 +

f 1b2 2

d,

where x 1  a and x n  x n1  ¢x or x n  a  1n - 12≤x. This is called the Trapezoidal Rule.

4

3 dx

L2

6 dx

f(a)

f(a)

a kf 1x i2 = k a f 1x i2.

41. Use the Trapezoidal Rule and the interval subdivision of Exercise 23(a) to approximate the area under the graph of f 1x2 = 1>x 2 over the interval 31, 74. 42. Use the Trapezoidal Rule and the interval subdivision of Exercise 24(a) to approximate the area under the graph of f 1x2 = x 2 + 1 over the interval 30, 54.

TECHNOLOGY CONNECTION The exact area of a semicircle of radius r can be found using the formula A  12 Pr 2. Using this equation, compare the answers to Exercises 43 and 44 with the exact area. Note that most calculators do not show entire semicircles.

43. Approximate the area under the graph of f 1x2 = 225 - x 2 using 10 rectangles. 44. Approximate the area under the graph of g1x2 = 249 - x 2 using 14 rectangles.

Then show that, in general, n

n

i=1

i=1

a kf 1x i2 = k a f 1x i2,

for any constant k and any function f defined for all x i’s.

Answers to Quick Checks 1. (a) 225 ft; (b) 2700 ft 5

6

2. Total profit = $8925

3. (a) a 5i; (b) a 11i 4. 112 5. 368 6. (a) 7.5; (b) 12 i=1

i=3

410

• Integration

CHAPTER 4

Area and Definite Integrals

4.3

In Sections 4.1 and 4.2, we considered the relationship between the area under the graph of a function f and the antiderivative of f. We have yet to establish the general rule that the antiderivative of a function f does in fact lead to the exact area under the graph of f. As we will see, we can use the antiderivative of a function to determine the exact area under the graph of the function. This process is called integration.

OBJECTIVES • Find the area under the graph of a nonnegative function over a given closed interval.

The Fundamental Theorem of Calculus

The area under the graph of a nonnegative continuous function f over an interval 3a, b4 is determined as an area function A, which is an antiderivative of f; that is, d A1x2 = f1x2. We have established this fact for a few cases in which f was a constant dx or linear function by using geometry formulas for areas of a rectangle and a triangle. When the graph of f is a curve, we can approximate the area underneath the graph using a Riemann sum, which suggests a general method for calculating the area underneath the graph of any nonnegative continuous function f. The following table summarizes some of the area functions we determined by geometry.

• Evaluate a definite integral. • Interpret an area below the horizontal axis. • Solve applied problems involving definite integrals.

Function

Area Function

f1x2 = 2x

A1x2 = x 2

Technology Connection (p. 395)

f 1x2 = 3

A1x2 = 3x

Technology Connection (p. 395)

f 1x2 = 3x 2

A1x2 = x 3

Technology Connection (p. 395)

f 1x2 = k

A1x2 = kx

Section 4.2

f 1x2 = mx

A1x2 = 12 mx 2

Section 4.2

You may have noticed that each time the derivative of the area function, A1x2, is f1x2. Is this always the case? We answer this by letting A1x2 represent the area under a nonnegative continuous function f over the interval 30, x4. To find A¿1x2, we use the definition of the derivative:

y or f(x) y = f (x)

A¿1x2 = lim

A(x)

A1x + h2 - A1x2

h:0

x

Text Reference

x

h

.

Since A1x + h2 is the area under f over the interval 30, x + h4, it follows that A1x + h2 - A1x2 is the area under f between x and x + h. y or f (x)

y or f (x) y = f (x)

y = f (x)

A(x + h)

A(x + h) − A(x)

x x+h

x

x x+h

x

4.3

• Area and Definite Integrals

411

As h approaches zero, the area A1x + h2 - A1x2 approaches the area of a rectangle with width h and height f1x2. That is, A1x + h2 - A1x2 L h # f1x2.

Thus, and

A1x + h2 - A1x2 h lim

L f1x2,

A1x + h2 - A1x2

h:0

Dividing both sides by h

= lim f1x2 = f1x2, h:0

h

which demonstrates that A¿1x2 = f1x2. We have proved the following remarkable result.

THEOREM 4

Let f be a nonnegative continuous function over an interval 30, b4, and let A1x2 be the area between the graph of f and the x-axis over the interval 30, x4, with 0 6 x 6 b. Then A1x2 is a differentiable function of x and A¿1x2 = f1x2.

Theorem 4 resolves the question posed earlier: Yes, the derivative of the area function is always the function under which the area is being calculated. However, Theorem 4 holds only for the interval 30, x4. How can we adapt Theorem 4 to apply when f is defined over any interval 3a, b4? Referring to the graph below, we see that the area over 3a, b4 is the same as the area over 30, b4 minus the area over 30, a4 or A1b2 - A1a2. y

y = f (x) A(b) − A(a)

A(a) a

Area = A1b2 - A1a2 = F1b2 + C - 1F1a2 + C2 = F1b2 - F1a2.

f(x) = 15 x2 + 3

9

x

In Section 4.1, we also found that a function’s antiderivatives can differ only in their constant terms. Thus, if F1x2 is another antiderivative of f1x2, then A1x2 = F1x2 + C, for some constant C, and

f(x) 10

b

8

This result tells us that as long as an area is computed by substituting an interval’s endpoints into an antiderivative and then subtracting, any antiderivative—and any choice of C—can be used. It generally simplifies computations to choose 0 as the value of C.

7 6

■ EXAMPLE 1 Find the area under the graph of f1x2 = 15 x 2 + 3 over the interval

5

32, 54.

4 3

Solution Although making a drawing is not required, doing so helps us visualize the

2

problem. Note that every antiderivative of f1x2 = 15 x 2 + 3 is of the form

1 0

1

2

3

4

5

6

x

F1x2 =

1 3 15 x

+ 3x + C.

412

CHAPTER 4

• Integration

For simplicity, we set C = 0, so

Area over 32, 54 = F152 - F122 8 = 125 15 + 15 - A 15 + 6 B = 16 45.

Although it is possible to express the area under a curve as the limit of a Riemann sum, as we did in Section 4.2, it is usually much easier to work with antiderivatives.

To find the area under the graph of a nonnegative continuous function f over the interval 3a, b4: 1. Find any antiderivative F1x2 of f1x2. Let C = 0 for simplicity. 2. Evaluate F1x2 at x = b and x = a, and compute F1b2 - F1a2. The result is the area under the graph over the interval 3a, b4.

y y = f(x) Area = F(b) – F(a)

a

b

x

■ EXAMPLE 2 Find the area under the graph of y = x 2 + 1 over the interval 3- 1, 24. Solution In this case, f1x2 = x 2 + 1, with a = - 1 and b = 2. 1. Find any antiderivative F1x2 of f1x2. We choose the simplest one: F1x2 =

x3 + x. 3

2. Substitute 2 and - 1, and find the difference F122 - F1- 12: F122 - F1- 12 = c

y 6

8 -1 = + 2 - c - 1d 3 3

Quick Check 1

O

Refer to the function and graph in Example 2. a) Calculate the area over the interval 30, 54. b) Calculate the area over the interval 3- 2, 24. c) Can you suggest a shortcut for part (b)?

1- 123 23 + 2d - c + 1- 12 d 3 3

5 4

8 1 = + 2 + + 1 3 3 = 6. We can make a partial check by counting the squares and parts of squares shaded on the graph to the right.

3 2 1 –3 –2 –1

1

2

3

O Quick Check 1

TECHNOLOGY CONNECTION

Using iPlot to Find the Area under a Graph The iPlot app can be used on your iPhone or iPad to evaluate definite integrals and find the area under a continuous nonnegative function over a closed interval. Let’s find the area

under the graph of the function considered in Example 2, f 1x2 = x 2 + 1, over the interval 3- 1, 24. After opening iPlot, touch the Functions icon at the bottom of the screen. Press + in the upper right. Then enter the function

x

4.3

• Area and Definite Integrals

Using iPlot to Find the Area under a Graph (continued) as X^2+1. Press Done in the upper right and then Plot to obtain the graph (Fig. 1). To find the area under the curve over the interval 3- 1, 24, press Integ (second from the right at the bottom) until it changes color. You may have to press it firmly. Then touch the screen, and you will see a tracing cursor (Fig. 2). Locate the cursor as close as you can to the lower bound of the interval. Then firmly press Apply (in the lower right corner). The screen will glow and then a display like that in Fig. 3 will appear. Touch the number at the top, and change it to - 1. Then touch the upper bound number, and change it to 2; the screen will look like Fig. 4. Next, press OK firmly. The complete curve with the shaded area and the answer, 6, will be displayed (Fig. 5). Caution! This app seems to crash frequently. You may need to start over. Hopefully, an update will improve the app’s stability. iPlot will also evaluate definite integrals over intervals where the function is not nonnegative. For example, 2 2 2 11 1x - 32dx = - 3 L - 0.666667, as seen in Fig. 6. As we shall see later in this section, since there is more area below the x-axis than above, the result is negative.

FIGURE 3

FIGURE 4

FIGURE 5

FIGURE 1

FIGURE 6

EXERCISES Evaluate each definite integral. 2

1.

L-2

14 - x 22 dx

6

3. FIGURE 2

ln x

L1 x 15

5.

L4

2

0

2.

L-1

1x 3 - 3x + 12 dx

2

dx

4.

4

L-8 11 + e x22

10.002x 4 - 0.3x 2 + 4x - 72 dx

dx

413

414

CHAPTER 4 y

• Integration

y = x3

Number of kilowatts generated per day

125

■ EXAMPLE 3 Let y = x 3 represent the number of kilowatts (kW) generated by a new power plant each day, x days after going on line. Find the area under the graph of y = x 3 over the interval 30, 54 and interpret the significance of the area.

Solution In this case, f1x2 = x 3, a = 0, and b = 5. 1. Find any antiderivative F1x2 of f1x2. We choose the simplest one: F1x2 = 0

x 5 Number of days since power plant went on line

x4 . 4

2. Substitute 5 and 0, and find the difference F152 - F102: F152 - F102 =

54 04 625 = = 156 14 . 4 4 4

The area represents the total number of kilowatts generated during the first 5 days. Note that kW/day # days = kW. The difference F1b2 - F1a2 has the same value for all antiderivatives F of a function f whether the function is nonnegative or not. It is called the definite integral of f from a to b.

DEFINITION

Let f be any continuous function over the interval 3a, b4 and F be any antiderivative of f. Then the definite integral of f from a to b is b

La

f1x2 dx = F1b2 - F1a2.

Evaluating definite integrals is called integration. The numbers a and b are known as the limits of integration. Note that this use of the word limit indicates an endpoint of an interval, not a value that is being approached, as you learned in Chapter 1.

■ EXAMPLE 4 Evaluate:

b

x 2 dx.

La

Solution Using the antiderivative F1x2 = x 3>3, we have b

La

x 2 dx =

b3 a3 - . 3 3

It is convenient to use an intermediate notation: b

La

f1x2 dx = 3F1x24ba = F1b2 - F1a2,

where F1x2 is an antiderivative of f1x2.

■ EXAMPLE 5 Evaluate each of the following: 4

a)

L-1

1x 2 - x2 dx;

3

b)

L0

e x dx;

e

c)

L1

a1 + 2x -

1 b dx (assume x 7 0). x

4.3

Solution 4

a)

1x 2 - x2 dx = c

L- 1

415

1- 123 1- 122 x3 x2 4 43 42 d = a b - a b 3 2 -1 3 2 3 2

= ¢ =

• Area and Definite Integrals

64 16 -1 1 - ≤ ≤ - ¢ 3 2 3 2

64 1 1 - 8 + + = 14 16 3 3 2

3

b) Quick Check 2 4

a) b) c)

12x - 3x2 dx; 3

L2 L0

e

c)

L1

a 1 + 2x -

ln 4

2e x dx; 5

x - 1 dx. x L1

1 b dx = x = = = =

3x + x 2 - ln x4e1

We assume x 7 0.

1e + e - ln e2 - 11 + 12 - ln 12 1e + e 2 - 12 - 11 + 1 - 02 e + e2 - 1 - 1 - 1 e + e2 - 3 2

O Quick Check 2 The fact that we can express the integral of a function either as a limit of a sum or in terms of an antiderivative is so important that it has a name: the Fundamental Theorem of Integral Calculus.

The Fundamental Theorem of Integral Calculus

If a continuous function f has an antiderivative F over 3a, b4, then b

n

lim f1x i2¢x = n: q a i=1

La

f1x2 dx = F1b2 - F1a2.

It is helpful to envision taking the limit as stretching the summation sign, © , into something resembling an S (the integral sign) and redefining ¢x as dx. Because ¢x is used in the limit, dx appears in the integral notation.

More on Area When we evaluate the definite integral of a nonnegative function, we get the area under the graph over an interval.

■ EXAMPLE 6 Suppose that y is the profit per mile traveled and x is the number of

miles traveled, in thousands. Find the area under y = 1>x over the interval 31, 44 and interpret the significance of this area.

Solution 4

dx = 3ln x441 = ln 4 - ln 1 x L1 = ln 4 - 0 L 1.3863

Considering the units, 1dollars>mile2 # miles = dollars, we see that the area represents a total profit of $1386.30 when the miles traveled increase from 1000 to 4000 miles.

y

Profit per mile (in dollars)

Evaluate each of the following:

e x dx = 3e x430 = e 3 - e 0 = e 3 - 1

L0

y = –1– x 1

1 4 Number of miles traveled (in thousands)

x

O

416

CHAPTER 4

• Integration

■ EXAMPLE 7 Find the area under y = 1>x 2 over the interval 31, b4. Solution b

y

b

dx = x - 2 dx 2 x L1 L1 x -2+1 b = c d -2 + 1 1 = c

y = ––1–2 x 1

-1 b

b

x 1 d = c- d x 1 -1 1

1

1 1 = a- b - a- b b 1 1 = 1 b

b

x

Now let’s compare the definite integrals of the functions y = x 2 and y = - x 2: 2

x 2 dx = c

L0

=

3 2

0 −1

L0 3

2 0 8 = 3 3 3

c

1 −1

3

- x 2 dx = c = -

x3 2 d 3 0

23 03 8 + = 3 3 3

y y The graphs of the functions y = x2 0 1 2 2 2 y = x and y = - x are reflections 4 x of each other across the x-axis. A –1 3 Thus, the shaded areas are the same, 8 3 . The evaluation procedure for –2 2 y = - x 2 gave us - 83. This illustrates that for negative-valued functions, –3 1 A the definite integral gives us the op–4 posite of the area between the curve x y = – x2 0 1 2 and the x-axis. Now let’s consider f1x2 = x 2 - 1 over the interval 30, 24. It has both positive and negative values. We apply the preceding evaluation procedure, even though the function values are not all nonnegative. We do so in two ways. First, let’s use the fact that for any a, b, c, if a 6 b 6 c, then

y f (x) = x2 − 1

2

x3 2 d 3 0

1

2

x

La

b

f1x2 dx =

La

c

f1x2 dx +

Lb

The area from a to b plus the area from b to c is the area from a to c.

f1x2 dx.

(We will consider this property of integrals again in Section 4.4.) Note that 1 is the x-intercept in 30, 24. 2

L0

1

1x 2 - 12 dx =

1x 2 - 12 dx +

2

1x 2 - 12 dx

L0 L1 1 2 x3 x3 = c - xd + c - xd 3 3 0 1 = ca

13 03 23 13 - 1b - a - 0b d + c a - 2b - a - 1b d 3 3 3 3

= -

c

1 - 1d 3

⎫⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎬ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎢ ⎭

⎫⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎬ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎢ ⎭

=

+

c

8 1 - 2 + 1d 3 3

2 4 2 + = . 3 3 3

This shows that the area above the x-axis exceeds the area below the x-axis by 23 unit.

4.3

• Area and Definite Integrals

417

Now let’s evaluate the original integral in another, more direct, way: 2

L0

1x 2 - 12 dx = c

2 x3 - xd 3 0

= a

23 03 - 2b - a - 0b 3 3

= a

8 2 - 2b - 0 = . 3 3

The definite integral of a continuous function over an interval is the sum of the areas above the x-axis minus the sum of the areas below the x-axis.

■ EXAMPLE 8 Consider 1-12 1- x 3 + 3x - 12 dx. Predict the sign of the result y 3

y = −x3 + 3x − 1

2

by examining the graph, and then evaluate the integral. Solution From the graph, it appears that there is considerably more area below the x-axis than above. Thus, we expect that 2

1- x 3 + 3x - 12 dx 6 0. L- 1

1 −2

−1

1

2

x

Evaluating the integral, we have

−1 −2

2

−3

L- 1

1- x 3 + 3x - 12 dx = c = a-

Quick Check 3

O

Let f1x2 = x - x . a) Predict the sign of the 2 value of 10 f1x2 dx by examining the graph. b) Evaluate this integral. 4

2 x4 3 + x2 - x d 4 2 -1

1- 12 24 3 3 + # 2 2 - 2b - a + 1- 122 - 1- 12b 4 2 4 2 4

= 1- 4 + 6 - 22 - a -

2

1 3 + + 1b = 0 - 2 14 4 2

= - 2 14. As a partial check, we note that the result is negative, as predicted.

O Quick Check 3

TECHNOLOGY CONNECTION

Approximating Definite Integrals

2 –1

( –X 3 + 3X − 1 ) dX –2.25

There are two methods for evaluating definite integrals with a calculator. Let’s consider the function from Example 8: f 1x2 = - x 3 + 3x - 1.

Method 1: fnInt First, we select fnInt from the MATH menu. Next, we enter the function, the variable, and the endpoints of the interval over which we are integrating. The calculator returns the same value for the definite integral as we found in Example 8.

Method 2: f (x) dx We first graph y1 = - x 3 + 3x - 1. Then we select f(x)dx from the CALC menu and enter the lower and upper limits of integration. The calculator shades the area and returns (continued)

418

CHAPTER 4

• Integration

EXERCISES

the same value for the definite integral as found in Example 8. 6

Evaluate each definite integral. 2

1x 2 - 12 dx L- 1

6

Y1 = –X^3 + 3X – 1

1.

Y1 = –X^3 + 3X – 1

3

–3

3

–3

Lower Limit? X = –1

3

2.

1x 3 - 3x + 12 dx L- 2

Upper Limit? X=2 –6

6

3.

–6

L1 x 2

6

2

4. –3

ln x

dx 4

L- 8 11 + e x22

dx

10

10.002x 4 - 0.3x 2 + 4x - 72 dx L- 10

3

5. f(x)dx = –2.25 –6

Applications Involving Definite Integrals Determining Total Profit

■ EXAMPLE 9 Business: Total Profit from Marginal Profit. Northeast Airlines determines that the marginal profit resulting from the sale of x seats on a jet traveling from Atlanta to Kansas City, in hundreds of dollars, is given by P¿1x2 = 1x - 6.

60

P1602 =

L0

=

Quick Check 4 Business. Referring to Example 9, find the total profit of Northeast Airlines when 140 seats are sold.

P¿1x2 dx 60

A 2x - 6 B dx

L0 60 2 = c x 3>2 - 6x d 3 0 L - 50.1613.

y Marginal profit (in hundreds of dollars per seat)

Find the total profit when 60 seats are sold. Solution We integrate to find P1602:

6 4

P′(x) = x − 6

2 0 −2 −4 −6

10

20

30

40

50

60

70

x

Number of seats sold P(60) = ∫

60

0

( x − 6) dx

Using a calculator

When 60 seats are sold, Northeast’s profit is - $5016.13. That is, the airline will lose $5016.13 on the flight.

O Quick Check 4 Finding Velocity and Distance from Acceleration Recall that the position coordinate at time t of an object moving along a number line is s1t2. Then s¿1t2 = v1t2 = the velocity at time t, s–1t2 = v¿1t2 = a1t2 = the acceleration at time t.

O

4.3

419

• Area and Definite Integrals

■ EXAMPLE 10 Physical Science: Distance. Suppose that v1t2 = 5t 4 and s102 = 9. Find s1t2. Assume that s1t2 is in feet and v1t2 is in feet per second.

Solution We first find s1t2 by integrating: s1t2 =

L

v1t2 dt =

L

5t 4 dt = t 5 + C.

Next we determine C by using the initial condition s102 = 9, which is the starting position for s at time t = 0: s102 = 05 + C = 9 C = 9. Thus, s1t2 = t 5 + 9.

■ EXAMPLE 11 Physical Science: Distance. Suppose that a1t2 = 12t 2 - 6, with v102 = the initial velocity = 5, and s102 = the initial position = 10. Find s1t2, and graph a1t2, v1t2, and s1t2.

Solution 1. We first find v1t2 by integrating a1t2: v1t2 =

L

a1t2 dt 112t 2 - 62 dt

=

L = 4t 3 - 6t + C1.

The condition v102 = 5 allows us to find C1: v102 = 4 # 03 - 6 # 0 + C1 = 5 C1 = 5.

Thus, v1t2 = 4t 3 - 6t + 5. 2. Next we find s1t2 by integrating v1t2:

a

a(t) = 12t 2 – 6

30 20 10 –6 –4 –2 –10

2

4

6

t

–20 –30

v(t) = 4t 3 – 6t + 5 v 30 20

s1t2 = =

L

v1t2 dt 14t 3 - 6t + 52 dt

L = t 4 - 3t 2 + 5t + C2.

10 –6 –4 –2 –10

2

4

6

t

–20 –30

The condition s102 = 10 allows us to find C2: s102 = 04 - 3 # 02 + 5 # 0 + C2 = 10 C2 = 10.

Thus, s1t2 = t 4 - 3t 2 + 5t + 10.

s(t) = t 4 – 3t 2 + 5t + 10 s 30 20 10 –6 –4 –2 –10 –20 –30

2

4

6

t

420

• Integration

CHAPTER 4

■ EXAMPLE 12 Physical Science: Total Distance Traveled. A particle starts out from the origin. Its velocity, in miles per hour, is given by v1t2 = 2t + t, where t is the number of hours since the particle left the origin. How far does the particle travel during the second, third, and fourth hours (from t = 1 to t = 4)? Solution Recall that velocity, or speed, is the rate of change of distance with respect to time. In other words, velocity is the derivative of the distance function, and the distance function is an antiderivative of the velocity function. To find the total distance traveled from t = 1 to t = 4, we evaluate the integral

v(t) Velocity (in miles per hour)

8 7 6

v(t) = t + t

5 4

A 2t + t B dt.

4

A 2t + t B dt =

L1

3

We have

2 1 0

4

1

2

3

4

5

6

L1

t

Time (in hours)

=

4

L1

1t 1>2 + t2 dt

C 23 t 3>2 + 12 t 2 D 41

= 23 # 4 3>2 + 12 # 4 2 16 2 1 = 16 3 + 2 - 3 - 2 15 = 14 3 + 2 73 = 6 = 12 16 mi.

Quick Check 5

O

Use the grid in the graph in Example 12 to support the claim that the particle travels between 6 and 7 mi during the fifth hour (from t = 4 to t = 5). Then calculate the actual distance traveled during that hour.

A 23 # 13>2 +

1 2

# 12 B

As a check, we can count shaded squares and parts of squares on the graph at the left. As another partial check, we can observe that the units used are 1mi>hr2 # hr = mi.

O Quick Check 5

Velocity (in feet per second)

y 60

■ EXA MPLE 13 Physical Science: Braking Distance. The driver of a vehicle

traveling at 40 mi> hr (58.67 ft> sec) applies the brakes, softly at first, then harder, coming to a complete stop after 7 sec. The velocity as a function of time is modeled by the function v1t2 = - 1.197t 2 + 58.67, where v is in feet per second, t is in seconds, and 0 … t … 7. How far did the vehicle travel while the driver was braking?

v(t) = −1.197t 2 + 58.67

40

20

0

7

t

Solution The distance traveled is given by the definite integral of v1t2: 7

Time (in seconds)

Quick Check 6 Suppose that the driver in Example 13 braked to a stop in 7 sec but did so “linearly,” that is, slowed from 40 mi> hr to a stop at a constant rate of deceleration. Find the braking distance in feet.

L0

1- 1.197t 2 + 58.672 dt = c -

7 1.197 3 t + 58.67t d 3 0

1.197 1723 + 58.67172 - 0 3 = 273.83 ft. = -

This is nearly the length of a football field! In the graph of v, the shaded area represents the distance the vehicle traveled during the 7 sec.

O Quick Check 6

O

421

Exercise Set 4.3

Section Summary • The exact area between the x-axis and the graph of the nonnegative continuous function y = f1x2 over the interval 3a, b4 is found by evaluating the definite integral b

f1x2 dx = F1b2 - F1a2,

La

where F is an antiderivative of f. • If a function has areas both below and above the x-axis, the definite integral gives the net total area, or the

difference between the sum of the areas above the x-axis and the sum of the areas below the x-axis. If there is more area above the x-axis than below, the definite integral will be positive. If there is more area below the x-axis than above, the definite integral will be negative. If the areas above and below the x-axis are the same, the definite integral will be 0.

EXERCISE SET

4.3 7. y = x 3; 30, 14

Find the area under the given curve over the indicated interval.

2. y = 5; 31, 34 6 5 4 3 2 1

0

y=4

1 2 3 4

x

0

3. y = 2x; 31, 34

y=5

4. y = x 2; 30, 34

9. y = 4 - x 2; 3- 2, 24 11. y = e x; 30, 34

9 y = x2 8 7 6 5 4 3 2 1

x

13. y =

5. y = x ; 30, 54

6. y = x ; 30, 24

y

y

−2 −1 0 1 2 3 4 5 6

1

2

x

−1

30 25 20 15 10 5 −2 −1

12. y =

2 ; 31, 44 x

3 ; 31, 64 x

x

15.

16.

Time (in days)

y = x3

17. 1 2 3

10. y = e x; 30, 24

In each of Exercises 15–24, explain what the shaded area represents.

3

x

−1

14. y = x 2 - 4x; 3- 4, - 24

−3 −2 −1 0 1 2 3 4 2

y = x2

−2

x

1

x

y

30 25 20 15 10 5

0.5

y = 1 − x2

−2

1 2 3 4

8 7 y = 2x 6 5 4 3 2 1 1 2 3 4

1

0.5 −0.5

y

0

2

y = x3

x

t

Velocity (in miles per hour)

6 5 4 3 2 1

1

18.

Time t (in hours)

Number of marriages per year

y

Cost per day (in dollars)

y

y

y

Number of kilowatts used per hour

1. y = 4; 31, 34

8. y = 1 - x 2; 3- 1, 14

Time (in hours)

Time (in years)

t

t

23.

Marginal cost (per unit)

Marginal revenue (in dollars per unit)

b)

a

1.5

t

25. y = x 3; 30, 24

L0

1x 2 - x2 dx

L0

y 2

y = x2 – x

– 0.5

0.5

1.0

1.5 x 2

1

t Time (in hours)

1

37.

b

1x 3 - 3x2 dx

L-1

38.

- 2e 3x dx

L0 y

2

b 0

29. y = 5 - x 2; 3 - 1, 24

30. y = e x; 3- 2, 34

31. y = e x; 3- 1, 54

32. y = 2x +

1 x

2

; 31, 44

In Exercises 33 and 34, determine visually whether 1ab f1x2 dx is positive, negative, or zero, and express 1ab f1x2 dx in terms of the area A. Explain your result. f A

x

–1

– 0.5

y

28. y = 2 - x - x 2; 3- 2, 14

x

1

1 –2

–1

–2 2

x

−1 −2

y = – 2e3x

y = x 3 – 3x

39–42. Check the results of each of Exercises 35–38 using a graphing calculator. Evaluate.

a

A

b

3

x

43.

L1

45.

f

2

13t 2 + 72 dt

44.

A 2x - 1 B dx

46.

12x 2 - 3x + 72 dx

48.

4

L1

47. b x

a A

8

L1

y

51. f A

La

A

b 1 2 2x

dx

b

b x

a

L-2

53.

La

A

e 2t dt b

54.

La

1- x 2 + 4x - 52 dx

3

e t dt

L-5

3 x - 2 B dx A2

3

2

A

49.

A

L-2

14t 3 - 12 dt

L1

5

A

34. a)

36.

y 0.5

26. y = x 4; 30, 14

27. y = x 2 + x + 1; 32, 34

b)

2

1x - x 22 dx

1

Find the area under the graph of each function over the given interval.

y

b x

A

y = x – x2

t Time (in minutes)

y

A f

35. Time (in days)

v Volume of blood (in cubic centimeters)

A

Evaluate. Then interpret the result in terms of the area above and/or below the x-axis.

Sales on tth day

22.

24.

33. a)

y

x Number of units produced

x Number of units produced

Concentration of a drug (in milligrams per cubic centimeter)

21.

• Integration

20.

Rate of memory (in words per minute)

19.

CHAPTER 4

Number of orders per hour

422

- e t dt

50. 52.

L-2 La

e -t dt

b 1 3 5x

dx

Exercise Set 4.3

L1

ax +

1 b dx x

e

56.

L1

ax -

1 b dx x

2

57.

L0

22x dx

(Hint: Simplify first.)

27

58.

L0

23x dx

Credit market debt. The annual rate of change in the national credit market debt (in billions of dollars per year) can be modeled by the function D¿1t2  857.98  829.66t  197.34t 2  15.36t 3, where t is the number of years since 1995. (Source: Federal Reserve System.) Use the preceding information for Exercises 65 and 66. 65. By how much did the credit market debt increase between 1996 and 2000?

APPLICATIONS Business and Economics 59. Business: total profit. Pure Water Enterprises finds that the marginal profit, in dollars, from drilling a well that is x feet deep is given by 5

P¿1x2 = 2x. Find the profit when a well 250 ft deep is drilled. 60. Business: total revenue. Sally’s Sweets finds that the marginal revenue, in dollars, from the sale of x pounds of maple-coated pecans is given by R¿1x2 = 6x -1>6. Find the revenue when 300 lb of maple-coated pecans are produced. 61. Business: increasing total cost. Kitchens-to-Please Contracting determines that the marginal cost, in dollars per foot, of installing x feet of kitchen countertop is given by C¿1x2 = 8x -1>3. Find the cost of installing an extra 14 ft of countertop after 50 ft have already been ordered. 62. Business: increasing total profit. Laso Industries finds that the marginal profit, in dollars, from the sale of x digital control boards is given by P¿1x2 = 2.6x 0.1. A customer orders 1200 digital control boards and later increases the order to 1500. Find the extra profit resulting from the increase in order size. 63. Accumulated sales. A company estimates that its sales will grow continuously at a rate given by the function S¿1t2 = 20e t, where S¿1t2 is the rate at which sales are increasing, in dollars per day, on day t. a) Find the accumulated sales for the first 5 days. b) Find the sales from the 2nd day through the 5th day. (This is the integral from 1 to 5.) 64. Accumulated sales. Raggs, Ltd., estimates that its sales will grow continuously at a rate given by the function S¿1t2 = 10e , t

where S¿1t2 is the rate at which sales are increasing, in dollars per day, on day t. a) Find the accumulated sales for the first 5 days. b) Find the sales from the 2nd day through the 5th day. (This is the integral from 1 to 5.)

66. By how much did the credit market debt increase between 1999 and 2005? Industrial learning curve. A company is producing a new product. Due to the nature of the product, the time required to produce each unit decreases as workers become more familiar with the production procedure. It is determined that the function for the learning process is 1 T1x2  2  0.3a b, x where T(x) is the time, in hours, required to produce the xth unit. Use this information for Exercises 67 and 68. 67. Find the total time required for a new worker to produce units 1 through 10; units 20 through 30. 68. Find the total time required for a new worker to produce units 1 through 20; units 20 through 40.

Social Sciences Memorizing. The rate of memorizing information initially increases. Eventually, however, a maximum rate is reached, after which it begins to decrease. Memory rate (in words per minute)

e

55.

423

0

10 t Time (in minutes)

69. Suppose that in a memory experiment the rate of memorizing is given by M¿1t2 = - 0.009t 2 + 0.2t, where M¿1t2 is the memory rate, in words per minute. How many words are memorized in the first 10 min (from t = 0 to t = 10)? 70. Suppose that in another memory experiment the rate of memorizing is given by M¿1t2 = - 0.003t 2 + 0.2t, where M¿1t2 is the memory rate, in words per minute. How many words are memorized in the first 10 min (from t = 0 to t = 10)? 71. See Exercise 69. How many words are memorized during minutes 10–15? 72. See Exercise 70. How many words are memorized during minutes 10–17?

424

CHAPTER 4

• Integration

Life and Physical Sciences Find s(t). 73. v1t2 = 3t 2, s102 = 4 74. v1t2 = 2t, s102 = 10 Find v(t). 75. a1t2 = 4t, v102 = 20 76. a1t2 = 6t, v102 = 30 Find s(t). 77. a1t2 = - 2t + 6, with v102 = 6 and s102 = 10 78. a1t2 = - 6t + 7, with v102 = 10 and s102 = 20 79. Physics. A particle is released as part of an experiment. Its speed t seconds after release is given by v1t2 = - 0.5t 2 + 10t, where v1t2 is in meters per second. a) How far does the particle travel during the first 5 sec? b) How far does it travel during the second 5 sec? 80. Physics. A particle is released during an experiment. Its speed t minutes after release is given by v1t2 = - 0.3t 2 + 9t, where v1t2 is in kilometers per minute. a) How far does the particle travel during the first 10 min? b) How far does it travel during the second 10 min? 81. Distance and speed. A motorcycle accelerates at a constant rate from 0 mph 1v102 = 02 to 60 mph in 15 sec. a) How fast is it traveling after 15 sec? b) How far has it traveled after 15 sec? (Hint: Convert seconds to hours.) 82. Distance and speed. A car accelerates at a constant rate from 0 mph to 60 mph in 30 sec. a) How fast is it traveling after 30 sec? b) How far has it traveled after 30 sec?

87. Distance. A car accelerates at a constant rate from 0 to 60 mph in min. How far does the car travel during that time? 88. Distance. A motorcycle accelerates at a constant rate from 0 to 50 mph in 15 sec. How far does it travel during that time? 89. Physics. A particle starts out from the origin. Its velocity, in miles per hour, after t hours is given by v1t2 = 3t 2 + 2t. How far does it travel from the 2nd hour through the 5th hour (from t = 1 to t = 5)? 90. Physics. A particle starts out from the origin. Its velocity, in miles per hour, after t hours is given by v1t2 = 4t 3 + 2t. How far does it travel from the start through the 3rd hour (from t = 0 to t = 3)?

SYNTHESIS 91. Accumulated sales. Bluetape, Inc. estimates that its sales will grow continuously at a rate given by S¿1t2 = 0.5e t, where S¿1t2 is the rate at which sales are increasing, in dollars per day, on day t. On what day will accumulated sales first exceed $10,000? 92. Total pollution. A factory is polluting a lake in such a way that the rate of pollutants entering the lake at time t, in months, is given by N¿1t2 = 280t 3>2, where N is the total number of pounds of pollutants in the lake at time t.

83. Distance and speed. A bicyclist decelerates at a constant rate from 30 km/hr to a standstill in 45 sec. a) How fast is the bicyclist traveling after 20 sec? b) How far has the bicyclist traveled after 45 sec? 84. Distance and speed. A cheetah decelerates at a constant rate from 50 km> hr to a complete stop in 20 sec. a) How fast is the cheetah moving after 10 sec? b) How far has the cheetah traveled after 20 sec? 85. Distance. For a freely falling object, a1t2 = - 32 ft>sec2, v102 = initial velocity = v0 1in ft>sec2, and s102 = initial height = s0 1in ft2. Find a general expression for s(t) in terms of v0 and s0. 86. Time. A ball is thrown upward from a height of 10 ft, that is, s102 = 10, at an initial velocity of 80 ft> sec, or v102 = 80. How long will it take before the ball hits the ground? (See Exercise 85.)

a) How many pounds of pollutants enter the lake in 16 months? b) An environmental board tells the factory that it must begin cleanup procedures after 50,000 lb of pollutants have entered the lake. After what length of time will this occur?

TECHNOLOGY CONNECTION

Evaluate. x - x -1

3 2

5 5

x - 1 93. dx L2 x - 1

94.

16

95.

1x - 12 2x dx

L4

8 3

97.

2x - 1

L1

3

99.

L2

L0

dx

L0

1 3

107.

1x + 223 dx

t - t t3

101.

L1 9

103.

dt

L-1.2

1x 3 - 9x 2 + 27x + 502 dx

1.4

1x 4 + 4x 3 - 36x 2 - 160x + 3002 dx

L-8 2

24 - x 2 dx

L-2 1

3

1t + 123 dt

L1

106.

A t + 23 B A t - 23 B dt

3 5

102.

6.3

1

108.

1

100.

Evaluate.

dx

x + 8 98. dx L0 x + 2

2

2x 5

96.

x2

L1

L4

ax t + 1 2t

109.

2

1 b dx x

L-1

13 + 21 - x 22 dx

8

110. dt

L0

x - 4

2

105.

L1

1ln x - e x2 dx = c

L2 x - 3 2

• Solve applied problems involving definite integrals.

L-10 x + 4

dx

a

Lb

f 1x2 dx.

1 28 1. (a) 46 23, or 140 3 ; (b) 9 3 , or 3 ; (c) integrate from 0 to 2, then double the result 2. (a) 102; (b) 6; (c) 4 - ln 5 L 2.39 3. (a) Positive; (b) 56 15 4. Approximately $26,433.49 5. Approximately 6.62 mi 6. Approximately 205.35 ft

1 - e 2 b - 11 - e 12 2 1 = e - e2 2

• Use properties of definite integrals to find the area between curves.

La

f 1x2dx = -

8 2

Answers to Quick Checks

= a

OBJECTIVES

L-2

10

113.

b

114. Prove that

2 1 - ex d x 1

4.4

111.

dx

2

2 1 1 1x 2 + x + 12 dx = c x 3 + x 2 + x d 3 2 L1 1 1# 3 1# 2 = a 2 + 2 + 2b 3 2 20 = 3

x 2>3 A 52 - x B dx

2

x1x - 524 dx 4 2

112.

Find the error in each of the following. Explain. 104.

425

• Properties of Definite Integrals

4.4

Properties of Definite Integrals The Additive Property of Definite Integrals We have seen that the definite integral c

f1x2 dx

La

can be regarded as the area under the graph of y = f1x2 Ú 0 over the interval 3a, c4. Thus, if b is such that a 6 b 6 c, the above integral can be expressed as a sum. This additive property of definite integrals is stated in the following theorem.

• Determine the average value of a function.

THEOREM 5 For a 6 b 6 c, La

b

f1x2 dx =

f

y

c

La

c

f1x2 dx +

Lb

f1x2 dx.

For any number b between a and c, the integral from a to c is the integral from a to b plus the integral from b to c.

a

b

c

x

426

• Integration

CHAPTER 4

Theorem 5 is especially useful when a function is defined piecewise, in different ways over different subintervals.

■ EXAMPLE 1 Find the area under the graph of y = f1x2 from - 4 to 5, where f1x2 = e

9, for x 6 3, x 2, for x Ú 3.

Solution 5

L-4

3

f1x2 dx =

L-4 3

= =

Quick Check 1

L-4

y

5

f1x2 dx +

L3

f1x2 dx f

5

9 dx +

93x43-4

L3 x3 5 + c d 3 3

x 2 dx 9

Find the area under the graph of y = g1x2 from - 3 to 6, where

= 913 - 1- 422 + a

x 2, for x … 2, g1x2 = e 8 - x, for x 7 2.

= 95 23

−4 −3 −2 −1

53 33 b 3 3

1

2

3

4

5

O Quick Check 1

O

■ EXAMPLE 2 Evaluate each definite integral: 4

a)

L-3

3

ƒ x ƒ dx;

b)

L0

ƒ 1 - x 2 ƒ dx.

Solution a) The absolute-value function f1x2 = ƒ x ƒ is defined piecewise as follows: f1x2 = ƒ x ƒ = e y 4

f(x) = | x |

L-3

2

0

ƒ x ƒ dx =

−2

−1

1

2

3

4

4

1- x2 dx +

x dx L-3 L0 x2 0 x2 4 = c- d + c d 2 -3 2 0 2 1322 0 42 02 = a- abb + a b 2 2 2 2 9 16 25 = + = . 2 2 2

1 −3

See Example 8 in Section R.5.

Therefore,

4 3

- x, for x 6 0, x, for x Ú 0.

x

As a check, this definite integral can also be evaluated using geometry, since the two regions are triangles. b) The graph of f1x2 = ƒ 1 - x 2 ƒ is the graph of y = 1 - x 2, where any portion of the graph below the x-axis (that is, where y 6 0) is reflected above the x-axis.

−3

−2

y

y

2

8

−1

1 −2

2

3

6

x

4

y = 1 – x2

−4 −6

y = | 1 – x 2|

2 −3

−2

−1

1

2

3

x

x

• Properties of Definite Integrals

4.4

427

We see that y is negative when x 6 - 1 or x 7 1. Therefore, the function is defined piecewise: f1x2 = ƒ 1 - x 2 ƒ = e

The definite integral of f1x2 = ƒ 1 - x 2 ƒ over the interval 30, 34 is treated as the sum of two definite integrals, those of y = 1 - x 2 over the interval 30, 14 and of y = x 2 - 1 over the interval 31, 34:

y 8 6

y = |1 – x 2|

3

4

L0

2 −3

−2

−1

1

2

x 2 - 1, for x 6 - 1 or x 7 1, See Exercises 101–108 in Section 2.1. 1 - x 2, for - 1 … x … 1.

1

ƒ 1 - x 2 ƒ dx =

L0

= cx -

x

3

11 - x 22 dx +

= a1 = Quick Check 2 7 Evaluate 10 ƒ 2x - 1 ƒ dx.

3

L1

1x 2 - 12 dx

3 1 3 1 1 x d + c x3 - x d 3 3 0 1

1# 3 1 1 1 1 b - a0 - # 03 b + a # 33 - 3b - a # 13 - 1b 3 3 3 3

22 . 3

O Quick Check 2 The Area of a Region Bounded by Two Graphs

y y = f(x) y = g(x)

Suppose that we want to find the area of a region bounded by the graphs of two function, y = f1x2 and y = g1x2, as shown at the left. Note that the area of the desired region A is the area of A2 minus that of A1. g

A

a

b

f

g

f

A

x

A2 A1 a

b

a

b

Thus, b

A =

b

f1x2 dx A2

La

g1x2 dx,

⎫ ⎢ ⎬ ⎢ ⎭

⎫ ⎢ ⎬ ⎢ ⎭

La

-

A1

b

or

A =

La

3f1x2 - g1x24 dx.

In general, we have the following theorem.

a

b

O

428

CHAPTER 4

• Integration

THEOREM 6 y

b

La

b ⎣

⎣

Let f and g be continuous functions and suppose that f1x2 Ú g1x2 over the interval 3a, b4. Then the area of the region between the two curves, from x = a to x = b, is

f(x) – g(x)⎣ dx

a ⎣

f(x)

3f1x2 - g1x24 dx.

g(x)

a

b

x

■ EXAMPLE 3 Find the area of the region bounded by the graphs of f1x2 = 2x + 1 and g1x2 = x 2 + 1.

Solution First, we make a reasonably accurate sketch, as in the figure at the right, to determine which is the upper graph. To calculate the points of intersection, we set f1x2 equal to g1x2 and solve. f1x2 2x + 1 0 0 x = 0

= g1x2 = x2 + 1 = x 2 - 2x = x1x - 22 or x = 2

L0

1

2

L0

0

12x - x 22 dx x3 2 d 3 0

= a2 2 -

23 03 b - a02 b 3 3

4 . 3

8 3

g(x) = x 2 + 1

(0, 1)

= cx 2 -

=

O Quick Check 3

(2, 5)

4

A

= 4 Find the area of the region bounded by the graphs of y = 1x and y = 13 x.

f(x) = 2x + 1

2

312x + 12 - 1x 2 + 124 dx =

Quick Check 3

5

3

The graphs intersect at x = 0 and x = 2. We see that, over the interval 30, 24, f is the upper graph. We now compute the area as follows: 2

y

1

2

3

x

O

429

• Properties of Definite Integrals

4.4

TECHNOLOGY CONNECTION To find the area bounded by two graphs, such as y1 = - 2x - 7 and y2 = - x 2 - 4, we graph each function and use INTERSECT from the CALC menu to determine the points of intersection. 6

Next, we use fnInt from the MATH menu. 3 –1

( –X 2 − 4 ) − ( –2X − 7 ) dX 10.66666667

6

–4

4

–4

4

The area bounded by the two curves is about 10.7. Intersection X = –1

Y = –5 –16

Intersection X=3

Y = –13

EXERCISES

–16

1. Find the area of the region in Example 3 using this

approach. 2. Use iPlot to find the same area.

■ EXAMPLE 4 Find the area of the region bounded by y = x 4 - 3x 3 - 4x 2 + 10,

y = 40 - x 2,

x = 1,

and

x = 3.

Solution

First, we make a reasonably accurate sketch, as in the figure below, to ensure that we have the correct configuration. Note that over 31, 34, the upper graph is y = 40 - x 2. Thus, 40 - x 2 Ú x 4 - 3x 3 - 4x 2 + 10 over 31, 34. y 50 40 30

y = 40 – x2

20 10 ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺10

1

2

3 4

5

6

7

8

x

⫺20 ⫺30

y = x4 – 3x 3 – 4x 2 + 10

By subtracting a negative number (the area between the red curve and the x-axis), we are adding areas. The limits of integration are stated, so we can compute the area as follows: 3

L1

3140 - x 22 - 1x 4 - 3x 3 - 4x 2 + 1024 dx 3

=

L1

1- x 4 + 3x 3 + 3x 2 + 302 dx

3 x5 3 + x 4 + x 3 + 30x d 5 4 1 35 3# 4 15 3 = a+ 3 + 33 + 30 # 3b - a + # 14 + 13 + 30 # 1b 5 4 5 4

= c-

= 97.6.

430

CHAPTER 4

• Integration

An Environmental Application

■ EXAMPLE 5 Life Science: Emission Control. A clever college student develops an engine that is believed to meet all state standards for emission control. The new engine’s rate of emission is given by

where E1t2 is the emissions, in billions of pollution particulates per year, at time t, in years. The emission rate of a conventional engine is given by C1t2 = 9 + t . 2

The graphs of both curves are shown at the right. a) At what point in time will the emission rates be the same? b) What reduction in emissions results from using the student’s engine?

Rate of emissions

E1t2 = 2t 2,

18 C(t) 9 E(t)

0

Solution

1

2 3 Years of use

t

a) The rate of emission will be the same when E1t2 = C1t2, or

Quick Check 4 Two rockets are fired upward. The first rocket’s velocity is given by the function v11t2 = 4t; the second rocket’s velocity is given by the 1 2 t . In both function v21t2 = 10 cases, t is in seconds and velocity is in feet per second. a) When the two rockets’ velocities are the same, how far ahead (in feet) is the first rocket? b) After how many seconds will the second rocket catch up to the first, and how far away will they be from the starting point?

2t 2 = 9 + t 2 t2 - 9 = 0 1t - 321t + 32 = 0 t = 3 or t = - 3. Since negative time has no meaning in this problem, the emission rates will be the same when t = 3 yr. b) The reduction in emissions is represented by the area of the shaded region in the figure above. It is the area between C1t2 = 9 + t 2 and E1t2 = 2t 2, from t = 0 to t = 3, and is computed as follows: 3

319 + t 2 - 2t 4 dt = 2

L0

3

2

L0

19 - t 22 dt

= c9t -

t3 3 d 3 0

= a9 # 3 -

33 03 b - a9 # 0 b 3 3

= 27 - 9 = 18 billion pollution particulates.

O Quick Check 4 Average Value of a Continuous Function Another important use of the area under a curve is in finding the average value of a continuous function over a closed interval. Suppose that T = f1t2 is the temperature at time t recorded at a weather station on a certain day. The station uses a 24-hr clock, so the domain of the temperature function is the interval 30, 244. The function is continuous, as shown in the following graph.

O

• Properties of Definite Integrals

4.4

431

T f(12) f(16) f(20) f(8) f(0) f(4) 0

4

8

12

16

20

24

t

To find the average temperature for the day, we might take six temperature readings at 4-hr intervals, starting at midnight: T0 = f102, T1 = f142, T2 = f182, T3 = f1122, T4 = f1162, T5 = f1202. The average reading would then be the sum of these six readings divided by 6: Tav =

T0 + T1 + T2 + T3 + T4 + T5 6

.

This computation of the average temperature has limitations. For example, suppose that it is a hot summer day, and at 2:00 in the afternoon (hour 14 on the 24-hr clock), there is a short thunderstorm that cools the air for an hour between our readings. This temporary dip would not show up in the average computed above. What can we do? We could take 48 readings at half-hour intervals. This should give us a better result. In fact, the shorter the time between readings, the better the result should be. It seems reasonable that we might define the average value of T over 30, 244 to be the limit, as n approaches q , of the average of n values: Average value T = lim a n: q

1 n 1 n Ti b = lim a a f1ti2b. a n i=1 n: q n i=1

Note that this is not too far from our definition of an integral. All we need is to get ¢t, which is 124 - 02>n, or 24>n, into the summation. We accomplish this by multiplying 1 # by 1, writing 1 as ¢t: ¢t 1 #1 n f1ti2 ¢tb n : q ¢t n ia =1 n 1 n = lim a # a f1ti2 ¢tb n : q 24 n i = 1

Average value of T = lim a

=

n 1 lim a f1ti2 ¢t 24 n : q i = 1

=

1 f1t2 dt. 24 L0

¢t =

24 1 n , so = n ¢t 24

24

DEFINITION

Let f be a continuous function over a closed interval 3a, b4. Its average value, yav, over 3a, b4 is given by b

yav =

1 f1x2 dx. b - a La

432

CHAPTER 4

• Integration

y

Let’s consider average value in another way. If we multiply both sides of

f

b

yav =

1 f1x2 dx b - a La

by b - a, we get yav

a

1b - a2yav = b x

b–a

TECHNOLOGY CONNECTION

b

La

f1x2 dx.

Now the expression on the left side is the area of a rectangle of length b - a and height yav. The area of such a rectangle is the same as the area under the graph of y = f1x2 over the interval 3a, b4, as shown in the figure at the left.

■ EXAMPLE 6 Find the average value of f1x2 = x 2

EXERCISE

1. Graph f 1x2 = x . Compute

y 4

over the interval 30, 24.

4

the average value of the function over the interval 30, 24, using the method of Example 6. Then use that value, yav, and draw a graph of it as a horizontal line using the same set of axes. What does this line represent in comparison to the graph of f 1x2 = x 4 and its associated area?

Solution The average value is

3

f(x) = x 2

2

1 x3 2 1 x 2 dx = c d 2 - 0 L0 2 3 0 3

2 3

1 2 0 a b 2 3 3 1 8 4 1 = # = , or 1 . 2 3 3 3

=

yav

1

0

1

2

x

Note that although the values of f1x2 increase from 0 to 4 over 30, 24, we do not expect the average value to be 2 (which is half of 4), because we see from the graph that f1x2 is less than 2 over more than half the interval. v 80 Speed (in miles per hour)

yav

■ EXAMPLE 7 Rico’s speed, in miles per hour, t minutes after entering the freeway, v(t) = −

1 3 t 200

+

3 2 t 20

−

3 t 8

+ 60, t ≤ 30

70

is given by v1t2 = -

60

From 5 min after entering the freeway to 25 min after doing so, what was Rico’s average speed? How far did he travel over that time interval?

50 40

Solution The average speed is

30

25

20 10 0

1 3 3 2 3 t + t - t + 60, t … 30. 200 20 8

5

10 15 20 25 Time (in minutes)

30

t

1 1 3 3 2 3 at + t - t + 60b dt 25 - 5 L5 200 20 8 25 1 1 4 1 3 3 2 = ct + t t + 60t d 20 800 20 16 5

Quick Check 5 The temperature, in degrees Fahrenheit, in Minneapolis on a winter’s day is modeled by the function f1x2 = - 0.012x 3 + 0.38x 2 - 1.99x - 10.1, where x is the number of hours from midnight 10 … x … 242. Find the average temperature in Minneapolis during this 24-hour period.

1 1 # 4 1 # 3 3 # 2 ca25 + 25 25 + 60 # 25b 20 800 20 16

=

- a=

1 # 4 1 # 3 3 # 2 5 + 5 5 + 60 # 5b d 800 20 16

1 53,625 9625 a b = 68 34 mph. 20 32 32

To find how far Rico traveled over the time interval 35, 254, we first note that t is given in minutes, not hours. Since 25 min - 5 min = 20 min is 13 hr, the distance traveled over 35, 254, is 1 3

# 68 34

O Quick Check 5

= 22 11 12 mi.

O

Exercise Set 4.4

Section Summary • The additive property of definite integrals states that a definite integral can be expressed as the sum of two (or more) other definite integrals. If f is continuous on 3a, c4 and we choose b such that a 6 b 6 c, then c

La

b

f1x2 dx =

La

c

f1x2 dx +

• The average value of a continuous function f over an interval 3a, b4 is b

yav =

1 f1x2 dx. b - a La

f1x2 dx.

Lb

• The area of a region bounded by the graphs of two functions, f1x2 and g1x2, where f1x2 Ú g1x2 over an interval 3a, b4, is b

A =

La

3f1x2 - g1x24 dx.

EXERCISE SET

4.4 Find the area under the graph of f over the interval 71, 58. 1. f 1x2 = e

15. f 1x2 = x 2 - x - 5, g1x2 = x + 10

16. f 1x2 = x 2 - 7x + 20, g1x2 = 2x + 6

2x + 1, for x … 3, 10 - x, for x 7 3

Find the area of the shaded region.

17. f 1x2 = 2x + x 2 - x 3, g1x2 = 0

x + 5, for x … 4, 2. f 1x2 = b 11 - 12 x, for x 7 4

y 3

Find the area under the graph of g over the interval 72, 38. 3. g1x2 = e

f

x 2 + 4, for x … 0, 4 - x, for x 7 0

2 1

- x 2 + 5, for x … 0, 4. g1x2 = e x + 5, for x 7 0

g

Find the area under the graph of f over the interval 76, 48. - x 2 - 6x + 7, 5. f 1x2 = b 3 2 x - 1,

for x 6 1, for x Ú 1

- x - 1, 6. f 1x2 = e - x 2 + 4x + 5,

for x 6 - 1, for x Ú - 1

−1

1

18. f 1x2 = x 3 + 3x 2 - 9x - 12, g1x2 = 4x + 3

15 10 5 –6

4

–4

L0

8.

L-1

2

9.

L0

(–5, –17)

4

ƒ x 3 - 1 ƒ dx

10.

L-3

1 2 3 4 x

–5 –10 –15 –20

ƒ 3x - 2 ƒ dx ƒ x 3 ƒ dx

19. f 1x2 = x 4 - 8x 3 + 18x 2, g1x2 = x + 28 y

In Exercises 11–16, determine the x-values at which the graphs of f and g cross. If no such x-values exist, state that fact. 11. f 1x2 = 9, g1x2 = x

2

12. f 1x2 = 8, g1x2 = 12 x 2

13. f 1x2 = 7, g1x2 = x 2 - 3x + 2

14. f 1x2 = - 6, g1x2 = x 2 + 3x + 13

g

(3, 15)

–2

1

ƒ x - 3 ƒ dx

f

y (–1, –1) 20

Find the area represented by each definite integral. 7.

x

2

f

50 40

g (4, 32)

30

(–1, 27) 20 10 –2 –1 –10

1

2

3

4

5 x

433

434

• Integration

CHAPTER 4

20. f 1x2 = 4x - x 2, g1x2 = x 2 - 6x + 8

a) Find the total profit from t = 0 to t = 10 (the first 10 days). Note:

y

T

4

3R¿1t2 - C¿1t24 dt. L0 b) Find the average daily profit for the first 10 days 1from t = 0 to t = 102. P1T2 = R1T2 - C1T2 =

g

f

3 2 1 –1

1

2

3

4

50. Total and average daily profit. Great Green, Inc., determines that its marginal revenue per day is given by

5 x

–1

Find the area of the region bounded by the graphs of the given equations. 21. y = x, y = x 3, x = 0, x = 1 22. y = x, y = x 4

23. y = x + 2, y = x 2

24. y = x 2 - 2x, y = x

25. y = 6x - x 2, y = x

26. y = x 2 - 6x, y = - x

27. y = 2x - x 2, y = - x

28. y = x 2, y = 2x

29. y = x, y = 2x

30. y = 3, y = x, x = 0

31. y = 5, y = 2x, x = 0

32. y = x 2, y = x 3

33. y = 4 - x 2, y = 4 - 4x

4

34. y = x 2 + 1, y = x 2, x = 1, x = 3 35. y = x + 3, y = x , x = 1, x = 2 2

2

36. y = 2x 2 - x - 3, y = x 2 + x 37. y = 2x 2 - 6x + 5, y = x 2 + 6x - 15 Find the average value over the given interval. 38. y = 2x 3; 3- 1, 14

40. y = e x; 30, 14

42. y = x 2 - x + 1; 30, 24

39. y = 4 - x 2; 3- 2, 24

41. y = e -x; 30, 14

43. f 1x2 = x 2 + x - 2; 30, 44

R¿1t2 = 75e t - 2t, R102 = 0, where R1t2 is the total accumulated revenue, in dollars, on the tth day. The company’s marginal cost per day is given by C¿1t2 = 75 - 3t, C102 = 0, where C1t2 is the total accumulated cost, in dollars, on the tth day. a) Find the total profit from t = 0 to t = 10 (see Exercise 49). b) Find the average daily profit for the first 10 days. 51. Accumulated sales. ProArt, Inc., determines that its weekly online sales, S1t2, in hundreds of dollars, t weeks after online sales began, can be estimated by S1t2 = 9e t. Find the average weekly sales for the first 5 weeks after online sales began. 52. Accumulated sales. Music Manager, Ltd., estimates that monthly revenue, R1t2, in thousands of dollars, attributable to its Web site t months after the Web site was launched, is given by R1t2 = 0.5e t. Find the average monthly revenue attributable to the Web site for its first 4 months of operation.

44. f 1x2 = mx + 1; 30, 24

53. Refer to Exercise 51. Find ProArt’s average weekly online sales for weeks 2 through 5 1t = 1 to t = 52.

46. f 1x2 = x n, n Z 0; 30, 14

54. Refer to Exercise 52. Find the average monthly revenue from Music Manager’s Web site for months 3 through 5 1t = 2 to t = 52.

45. f 1x2 = 4x + 5; 30, a4

47. f 1x2 = x n, n Z 0; 31, 24 48. f 1x2 =

n ; 31, 54 x

APPLICATIONS Business and Economics 49. Total and average daily profit. Shylls, Inc., determines that its marginal revenue per day is given by R¿1t2 = 100e t, R102 = 0, where R1t2 is the total accumulated revenue, in dollars, on the tth day. The company’s marginal cost per day is given by C¿1t2 = 100 - 0.2t, C102 = 0, where C1t2 is the total accumulated cost, in dollars, on the tth day.

Social Sciences 55. Memorizing. In a memory experiment, Alice is able to memorize words at the rate given by m¿1t2 = - 0.009t 2 + 0.2t 1words per minute2.

In the same memory experiment, Ben is able to memorize words at the rate given by

M¿1t2 = - 0.003t 2 + 0.2t 1words per minute2.

a) Who has the higher rate of memorization? b) How many more words does that person memorize from t = 0 to t = 10 (during the first 10 min of the experiment)? c) Over the first 10 min of the experiment, on average, how many words per minute did Alice memorize?

Exercise Set 4.4

d) Over the first 10 min of the experiment, on average, how many words per minute did Ben memorize?

Calves,” Journal of Veterinary Pharmacology and Therapeutics, Vol. 25, 299–309 (2002).) C

56. Results of studying. Antonio’s score on a test is given by where s1t2 is his score after t hours of studying. Bonnie’s score on the same test is given by S1t2 = 10t, 0 … t … 10, where S1t2 is her score after t hours of studying. a) For 0 6 t 6 10, who will have the higher test score? b) Find the average value of s1t2 over the interval 37, 104, and explain what it represents. c) Find the average value of S1t2 over the interval 36, 104, and explain what it represents. d) Assuming that both students have the same study habits and are equally likely to study for any number of hours, t, in 30, 104, on average, how far apart will their test scores be? 57. Results of practice. A keyboarder’s speed over a 5-min interval is given by W1t2 = - 6t 2 + 12t + 90, t in 30, 54,

where W1t2 is the speed, in words per minute, at time t. y

W(t) = – 6t2 + 12t + 90, 0≤t≤5

80

50 Concentration (in mg/mL)

s1t2 = t , 0 … t … 10, 2

435

40 C(t) = 42.03e – 0.01050t

30 20 10 0

20

40

60 80 Hours

100 120

t

a) Given that this model is accurate for 0 … t … 120, what is the initial dosage? b) What is the average amount of phenylbutazone in the calf’s body for the time between 10 and 120 hours? 60. New York temperature. For any date, the average temperature on that date in New York can be approximated by the function T1x2 = 43.5 - 18.4x + 8.57x 2 - 0.996x 3 + 0.0338x 4, where T represents the temperature in degrees Fahrenheit, x = 1 represents the middle of January, x = 2 represents the middle of February, and so on. (Source: www. worldclimate.com.) Compute the average temperature in New York over the whole year to the nearest degree.

60

20 1

2

3

4

5

t

a) Find the speed at the beginning of the interval. b) Find the maximum speed and when it occurs. c) Find the average speed over the 5-min interval. 58. Average population. The population of the United States can be approximated by

Temperature (in degrees Fahrenheit)

40

T 80 70 60 50 40 30 20 10 0

P1t2 = 282.3e 0.01t, where P is in millions and t is the number of years since 2000. (Source: Population Division, U.S. Census Bureau.) Find the average value of the population from 2001 to 2005.

Natural and Life Sciences 59. Average drug dose. The concentration, C, of phenylbutazone, in micrograms per milliliter (μg/mL), in the plasma of a calf injected with this anti-inflammatory agent is given approximately by C1t2 = 42.03e -0.01050t, where t is the number of hours after the injection and 0 … t … 120. (Source: A. K. Arifah and P. Lees, “Pharmacodynamics and Pharmacokinetics of Phenylbutazone in

T(x) = 43.5 – 18.4x + 8.57x 2 – 0.996x 3 + 0.0338x 4

2

4

8 6 10 Mid-month

12

x

61. Outside temperature. The temperature over a 10-hr period is given by f 1t2 = - t 2 + 5t + 40, 0 … t … 10.

a) Find the average temperature. b) Find the minimum temperature. c) Find the maximum temperature. 62. Engine emissions. The emissions of an engine are given by E1t2 = 2t 2, where E1t2 is the engine’s rate of emission, in billions of pollution particulates per year, at time t, in years. Find the average emissions from t = 1 to t = 5.

436

• Integration

CHAPTER 4

SYNTHESIS

The total blood flow, Q, is given by R

Find the area of the region bounded by the given graphs.

Q =

L0 Find Q.

-2

63. y = x , y = x , x = 5 2

64. y = e x, y = e -x, x = - 2

71. Solve for K, given that

65. y = x + 6, y = - 2x, y = x 3

2

66. y = x , y = x , x = - 1 2

3

L1

68. Find the area bounded by y = 3x - 20x , the x-axis, and the first coordinates of the relative maximum and minimum values of the function. 5

313x 2 + 5x2 - 13x + K24 dx = 6.

TECHNOLOGY CONNECTION

67. x + 2y = 2, y - x = 1, 2x + y = 7 3

Find the area of the region enclosed by the given graphs. 72. y = x 2 + 4x, y = 216 - x 2

69. Find the area bounded by y = x 3 - 3x + 2, the x-axis, and the first coordinates of the relative maximum and minimum values of the function.

73. y = x24 - x 2, y =

- 4x x + 1 2

, x = 0, x = 2

74. y = 2x 2 + x - 4, y = 1 - x + 8x 2 - 4x 4

70. Life science: Poiseuille’s Law. The flow of blood in a blood vessel is faster toward the center of the vessel and slower toward the outside. The speed of the blood is given by

75. y = 21 - x 2, y = 1 - x 2, x = - 1, x = 1 76. Consider the following functions: f 1x2 = 3.8x 5 - 18.6x 3, g1x2 = 19x 4 - 55.8x 2.

p V = 1R 2 - r 22, 4Lv

a) Graph these functions in the window 3- 3, 3, - 80, 804, with Yscl = 10. b) Estimate the first coordinates a, b, and c of the three points of intersection of the two graphs. c) Find the area between the curves on the interval 3a, b4. d) Find the area between the curves on the interval 3b, c4.

where R is the radius of the blood vessel, r is the distance of the blood from the center of the vessel, and p, v, and L are physical constants related to the pressure and viscosity of the blood and the length of the blood vessel. If R is constant, we can think of V as a function of r: V1r2 =

2p # V1r2 # r # dr.

p 1R 2 - r 22. 4Lv

Answers to Quick Checks 2 1 9 1 2 85 1. 83 3 , or 27 3 2. 2 , or 42 2 3. 2 , or 4 2 4. (a) 1066 3 ft; (b) 60 sec, 7200 ft 5. Approximately - 2.5°F

r R L

4.5 OBJECTIVES • Evaluate integrals using substitution. • Solve applied problems involving integration by substitution.

Integration Techniques: Substitution The following formulas provide a basis for an integration technique called substitution.

A. B. C.

L L

u r du =

ur + 1 + C, assuming r Z - 1 r + 1

e u du = e u + C

1 du = ln ƒ u ƒ + C; or Lu

1 du = ln u + C, u 7 0 Lu

(Unless noted otherwise, we will assume u 7 0.)

4.5

• Integration Techniques: Substitution

437

In the above formulas, the variable u represents a more complicated expression in terms of x. First, consider the integral 1 x 7 dx. We can carry out this integration using the Power Rule of Antidifferentiation: L

x 7 dx =

x8 x7+1 + C = + C, or 7 + 1 8

1 8 x + C. 8

But, what about an integral like 1 13x - 427 dx, whose integrand is more complicated? Suppose we thought the antiderivative was 13x - 428 8

+ C.

If we do a check by differentiating, we get 8#

1# 13x - 427 # 3 # dx. 8

This simplifies to

313x - 427, which is not 13x - 427,

though it is off only by the constant factor 3. Instead, let’s make this substitution: u = 3x - 4. Then du>dx = 3 and recalling our work on differentials (Section 2.6), we have du = 3 # dx, or

du = dx. 3

With substitution, our original integral, 1 13x - 427 dx, takes the form du du 13x - 427 dx = u7 # Substituting u for 3x - 4 and for dx 3 3 L L 1# 1 Factoring out the constant u7 du 3 3 L 1 # u8 By substitution formula A = + C 3 8 1 1 = # # 13x - 428 + C = 13x - 428 + C. 3 8 24 =

We leave it to the student to check that this is indeed the antiderivative. Note how this procedure reverses the Chain Rule. Recall the Leibniz notation, dy>dx, for a derivative. We gave specific definitions of the differentials dy and dx in Section 2.6. Recall that dy dx

= f ¿1x2

and

dy = f ¿1x2 dx.

We will make extensive use of this notation in this section.

■ EXAMPLE 1 For y = f1x2 = x 3, find dy. Solution We have dy dx so

= f ¿1x2 = 3x 2,

dy = f ¿1x2 dx = 3x 2 dx.

438

CHAPTER 4

• Integration

■ EXAMPLE 2 For u = F1x2 = x 2>3, find du. Solution We have du = F ¿1x2 = 23 x -1>3, dx so

du = F ¿1x2 dx = 23 x -1>3 dx.

■ EXAMPLE 3 For u = g1x2 = ln x, find du. Solution We have du 1 = g¿1x2 = , x dx so

Quick Check 1 Find each differential. a) For y = 2x, find dy. b) For u = x 2 - 3x, find du. 1 c) For y = 3 , find dy. x d) For u = 4x - 3, find du.

1 dx, or x

du = g¿1x2 dx =

dx . x

■ EXAMPLE 4 For y = f1x2 = e x , find dy. 2

Solution Using the Chain Rule, we have dy dx

so

= f ¿1x2 = e x # 2x, 2

dy = f ¿1x2 dx = e x # 2x dx. 2

O Quick Check 1 So far, the dx in L

f1x2 dx

has played no role in integration other than to indicate the variable of integration. Now it becomes convenient to make use of dx. Consider the integral 2

L

2xe x dx.

Finding an antiderivative may seem impossible. Yet, if we note that 2xe x = e x # 2x, we 2 2 see in Example 4 that f1x2 = e x is an antiderivative of f ¿1x2 = 2xe x . How might we find such an antiderivative directly? Suppose that we let u = x 2. Then 2

du = 2x, and du = 2x dx. dx If we substitute u for x 2 and du for 2x dx, we have 2

L

2xe x dx =

2

L

e x 2x dx =

L

e u du.

2

O

4.5

• Integration Techniques: Substitution

439

Since L

e u du = e u + C,

it follows that 2

L

2xe x dx =

e u du

L = eu + C 2 = e x + C.

In effect, we have used the Chain Rule in reverse. We can check the result by differentiating. The procedure is referred to as substitution, or change of variable. It can involve trial and error, but you will become more proficient the more you practice. If you try a substitution that doesn’t result in an integrand that can be easily integrated, try another substitution. While there are many integrations that cannot be carried out using substitution, any integral that fits formula A, B, or C on p. 436 can be evaluated with this procedure.

■ EXAMPLE 5 Evaluate: 1 3x 21x 3 + 1210 dx. Solution Note that 3x 2 is the derivative of x 3. Thus, L

3x 21x 3 + 1210 dx =

L

1x 3 + 1210 3x 2 dx

Substitution

u = x 3 + 1, du = 3x 2 dx

=

u 10 du L u11 = + C 11 1 = 11 1x 3 + 1211 + C.

“Reversing” the substitution

As a check, we differentiate: d dx

C 111 1x 3 + 1211 + C D =

+ 1210 # 3x 2 + 0

= 1x 3 + 1210 # 3x 2 = 3x 21x 3 + 1210.

Quick Check 2

O

Evaluate: 1 4x12x 2 + 323 dx.

11 3 11 1x

O Quick Check 2 ■ EXAMPLE 6 Evaluate:

2x dx . 2 L1 + x

Solution 2x dx du = 2 L1 + x Lu = ln u + C

Quick Check 3 Evaluate:

ex x dx. L1 + e

Substitution

Remember:

= ln 11 + x 22 + C

O Quick Check 3

u = 1 + x 2; du = 2x dx. du 1# = du. u L Lu

O

440

CHAPTER 4

• Integration

■ EXAMPLE 7 Evaluate:

2x dx . 2 2 L 11 + x 2

Solution 2x dx du = 2 2 2 L 11 + x 2 Lu =

Substitution

u - 2 du

L = - u -1 + C 1 = - + C u Quick Check 4

O

Evaluate:

6x 2 L 23 + 2x 3

dx.

u = 1 + x 2, du = 2x dx

1 = + C 1 + x2

t

Don’t forget to reverse the substitution after integrating.

O Quick Check 4 ■ EXAMPLE 8 Evaluate:

ln 13x2 dx . x L

Solution

Quick Check 5

1ln x22 Evaluate: dx. L x

ln 13x2 dx u = ln 13x2, = u du Substitution x 1 L L du = dx 2 x u = + C 2 1ln 13x222 = + C 2

O Quick Check 5

O

■ EXAMPLE 9 Evaluate: 1 xe x dx. 2

Solution

2

If we try u = x 2, we have du = 2x dx. We don’t have 2x dx in 1 xe x dx. We do have x dx, so we need a factor of 2. To provide this factor, we multiply by 1, in the form 12 # 2: 2

L

Quick Check 6 3 Evaluate: 1 x 2e 4x dx.

1 2 2xe x dx 2L 1 1 2 = e x 12x dx2 = e u du 2L 2L 1 1 2 = e u + C = e x + C. 2 2

xe x dx =

O Quick Check 6 With practice, you will be able to make certain substitutions mentally and just write down the answer. Example 10 illustrates one such case.

O

4.5

• Integration Techniques: Substitution

441

dx . Lx + 3

■ EXAMPLE 10 Evaluate: Solution

dx du Substitution u = x + 3, = du = 1 dx = dx u Lx + 3 L = ln u + C We assume x + 3 7 0. = ln 1x + 32 + C 1 ■ EXAMPLE 11 Evaluate: 10 5x2x 2 + 3 dx. Round to the nearest thousandth.

Solution We first find the indefinite integral and then evaluate that integral over 30, 14:

TECHNOLOGY CONNECTION

EXERCISE 1. Use a calculator to evaluate

L

1

2 10 5x2x + 3 dx.

5x2x 2 + 3 dx = 5

L

x2x 2 + 3 dx

5 2x2x 2 + 3 dx 2L 5 = 2x 2 + 3 2x dx 2L =

=

5 2u du 2L

=

5 u 1>2 du 2L

5 # 2 3>2 u + C 2 3 5 = 1x 2 + 323>2 + C. 3

Substitution

u = x 2 + 3, du = 2x dx

=

Reversing the substitution before evaluating with the bounds

Using 0 for the constant C, we have 1

1 5 5x2x 2 + 3 dx = c 1x 2 + 323>2 d 3 L0 0 5 1 = C 1x 2 + 323>2 D 0 3 5 = C 4 3>2 - 33>2 D 3

Quick Check 7

L 4.673.

Evaluate: 2 10

1x + 121x + 2x + 32 dx. 2

4

O Quick Check 7 In some cases, after a substitution is made, a further simplification can allow us to complete an integration:

O

442

CHAPTER 4

• Integration

■ EXAMPLE 12 Evaluate:

x dx. Assume that x + 2 7 0. Lx + 2

Solution We substitute u = x + 2 and du = dx. We observe that x = u - 2. The substitutions are made: u = x + 2, u - 2 x Substitution dx = du du = dx L u Lx + 2 x = u - 2 2 u - 2 2 u 2 = a1 - b du = = 1 u u u u u L = u - 2 ln u + C1 = x + 2 - 2 ln 1x + 22 + C1 = x - 2 ln 1x + 22 + C

Quick Check 8

O

x dx by 3 L 1x - 12 letting u = x - 1. Evaluate

Reversing the substitution C = C1 + 2

O Quick Check 8 Strategy for Substitution The following strategy may help in carrying out the procedure of substitution: 1. Decide which rule of antidifferentiation is appropriate. a) If you believe it is the Power Rule, let u be the base (see Examples 5, 7, 8, and 11). b) If you believe it is the Exponential Rule (base e), let u be the expression in the exponent (see Example 9). c) If you believe it is the Natural Logarithm Rule, let u be the denominator (see Examples 6 and 10). 2. Determine du. 3. Inspect the integrand to be sure the substitution accounts for all factors. You may need to insert constants (see Examples 9 and 11) or make an extra substitution (see Example 12). 4. Perform the antidifferentiation. 5. Reverse the substitution. If there are bounds, use them to evaluate the integral after the substitution has been reversed. 6. Always check your answer by differentiation.

Section Summary • Integration by substitution is the reverse of applying the Chain Rule of Differentiation. • The substitution is reversed after the integration has been performed.

• Results should be checked using differentiation.

Exercise Set 4.5

443

EXERCISE SET

4.5 Evaluate. Assume u>0 when ln u appears. (Be sure to check by differentiating!) 1. 3. 5.

18 + x 2 3x dx 3 5

L L L

2

1x 2 - 627x dx 13t 4 + 22t 3 dt

2 7. dx 1 + 2x L 9. 11. 13. 15. 17.

L

1ln x23

1 dx x

L

e

L

x 4e x dx te

-t2

10.

14.

dx 5

L

6.

12.

e dx x>3

4.

L

1x - 72 2x dx

L L

16. 18.

dt

L

1x 3 + 124x 2 dx 12t 5 - 32t 4 dt

1ln x27

L

1 dx x

39.

e dx e

x>2

45.

dx

x 3e x dx t e

dt

53.

dx 23. 1 L - x

dx 24. 4 L - x

55.

27. 28.

29.

L L L

32.

L 12 - x 2

40.

4 7

5

L

12x 21 + 6x 2 dx

42.

3

7

2ax + b dx be ax dx 3x 2 dx

L 11 + x 325 4

L

5x21 - x 2 dx

1 2

2xe x dx

L0

44.

3

L0

2

x1x 2 + 125 dx

L0

46. 48.

2x + 1

L1 x + x - 1 2

L0 L0

50.

2x + 3

L1 x 2 + 3x

dx

b

52.

L0

2e -2x dx

b

me -mx dx

54.

L0

ke -kx dx

3

1x - 622 dx

56.

L0

1x - 522 dx

0

3x 2 dx

L0 11 + x 2

58.

3 5

27

e 4x dx

L0

3

dx

e -x dx

2

57.

x1x 2 - 127 dx

L1

2

dt L0 1 + t

L0

3x 2e x dx

x 3 dx

L-1 12 - x 427 1

1x 4 + x 3 + x 22714x 3 + 3x 2 + 2x2 dx

59.

1x 3 - x 2 - x2913x 2 - 2x - 12 dx

61. Use a graphing calculator to check the results of any of Exercises 43–60.

e x dx x L4 + e

30.

ln x dx L x 1ln x22

L

x 3 dx

L

4

e t dt t L3 + e

x

L0

3

7x 21 + x 2 dx

60.

5

L0

12x21 - x 2 dx

Evaluate. Use the technique of Example 12. 62.

x dx Lx - 5

64.

x dx L 1 - 4x

65.

x + 3 dx Lx - 2

2

31.

38.

b

t 1t - 12 dt 2

L

P0e kt dt

L

b

dx 22. 1 + 7x L

26.

L

36.

4

49.

dx 21. 12 + 3x L

t1t - 12 dt

L

x 2ax 2 + b dx

4

47.

51.

25.

L x ln x 2

1

1 20. dx 2 + 8x L

5

dx

Evaluate.

1 19. dx 5 + 2x L

2

34.

1

2 -t3

L

37.

43.

4

L

35.

41.

7x

L

dx x L ln x

6

5 8. dx 5x + 7 L

3x

L

2.

2

33.

(Hint: Use the properties of logarithms.)

dx

63.

3x dx L 2x + 1

(Hint: u = x - 2.)

444 66. 67. 68. 69. 70.

CHAPTER 4

• Integration

2x + 3 dx L 3x - 2 L L L

73. Profit from marginal profit. A firm has the marginalprofit function

x 21x + 1210 dx

dP 9000 - 3000x = 2 . dx 1x - 6x + 1022

(Hint: u = x + 1.)

y

x 31x + 227 dx

800

9000 – 3000x dP = ——–—–——– — d x (x 2 – 6x + 10)2

600

x L 2x - 2

400

(Hint: u = x - 2.)

x 2 2x - 2 dx

200 2

dx

4

6

8

10 x

–200 –400 –600 –800

APPLICATIONS Find the total-profit function given that P = $1500 at x = 3.

Business and Economics 71. Demand from marginal demand. A firm has the marginal-demand function D¿1x2 =

- 2000x 225 - x

2

Social Sciences 74. Divorce rate. The divorce rate in the United States is approximated by

.

D1t2 = 100,000e 0.025t, where D1t2 is the number of divorces occurring at time t and t is the number of years measured from 1900. 9 That is, t = 0 corresponds to 1900, t = 98 365 corresponds to January 9, 1998, and so on. a) Find the total number of divorces from 1900 to 2005. Note that this is given by

y –1000

1

2

3

4

x

–2000 –3000 –4000 –5000

D'(x) =

– 2000x √ 25 – x 2

105

D1t2 dt. L0 b) Find the total number of divorces from 1980 to 2006. Note that this is given by

–6000 –7000

106

Find the demand function given that D = 13,000 when x = $3 per unit.

L80

72. Value of an investment. V. King Manufacturing buys a new machine for $250,000. The marginal revenue from the sale of products produced by the machine after t years is given by

SYNTHESIS

D1t2 dt.

Find the area of the shaded region. y

y

76.

R¿1t2 = 4000t.

2

The salvage value of the machine, in dollars, after t years is given by

y = x16 – x 2

1

V1t2 = 200,000 - 25,000e 0.1t.

75.

–2

The total profit from the machine, in dollars, after t years is given by Revenue Revenue Cost from from ≤ + ± ≤ P1t2 = ± of . sale of sale of P machine Q product machine The company knows that R102 = 0. a) Find P1t2. b) Find P1102.

8

y = – x4 – x 2

–1

1

2 x

4 –4 – 3 –2 –1

1 2 3 4 x

–4

–1

–8

–2

Evaluate. Assume u>0 when ln u appears. 77. 79.

dx L ax + b e 2t L 2t

dt

78. 80.

L

5x 21 - 4x 2 dx x2 3

L ex

dx

4.6

81.

83. 85.

87. 88.

89. 90.

1ln x299 x

L

1e + 22e dt t

L

82.

dx

t2 4 L2 2 + t3

t

84.

dt

86.

31ln x22 + 31ln x2 + 44 x

L x - 3

L 1x - 6x2 2

1>3

t4 + 8 t 2 + 2t

L 1t + 122 aHint:

dx

97.

L x 1ln x24 L

x 2 2x 3 + 1 dx

dx

x + 3 dx Lx + 1

93.

t - 5 dt Lt - 4

95.

dx e L + 1 x

dx, n Z - 1

dx L x ln x 3ln 1ln x24

99.

e -mx -mx dx L 1 - ae

100. 101.

dt

t 2 + 2t + 1 - 1

=

t + 2t + 1 2

= 1 -

1

1t + 122

.b

L L

9x17x 2 + 92n dx, n Z - 1 5x 212x 3 - 72n dx, n Z - 1

dx 2

dx; (b) du = 12x - 32 dx; 22x 3 (c) dy = - 4 dx; (d) du = 4 dx x 2. 14 12x 2 + 324 + C 3. ln 11 + e x2 + C 1 4x3 4. 223 + 2x 3 + C 5. 13 1ln x23 + C 6. 12 e + C 1 1 7. 16,080.8 8. + C x - 1 21x - 122 1. (a) dy =

(Hint: See Exercise 90.)

¢ Hint:

2 x + 3 = 1 + .≤ x + 1 x + 1 94.

¢ Hint:

32 f 1x24 dx = 3 f 1x242 + C.

Answers to Quick Checks

2

92.

x

L

L

1t + 12

L 1x + 32

1ln x2n

98.

dt

2

91.

e x - e -x x -x dx Le + e

102. Determine whether the following is a theorem:

t 2 + 2t

x + 6x

L t2

96.

dt

dx

t 3 ln 1t 4 + 82

L

e 1>t

• Integration Techniques: Integration by Parts

dx n, n Z - 1 L x1ln x2

1

e -x 1 = .≤ e + 1 1 + e -x x

4.6 OBJECTIVES • Evaluate integrals using the formula for integration by parts. • Solve applied problems involving integration by parts.

Integration Techniques: Integration by Parts Recall the Product Rule for differentiation: d dv du 1uv2 = u + v . dx dx dx Integrating both sides with respect to x, we get uv = =

dv du dx + v dx dx L L dx u

L

u dv +

L

v du.

Solving for 1 u dv, we get the following theorem.

445

446

CHAPTER 4

• Integration

THEOREM 7 L

The Integration-by-Parts Formula

u dv = uv -

L

v du

This equation can be used as a formula for integrating in certain situations— that is, situations in which an integrand is a product of two functions, and one of the functions can be integrated using the techniques we have already developed. For example, L

xe x dx

can be considered as L

x1e x dx2 =

L

u dv,

where we let u = x and dv = e x dx. In this case, differentiating u gives du = dx, and integrating dv gives v = e x.

We select C = 0 to obtain the simplest antiderivative.

Then the Integration-by-Parts Formula gives us u

L

dv

1x21e x dx2

u

=

v

1x21e x2

v

-

du

1e x21dx2

L = xe x - e x + C.

This method of integrating is called integration by parts. As always, to check, we can simply differentiate. This check is left to the student. Note that integration by parts, like substitution, is a trial-and-error process. In the preceding example, suppose that we had reversed the roles of x and ex. We would have obtained u = e x,

dv = x dx,

du = e x dx, u

and

L

v =

dv

1e x21x dx2

u

=

x2 , 2

1e x2a

v

v

du

x2 x2 b a b1e x dx2. 2 L 2

Now the integrand on the right is more difficult to integrate than the one with which we began. When we can integrate both factors of an integrand, and thus have a choice as to how to apply the Integration-by-Parts Formula, it can happen that only one (or maybe none) of the possibilities will work.

4.6

• Integration Techniques: Integration by Parts

447

Tips on Using Integration by Parts 1. If you have had no success using substitution, try integration by parts. 2. Use integration by parts when an integral is of the form L

f1x2 g1x2 dx.

Match it with an integral of the form L

u dv

by choosing a function to be u = f1x2, where f1x2 can be differentiated, and the remaining factor to be dv = g1x2 dx, where g1x2 can be integrated. 3. Find du by differentiating and v by integrating. 4. If the resulting integral is more complicated than the original, make some other choice for u and dv. 5. To check your result, differentiate.

Let’s consider some additional examples.

■ EXAMPLE 1 Evaluate: 1 ln x dx. Assume x 7 0.

Solution Note that 1 1dx>x2 = ln x + C, but we do not yet know how to find

1 ln x dx since we have not yet found a function whose derivative is ln x. Since we can differentiate ln x, we let u = ln x Then

du =

and dv = dx.

1 dx and x

v = x.

Using the Integration-by-Parts Formula gives u

dv

u

v

v

du

1 1ln x21dx2 = 1ln x2x x a dxb x L L = x ln x Quick Check 1 Evaluate: 1 xe 3x dx.

dx L = x ln x - x + C.

O Quick Check 1 ■ EXAMPLE 2 Evaluate: 1 x ln x dx. Solution Let’s examine several choices, as follows. Attempt 1:

We let

u = 1 and dv = x ln x dx. This will not work because we do not as yet know how to integrate dv = x ln x dx.

O

448

CHAPTER 4

• Integration

Attempt 2:

We let

u = x ln x Then

and dv = dx.

1 du = cxa b + 1ln x21 d dx and x

v = x

= 11 + ln x2 dx.

Using the Integration-by-Parts Formula, we have u

L

dv

u

v

v

1x ln x2 dx = 1x ln x2x = x 2 ln x -

L

L

du

x111 + ln x2 dx2

1x + x ln x2 dx.

This integral seems more complicated than the original, but we will reconsider it in Example 6. Attempt 3:

We let

u = ln x Then

du =

and dv = x dx.

1 dx and x

v =

x2 . 2

Using the Integration-by-Parts Formula, we have u

L

dv

u

v

x2 x ln x dx = ln x # 2

du

v

x2 1 x2 1 a dx b = ln x x dx x 2 2L L2

x2 x2 = ln x + C. 2 4 This choice of u and dv allows us to evaluate the integral.

■ EXAMPLE 3 Evaluate: 1 x25x + 1 dx. Solution We let u = x Then

and dv = 15x + 121>2 dx.

du = dx and

v =

2 15 15x

+ 123>2.

Note that we have to use substitution in order to integrate dv: L

15x + 121>2 dx =

w = 5x + 1, 1 1 Substitution 15x + 121>2 5 dx = w 1>2 dw dw = 5 dx 5L 5L 1 w 1>2 + 1 2 3>2 2 v = # 1 = w = 15x + 123>2. 5 2 + 1 15 15

Using the Integration-by-Parts Formula gives us u

L

dv

x A 25x + 1 dx B

u

v

2 = x # 15 15x + 123>2 -

=

2 15 x15x

+ 123>2 -

=

2 15 x15x

+ 123>2 -

v

L

#

2 15

15x + 12

3>2

2 2 15 25 15x 4 375 15x

du

dx

+ 125>2 + C

+ 125>2 + C.

• Integration Techniques: Integration by Parts

4.6

449

This integral may also be evaluated using a substitution similar to that shown in Example 12 of Section 4.5. We revisit the evaluation of this integral in Exercise 43 at the end of this section.

Quick Check 2

O

Evaluate: 1 2x23x - 2 dx.

O Quick Check 2 2 ■ EXAMPLE 4 Evaluate: 11 ln x dx.

Solution First, we find the indefinite integral (see Example 1). Next, we evaluate the definite integral: 2

ln x dx = 3x ln x - x421

L1

= 12 ln 2 - 22 - 11 # ln 1 - 12 = 2 ln 2 - 2 + 1 = 2 ln 2 - 1 L 0.386.

Repeated Integration by Parts In some cases, we may need to apply the Integration-by-Parts Formula more than once. y 0.6 0.5 0.4 0.3 0.2 0.1 0

■ EXAMPLE 5 Evaluate 107x 2e -x dx to find the area of the shaded region shown to

f(x) = x 2e–x

the left.

Solution We first let and dv = e -xdx.

u = x2 Then 1

7

10 x

du = 2x dx and

v = - e -x.

Using the Integration-by-Parts Formula gives u

L

TECHNOLOGY CONNECTION

dv

x 21e -x dx2

u

=

v

x 21- e -x2

= - x 2e -x +

EXERCISE 1. Use a calculator or iPlot to

evaluate 7

L0

v

-

L

L

du

- e -x12x dx2

(1)

2xe -x dx.

To evaluate the integral on the right, we can apply integration by parts again, as follows. We let

x 2e -x dx.

u = 2x Then

and dv = e -xdx. v = - e -x.

du = 2 dx and

Using the Integration-by-Parts Formula once again, we get u

L

dv

u

v

2x1e-xdx2 = 2x1- e-x2 -

v

L

du

- e-x12 dx2

= - 2xe -x - 2e -x + C.

(2)

When we substitute equation (2) into (1), the original integral becomes L

x 2e -x dx = - x 2e -x - 2xe -x - 2e -x + C = - e -x1x 2 + 2x + 22 + C.

Factoring simplifies the next step.

450

CHAPTER 4

• Integration

We now evaluate the definite integral: 7

L0

Quick Check 3 3

Evaluate:

x

L0 2x + 1

dx.

x 2e -xdx = 3- e -x1x 2 + 2x + 22470 = 3- e -7172 + 2172 + 224 - 3- e -0102 + 2102 + 224 = - 65e -7 + 2 L 1.94.

O Quick Check 3 Recurring Integrals Occasionally integration by parts yields an integral of the form 1 v du that is identical to the original integral. If we are alert and notice this when it occurs, we can find a solution of the original integral algebraically. Let’s use this approach and reconsider Example 2.

■ EXAMPLE 6 Evaluate 1 x ln x dx using the result of the second attempt in Example 2.

Solution For the second attempt in Example 2, we let u = x ln x so that

and dv = dx,

du = 11 + ln x2 dx and

v = x.

Let’s now work further with the result we abandoned earlier: u

L

dv

u

v

1x ln x2 dx = 1x ln x2x = x 2 ln x = x 2 ln x -

L L

v

L

x111 + ln x2 dx2

1x + x ln x2 dx x dx -

= x 2 ln x - 12 x 2 2

L L

du

L

L

x ln x dx

x ln x dx

Substituting the integral of each term in the sum The original integral is duplicated.

x ln x dx = x 2 ln x - 12 x 2

Adding

x ln x dx = 12 x 2 ln x - 14 x 2 + C.

Dividing both sides by 2

L

x ln x dx to both sides

Tabular Integration by Parts In situations like that in Example 5, we have an integral, L

f1x2 g1x2 dx,

for which f1x2 can be repeatedly differentiated easily to a derivative that is eventually 0. The function g1x2 can also be repeatedly integrated easily. In such cases, we can use integration by parts more than once to evaluate the integral.

O

• Integration Techniques: Integration by Parts

4.6

451

■ EXAMPLE 7 Evaluate: 1 x 3e x dx. Solution We use integration by parts repeatedly, watching for patterns: u dv

L

x 3e x dx

u v

=

v

x 3e x

-

= x 3e x -

du

ex3x2 dx

L L

3x 2e x dx.

This integral is simpler than the original.

(1)

To solve 1 3x 2e x dx, we select u = 3x 2 and dv = e x dx, so du = 6x dx and v = e x, u

and

L

v

v

3x e dx = 3x e 2 x

2 x

= 3x 2e x -

du

x

L

e 6x dx

L

6xe x dx.

This integral is the simplest so far.

(2)

To solve 1 6xe x dx, we select u = 6x and dv = e x dx, so du = 6 dx and v = e x, u v

and

6xe dx = 6xe x

L

v

x

du

x

L

e 6 dx

= 6xe x - 6

e x dx L = 6xe x - 6e x + C.

(3)

Combining equations (1), (2), and (3), we have L

x 3e x dx = x 3e x - a3x 2e x = x 3e x - 3x 2e x +

L

Substituting equation (2) into equation (1)

6xe x dxb

6xe x dx

L = x 3e x - 3x 2e x + 6xe x - 6e x + C.

Substituting equation (3)

As you can see, this approach can get complicated. Using tabular integration, as shown in the following table, can greatly simplify our work. f 1x2 and Repeated Derivatives x3 3x 2 6x 6 0

Sign of Product

g1x2 and Repeated Integrals

1+2 1- 2 1+2 1- 2

ex ex ex ex ex

We then add products along the arrows, making the alternating sign changes, and obtain the correct result: Quick Check 4 Evaluate: 1 x 4e 2x dx.

L

x 3e x dx = x 3e x - 3x 2e x + 6xe x - 6e x + C.

O Quick Check 4

O

452

CHAPTER 4

• Integration

Section Summary • The Integration-by-Parts Formula is the reverse of the Product Rule for differentiation: L

u dv = uv -

L

v du.

• The choices for u and dv should be such that the integral 1 v du is simpler than the original integral. If this does not turn out to be the case, other choices should be made. • Tabular integration is useful in cases where repeated integration by parts is necessary.

EXERCISE SET

4.6 Evaluate using integration by parts or substitution. Check by differentiating. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29.

4x

L L L L L L L L L L L L

4xe dx x 313x 22 dx xe 5x dx xe

-2x

dx

x 2 ln x dx x ln 2x dx ln 1x + 52 dx 1x + 22 ln x dx

L

6. 8. 10. 12. 14. 16. 18.

x2x + 2 dx

20.

x 3 ln 12x2 dx x 2e x dx x e dx 3 -2x

L

4.

1x - 12 ln x dx

2 2x

L

2.

x e

22. 24. 26.

dx

28.

1x 4 + 42e 3x dx

30.

3x

L L L L L L L L L L L L L

Evaluate using integration by parts. 2

31.

3xe dx

L1

6

x 212x2 dx

33.

2

x 2 ln x dx

L2

32.

L1

x 3 ln x dx 5

ln 1x + 82 dx

34.

L0

ln 1x + 72 dx

1

2xe 4x dx

35.

L0

xe x dx 1

xe

-x

36. dx

L0

1x 3 + 2x 2 + 32e -2x dx

8

x 3 ln x dx

37.

L0

x2x + 1 dx ln 3

x 2 ln x 3 dx ln 1x + 42 dx 1x + 12 ln x dx 1x - 22 ln x dx

38.

L0

x 2e 2x dx

39. Cost from marginal cost. A company determines that its marginal-cost function is given by C¿1x2 = 4x2x + 3. Find the total cost given that C1132 = $1126.40. 40. Profit from marginal profit. A firm determines that its marginal-profit function is given by P¿1x2 = 1000x 2e -0.2x.

x2x + 5 dx x 2 ln 15x2 dx

y 14,000

P'(x) = 1000x 2e–0.2x

12,000 10,000

1ln x22 dx

8000 6000 4000

x

-5

ln x dx

2000 20

40

60

x

5 4x

L L

x e dx 1x 3 - x + 12e -x dx

Find the total profit given that P = - $2000 when x = 0.

Exercise Set 4.6

Life and Physical Sciences 41. Electrical energy use. The rate at which electrical energy is used by the Ortiz family, in kilowatt-hours (kW-h) per day, is given by K1t2 = 10te -t, where t is time, in hours. That is, t is in the interval 30, 244.

In Exercises 45 and 46, both substitution and integration by parts are used to determine the indefinite integral. 45. Evaluate 1 e 2x dx by letting u = 2x. Note that x = u 2, so dx = 2u du. Make the substitutions and observe that the new integral (with variable u) can be evaluated using integration by parts.

Evaluate using integration by parts.

K(t) = 10te–t

47.

2

2

4

6

8

t

a) How many kilowatt-hours does the family use in the first T hours of a day 1t = 0 to t = T2? b) How many kilowatt-hours does the family use in the first 4 hours of the day? 42. Drug dosage. Suppose that an oral dose of a drug is taken. Over time, the drug is assimilated in the body and excreted through the urine. The total amount of the drug that has passed through the body in time T is given by T

49. 51. 52. 53. 54.

E1t2 dt,

L0 where E is the rate of excretion of the drug. A typical rate-of-excretion function is E1t2 = te -kt, where k 7 0 and t is the time, in hours. a) Find a formula for T

E1t2 dt.

L0 b) Find

ln x L 2x

48. 50.

dx

te t

L 1t + 122

dt

13t 2 - 48 5 L 2 4t + 7

dt

127x 3 + 83x - 2223x + 8 dx 6

L L L L

x 21ln x22 dx x n1ln x22 dx, n Z - 1 x n ln x dx, n Z - 1

x ne x dx = x ne x - n x n - 1e x dx. L L 56. Verify that for any positive integer n, 1ln x2n dx = x1ln x2n - n 1ln x2n - 1 dx. L L 57. Determine whether the following is a theorem: f1x2g1x2 dx = f1x2 dx # g1x2 dx. L L L Explain.

E1t2 dt, when k = 0.2 mg>hr.

58. Compare the procedures of differentiation and integration. Which seems to be the most complicated or difficult and why?

SYNTHESIS In Exercises 43 and 44, evaluate the given indefinite integral using substitution. Refer to Example 12 in Section 4.5 to review the technique. 43. Evaluate 1 x25x + 1 dx by letting u = 5x + 1 and du = 5 dx A so that dx = 15 du B and observing that u - 1 x = . Compare your answer to that found in 5 Example 3 of this section. Are they the same? (Hint: Simplify both forms of the answer into a common third form.) x

dx. L 2x - 3 a) Evaluate this integral using integration by parts. b) Evaluate it using the substitution u = x - 3 and observing that x = u + 3. c) Show algebraically that the answers from parts (a) and (b) are equivalent.

44. Consider

L

2x ln x dx

55. Verify that for any positive integer n,

10

L0

1

dx by letting u = 2x and L 1 + 2x following the procedure used in Exercise 45.

46. Evaluate

y 4

453

TECHNOLOGY CONNECTION 59. Use a graphing calculator to evaluate 10

L1

x 5 ln x dx.

Answers to Quick Checks 1 x 3x e - e 3x + C 3 9 4 4 2. x13x - 223>2 13x - 225>2 + C 3. 2 23 9 135 1 3 3 3 4. e 2x a x 4 - x 3 + x 2 - x + b + C 2 2 2 4 1.

454

CHAPTER 4

• Integration

4.7 OBJECTIVE • Evaluate integrals using a table of integration formulas.

Integration Techniques: Tables Tables of Integration Formulas You have probably noticed that, generally speaking, integration is more challenging than differentiation. Because of this, integral formulas that are reasonable and/or important have been gathered into tables. Table 1, shown below and inside the back cover of this book, is a brief example of such a table. Entire books of integration formulas are available in libraries, and lengthy tables are also available online. Such tables are usually classified by the form of the integrand. The idea is to properly match the integral in question with a formula in the table. Sometimes some algebra or a technique such as substitution or integration by parts may be needed as well as a table. TABLE 1

1. 2. 3. 4. 5.

L

Integration Formulas

x n dx =

xn+1 + C, n Z - 1 n + 1

dx = ln x + C, x 7 0 L x L L L

u dv = uv -

L

v du

e x dx = e x + C 1 # ax e + C a

e ax dx =

# e ax1ax - 12 + C a2 n x ne ax x n e ax dx = x n - 1e ax dx + C 7. a aL L 6.

8. 9. 10. 11. 12. 13. 14.

15.

16.

L

L L L L

xe ax dx =

1

ln x dx = x ln x - x + C 1ln x2n dx = x1ln x2n - n x n ln x dx = x n + 1 c a x dx = 1

L 2x 2 - a 2

La - x

dx = ln ƒ x + 2x 2 + a 2 ƒ + C dx = ln ƒ x + 2x 2 - a 2 ƒ + C

dx =

1 x - a ln ` ` +C 2a x + a

2

dx =

1 a + x ln ` ` +C 2a a - x

1 2

1 L x2a + x 2

1 ln x d + C, n Z - 1 n + 1 1n + 122

2

1 Lx - a

1ln x2n - 1 dx + C, n Z - 1

ax + C, a 7 0, a Z 1 ln a

L 2x 2 + a 2 1

2

L

2

dx = -

1 a + 2a 2 + x 2 ln ` ` +C a x

(continued)

4.7 TABLE 1

17.

• Integration Techniques: Tables

455

(continued)

1 L x2a 2 - x 2

dx = -

a + 2a 2 - x 2 1 ln ` ` +C x a

a x x a dx = 2 + - 2 ln ƒ a + bx ƒ + C a + bx b b b L x a 1 19. dx = 2 + 2 ln ƒ a + bx ƒ + C 2 1a + bx2 b 1a + bx2 b L 18.

20.

21. 22. 23. 24. 25. 26.

1 1 x dx = ln ` ` +C a a + bx L x1a + bx2 1 L x1a + bx2 L L L

2

1 x 1 + 2 ln ` ` +C a + bx a1a + bx2 a

dx =

2x 2 ; a 2 dx =

1 2

x2a + bx dx = x 2 2a + bx dx = x dx

L 2a + bx x 2 dx L 2a + bx

= =

2 3b 2

C x2x 2 ; a 2 ; a 2 ln ƒ x + 2x 2 ; a 2 ƒ D + C 2

15b 2 2

105b 3

2

Solution The integral

115b 2x 2 - 12abx + 8a 221a + bx23>2 + C

1bx - 2a22a + bx + C

15b 3

■ EXAMPLE 1 Evaluate:

13bx - 2a21a + bx23>2 + C

13b 2x 2 - 4abx + 8a 222a + bx + C

dx . L x13 - x2

dx fits formula 20 in Table 1: L x13 - x2

1 x 1 dx = ln ` ` + C. a a + bx L x1a + bx2 In the given integral, a = 3 and b = - 1, so we have, by the formula, dx 1 x = ln ` ` +C x13 x2 3 3 + 112x L Quick Check 1 Evaluate:

2x dx. 2 L 13 - 5x2

=

1 x ln ` ` + C. 3 3 - x

O Quick Check 1 ■ EXAMPLE 2 Evaluate:

5x dx. L 7x - 8

Solution We first factor 5 out of the integral. The integral then fits formula 18 in Table 1: x a x a dx = 2 + - 2 ln ƒ a + bx ƒ + C. a + bx b b b L

O

456

CHAPTER 4

• Integration

In the given integral, a = - 8 and b = 7, so we have, by the formula, 5x x dx = 5 dx L 7x - 8 L - 8 + 7x -8 x -8 = 5c 2 + - 2 ln ƒ - 8 + 7x ƒ d + C 7 7 7 x 8 -8 ln ƒ 7x - 8 ƒ d + C + + 49 7 49 5x 40 40 + + = ln ƒ 7x - 8 ƒ + C. 49 7 49 = 5c

Quick Check 2

O

Evaluate:

3 dx. 2 L 2x17 - 3x2

O Quick Check 2 ■ EXAMPLE 3 Evaluate: 1 216x 2 + 3 dx. Solution This integral almost fits formula 22 in Table 1: L

2x 2 ; a 2 dx =

1 2

C x2x 2 ; a 2 ; a 2 ln ƒ x + 2x 2 ; a 2 ƒ D + C.

But the coefficient of x 2 needs to be 1. To achieve this, we first factor out 16. Then we apply formula 22: L

216x 2 + 3 dx = =

216 A x L L

= 4

B dx

2

+

3 16

4 2x 2 +

3 16

dx

Using the properties of radicals; 216 = 4

3 16

dx

We have a 2 =

3 16

2x

3 16 ƒ

L

2x

2

+

Factoring

= 4 # 12 C x 2x 2 +

3 16

+

3 16

3 16

+

3 16

ln ƒ x +

= 2 C x 2x 2 +

ln ƒ x +

2x

2

2

+

+

3 16 ƒ

D +C

D + C.

In the given integral, a 2 = 3>16 and a = 23>4, though we did not need to use a in this form when applying the formula.

Quick Check 3

O

Evaluate: 1 x 2 28 + 3x dx.

O Quick Check 3 ■ EXAMPLE 4 Evaluate:

dx . L x - 25 2

Solution This integral fits formula 14 in Table 1: 1 1 x - a dx = ln ` ` + C. 2 2a x + a Lx - a 2

In the given integral, a 2 = 25, so a = 5. We have, by the formula, Quick Check 4 Evaluate:

in formula 22.

dx 1 x - 5 = ln ` ` + C. 10 x + 5 L x - 25 2

4 dx. L x - 11 2

O Quick Check 4

O

457

Exercise Set 4.7

■ EXAMPLE 5 Evaluate: 1 1ln x23 dx. Solution This integral fits formula 9 in Table 1: L

1ln x2n dx = x1ln x2n - n

L

1ln x2n - 1 dx + C, n Z - 1.

We must apply the formula three times: L

1ln x23 dx = x1ln x23 - 3

L

1ln x22 dx + C

= x1ln x23 - 3cx1ln x22 - 2

L

Formula 9, with n = 3

ln x dx d + C

= x1ln x23 - 3cx1ln x22 - 2ax ln x -

L

Applying formula 9 again, with n = 2

dxb d + C

= x1ln x2 - 3x1ln x2 + 6x ln x - 6x + C. 3

Quick Check 5 Evaluate: 1 x 4 ln x dx.

Applying formula 9 for the third time, with n = 1

2

O Quick Check 5

O

The Web site www.integrals.com can be used to find integrals. If you have access to the Internet, use this Web site to check Examples 1–5 or to do the exercises in the following set.

Section Summary • Tables of integrals or the Web site www.integrals.com can be used to evaluate many integrals.

• Some algebraic simplification of the integrand may be required before the correct integral form can be identified.

EXERCISE SET

4.7 Evaluate using Table 1. 1. 3. 5. 7. 9. 11. 13. 15.

L L

xe -3x dx

2.

6x dx

4.

1 L 25 - x

dx 2

x dx L3 - x 1 L x18 - x2 L L L

17.

6. 8.

dx 2

10.

ln 13x2 dx

12.

x 4 ln x dx

14.

x 3 ln x dx

16.

L

2xe 3x dx 1

dx

L 2x 2 - 9 1 L x24 + x x

L 11 - x2 L

2

dx

dx 2

2x 2 + 9 dx

4 ln a xb dx 5 L L L

x 3e -2x dx 5x 4 ln x dx

19. 21. 23. 25. 27.

29.

dx

18.

L 2x + 7 2

10 dx L x15 - 7x2 -5 L 4x - 1 2

L

dx

24m 2 + 16 dm - 5 ln x x3

L ex

L x -3

L

2

dx

dx

x21 + 2x dx

20. 22. 24. 26. 28.

30.

3 dx L x21 - x 2 2 dx 5x17x + 22 L L

29t 2 - 1 dt 3 ln x

L x2 L

dx

1ln x24 dx 3

L 24x 2 + 100 L

x22 + 3x dx

dx

458

• Integration

CHAPTER 4

APPLICATIONS

SYNTHESIS

Business and Economics

Evaluate using Table 1 or the Web site www.integrals.com.

31. Supply from marginal supply. A lawn machinery company introduces a new kind of lawn seeder. It finds that its marginal supply for the seeder satisfies the function

33.

S¿1x2 =

35.

100x

, 0 … x … 19, 120 - x22

where S is the quantity purchased when the price is x thousand dollars per seeder. Find the supply function, S1x2, given that the company will sell 2000 seeders when the price is 19 thousand dollars.

37.

8 L 3x - 2x 2

dx

dx L x 3 - 4x 2 + 4x - e -2x dx L 9 - 6e -x + e -2x

36. 38.

x dx L 4x - 12x + 9 2

L

e x 2e 2x + 1 dx 21ln x22 + 49

L

2x

Answers to Quick Checks

100x S'(x) = ——–—– (20 – x) 2 y

1. Using formula 19:

2 6 + ln ƒ 3 - 5x ƒ + C 2513 - 5x2 25

60

2. Using formula 21:

40

3 x 3 + ln ` ` +C 1417 - 3x2 98 7 - 3x

3. Using formula 24: 2 1135x 2 - 288x + 512218 + 3x23>2 + C 2835 1x - 21122 2 4. Using formula 14: 211 ln ` ` +C 11 x 2 - 11

80

20 5

10

15

20

x

Social Sciences 32. Learning rate. The rate of change of the probability that an employee learns a task on a new assembly line is given by p¿1t2 =

34.

1 t12 + t22

,

where p1t2 is the probability of learning the task after t months. Find p1t2 given that p = 0.8267 when t = 2. y 35 30 25 20 15

1 p'(t) = ——–— t(2 + t)2

10 5 0.05 0.1

0.15

0.2

t

5. Using formula 10:

x5 x2 ln x + C 5 25

dx

CHAPTER 4 SUMMARY KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 4.1 Antidifferentiation is the reverse of differentiation. A function F is an antiderivative of a function f if d F1x2 = f 1x2. dx

d 2 1x 2 = 2x. Antiderivadx tives of f1x2 = 2x have the form x 2 + C. These antiderivatives can also be expressed using the indefinite integral 1 2x dx = x 2 + C, where the function f1x2 = 2x is the integrand. F1x2 = x 2 is an antiderivative of f 1x2 = 2x since

Antiderivatives of a function f all differ by a constant C, called the constant of integration. An indefinite integral of a function is symbolized by f1x2 dx = F1x2 + C,

L where f 1x2 is called the integrand.

G1x2 = ln x is an antiderivative of g1x2 = The corresponding indefinite integral is 1 dx = ln x + C. Lx

We use four rules of antidifferentiation: • A1.

L

k dx = kx + C

xn + 1 + C, n Z - 1 A2. xn dx = n + 1 L 1 A3. dx = ln x + C, x 7 0 Lx 1 A4. e dx = e ax + C a L There are two common properties of indefinite integrals:

• • •

ax

#

P1. 1 3c f1x24dx = c 1 f1x2 dx P2. 1 3 f1x2 ; g1x24 dx

= 1 f1x2 dx ; 1 g1x2 dx.

1 d 1 1ln x2 = . since x x dx

• • • • •

L L L L

9 dx = 9x + C x 6 dx =

A1

1 7 x + C 7

A2

13x 4 + 4x - 52 dx = e 3x dx =

1 3x e + C 3

5 dx = 5 ln x + C Lx L

2x dx = 1

L x5

dx =

L L

x 1>2 dx =

x -5 dx =

3 5 x + 2x 2 - 5x + C 5

A2, P1, P2

A4 A3

2 3>2 x + C 3

A2

1 -4 1 x = - 4 + C -4 4x

A2

x2 - 1 1 1 dx = ax - b dx = x 2 - ln x + C x x 2 L L L

1x + 422 dx =

A2, A3, P2

1x 2 + 8x + 162 dx

L 1 3 = x + 4x 2 + 16x + C 3

A1, A2, P1, P2

(continued)

Chapter 4 Summary

459

460

CHAPTER 4

•

Integration

KEY TERMS AND CONCEPTS SECTION 4.1 (continued ) An initial condition is a point that is a solution of a particular antiderivative.

EXAMPLES Find 1 13x - 22 dx such that (1, 4) is a solution of the antiderivative. 3 We antidifferentiate: 13x - 22 dx = x 2 - 2x + C. Therefore, we 2 L 3 2 x - 2x + C. We are given 11, 42 as an initial condition, 2 so we substitute and solve for C: have F1x2 =

3 1122 - 2112 + C 2 3 - 2 + C 4 = 2 1 4 = - + C 2 9 C = . 2 4 =

Therefore, the particular antiderivative that meets the initial condition is 9 3 F1x2 = x 2 - 2x + . 2 2

SECTION 4.2 Physical Science. A jogger runs according to a velocity function v1t2 = 6, where t is in hours and v is in miles per hour. In 3 hr, the mi = 18 mi, jogger will have run 3 hr # 6 hr which is the area under the line representing the velocity function.

y Velocity (in miles/hour)

The area under the graph of a function can be interpreted in a meaningful way. The units of the area are determined by multiplying the units of the input variable by the units of the output variable.

v(t) = 6

6

A = 3· 6 = 18 miles

0

3

x

Time (in hours)

Business. A company’s marginal revenue is modeled by R¿1x2 = 0.37x,

y Revenue (in thousands of dollars per unit)

Common geometry formulas can sometimes be used to calculate the area.

R⬘(x) = 0.37x

where x is the number of units 1 (100)(37) R=− sold and R¿ is in thousands of 2 = 1850 dollars per unit. The total revenue from selling 100 units 0 100 x is the area under the marginal Number of units sold revenue function. 1 thousands of dollars Total revenue = 1100 units2a37 b 2 unit = 1850 thousand dollars, or $1,850,000.

Chapter 4 Summary

461

KEY TERMS AND CONCEPTS

EXAMPLES

Riemann summation uses rectangles to approximate the area under a curve. The more subintervals (rectangles) used, the more accurate the approximation of the area.

Approximate the area under the graph of f1x2 = - 13 x 2 + 3x over the interval 31, 94 using 4 subintervals. 9 - 1 Each subinterval will have width ¢x = = 2, with x i ranging 4 from x 1 = 1 to x 4 = 7. The area under the curve over 31, 94 is approxi-

If the number of subintervals is allowed to approach infinity, we have a definite integral, which represents the exact area under the graph of a continuous and nonnegative function f1x2 Ú 0 over an interval 3a, b4 :

mated as follows:

# # # # # a f 1x i2 ¢x = f 112 2 + f 132 2 + f 152 2 + f 172 2 4

i=1

20 # 14 # 8# 2 + 6#2 + 2 + 2 3 3 3 = 40. =

b

Exact area =

La

f1x2 dx .

f(x)

Thus, the area is approximately 40 square units.

7 6 5 4 3 2 1

f

1 2 3 4 5 6 7 8 9 10 x

SECTION 4.3 The Fundamental Theorem of Calculus tells us that the exact area under a continuous function f over an interval 3a, b4 is calculated directly using a definite integral: b

f 1x2 dx = F1b2 - F1a2.

The exact area under the graph of f1x2 = - 13 x 2 + 3x over the interval 31, 94 is 9

L1

¢ - x 2 + 3x ≤ dx = c - x 3 + 1 3

1 9

La The function F1x2 is any antiderivative of f 1x2 (we usually set the constant of integration equal to 0).

= ¢=

3 2 9 x d 2 1

1 3 1 3 1923 + 1922 ≤ - ¢ - 1123 + 1122 ≤ 9 2 9 2

352 1 = 39 . 9 9

f(x) 7 6 5 4 3 2 1

f

1 2 3 4 5 6 7 8 9 10 x

(continued)

462

CHAPTER 4

•

Integration

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 4.3 (continued ) The Fundamental Theorem of Calculus is true for all continuous functions over an interval 3a, b4. The definite integral gives the net area between the graph of a continuous function f and the x-axis over an interval 3a, b4 :

Evaluate the definite integral of f1x2 = x 2 - 1 over the interval 3- 1, 14. The function f is negative, therefore, the definite integral will be negative. 1

1 1 1x2 - 12 dx = c x3 - x d 3 -1 L-1

1 1 = a 1123 - 112b - a 1- 123 - 1- 12b 3 3 4 = - . 3

• If f is negative over an interval 3a, b4,

then the definite integral will be negative. • If f has more area above the x-axis than below it over an interval 3a, b4, then the definite integral will be positive. • If f has more area below the x-axis than above it over an interval 3a, b4, then the definite integral will be negative. • If f has equal areas above and below the x-axis over an interval 3a, b4, then the definite integral will be zero.

y 7 6 5 4 3 2 1 −1

−1

f(x) = x 2 − 1

1

2

3

x

Area over [−1, 1]

Evaluate the definite integral of f 1x2 = x 2 - 1 over the interval [0, 3]. The function f has more area above the x-axis than below. For the portion of the graph below the x-axis, we integrate from 0 to 1: 1

1 1 2 1x 2 - 12 dx = c x 3 - x d = - . 3 3 L0 0 For the portion of the graph above the x-axis, we integrate from 1 to 3: 3

3 1 1 20 1x 2 - 12 dx = c x 3 - x d = 19 - 32 - a - 1b = . 3 3 3 L1 1 We sum the two results: 3

3 20 18 1 2 1x 2 - 12 dx = c x 3 - x d = - + = = 6. 3 3 3 3 L0 0 Thus, the net area is 6. We can also integrate from 0 to 3 directly: 3

3 1 1x 2 - 12 dx = c x 3 - x d = 19 - 32 - 102 = 6. 3 L0 0

y 7 6 5 4 3 2 1 −1

−1

f(x) = x 2 − 1

1

2

Area over [0, 3]

3

x

Chapter 4 Summary

KEY TERMS AND CONCEPTS

463

EXAMPLES The function g1x2 = x - 2 over 30, 44 has equal areas below and above the x-axis. Therefore, 4

4 1 1x - 22 dx = c x 2 - 2x d = 8 - 8 = 0. 2 0 L0

y 2

g(x) = x − 2

1

1

2

3

x

4

−1 −2

Area over [0, 4]

SECTION 4.4 There are many useful properties of definite integrals.

The additive property is useful for piecewise-defined functions, which include the absolute-value function.

• Additive property: if a 6 b 6 c, we

3

have c

La

b

f1x2 dx =

La

L-2

c

f1x2 dx +

Lb

0

ƒ x ƒ dx =

L-2

1- x2 dx +

= 2 +

f1x2 dx.

• Area of a region bounded by two

=

curves: if f1x2 Ú g1x2 over an interval 3a, b4, then the area between the graphs of f and g from x = a to x = b is

L0

x dx

9 2

13 . 2 y 4

b

3f1x2 - g1x24 dx . La • Average value: the average value of a function f over an interval 3a, b4 is given by A =

3

f(x) = | x |

2 1

b

yav =

3

1 f1x2 dx. b - a La

−2

−1

1

3 x

2

Let f1x2 = x 2 and g1x2 = x + 2. Setting these expressions equal to one another and solving for x, we find that the curves intersect when x = - 1 and x = 2. Furthermore, we see that g1x2 = x + 2 is the “top” function. Therefore, the area between these curves is 2

A =

2 1 1 1x + 2 - x22 dx = c x2 + 2x - x3 d 2 3 L-1 -1

1 1 1 1 = ¢ 1222 + 2122 - 1223 ≤ - ¢ 1- 122 + 21- 12 - 1- 123 ≤ 2 3 2 3 =

9 . 2

y 4

y=x+2

3 2

y = x2

1 −2

−1

1

2

x

(continued)

464

CHAPTER 4

•

Integration

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 4.4 (continued ) The average value of y = x 3 over the interval 31, 44 is 4

yav =

4 1 1 1 x 3 dx = c x 4 d = 21 . 3 L1 12 4 1

y 60 50

y = x3

40 30

1 y av = 21− 4

20 10 1

2

3

4

x

SECTION 4.5 Integration by substitution is the reverse of applying the Chain Rule. We choose u, determine du, and rewrite the integrand in terms of u. It may be necessary to multiply the integrand by a constant and the entire integral by its reciprocal to obtain the correct form. Results should be checked using differentiation!

Evaluate 1 2x1x2 + 125 dx. We let u = x2 + 1, so du = 2x dx. Therefore, we have

L Evaluate

2x1x 2 + 125 dx =

L

u5 du =

1 6 1 u + C = 1x 2 + 126 + C . 6 6

1 dx. L 3x - 2

We let u = 3x - 2, so du = 3 dx. We multiply the integrand by 3 and the integral by 13 outside the integral: 1 1 1 1 1 dx = 3 dx = du 3 L 3x - 2 3 Lu L 3x - 2 1 1 = ln u + C = ln 13x - 22 + C, where 13x - 22 7 0. 3 3

SECTION 4.6 The Integration-by-Parts Formula is the reverse of the product rule for derivatives: L

u dv = uv -

L

v du

Evaluate 1 x 3 ln x dx. We let u = ln x and dv = x 3 dx. We have du =

1 1 dx and v = x 4 . x 4

Therefore,

L

x 3 ln x dx =

1 4 1 1 x ln x ¢ x 4 ≤ ¢ dx ≤ x 4 4 L

1 1 4 x ln x x 3 dx 4 4L 1 4 1 x + C. = x 4 ln x 4 16

=

(continued)

Chapter 4 Summary

KEY TERMS AND CONCEPTS

EXAMPLES

Tabular integration by parts is useful when this formula has to be applied more than once to evaluate an integral.

Evaluate 1 x 3e 2x dx.

465

This will involve repeated integrations by parts. Since x 3 eventually differentiates to 0 and e 2x is easily integrable, we use tabular integration by parts: f 1x2 and Repeated Derivatives

Sign of Product

g1x2 and Repeated Integrals

x3

1+2

e 2x

3x 2

1- 2

1 2

e 2x

6x

1+2

1 4

e 2x

6

1- 2

1 8

e 2x

1 16

e 2x

0

We multiply along the arrows, alternate signs, and simplify when possible. The antiderivative is L

1 3 2x 3 6 6 2x x e - x 2e 2x + xe 2x e 2 4 8 16 1 3 3 3 = e 2x a x 3 - x 2 + x - b + C. 2 4 4 8

x 3e 2x dx =

SECTION 4.7 Integration tables show formulas for evaluating many general forms of integrals.

Evaluate

3 L 2x + 64 2

dx.

We see that formula 12 from the table of integration formulas on pp. 454–455 is appropriate: 1

dx = ln ƒ x + 2x2 + a2 ƒ + C. L 2x + a2 We bring the constant 3 to the front of the integral, and we note that a 2 = 64, so a = 8 : 2

3 L 2x + 64 2

dx = 3

1 L 2x + 64 2

dx = 3 ln ƒ x + 2x 2 + 64 ƒ + C.

CHAPTER 4 REVIEW EXERCISES These review exercises are for test preparation. They can also be used as a practice test. Answers are at the back of the book. The blue bracketed section references tell you what part(s) of the chapter to restudy if your answer is incorrect.

REVIEW EXERCISES 11. Business: total cost. The marginal cost, in dollars, of producing the xth car stereo is given by C¿1x2 = 0.004x 2 - 2x + 500.

CONCEPT REINFORCEMENT

y Marginal cost (in dollars)

Classify each statement as either true or false. 1. Riemann sums are a way of approximating the area under a curve by using rectangles. [4.2] b

2. If a and b are both negative, then negative. [4.3]

La

f 1x2 dx is

3. For any continuous function f defined over 3- 1, 74, it follows that 2

7

Match each integral in column A with the corresponding antiderivative in column B. [4.1, 4.5] Column A

7. 8. 9. 10.

L

200 100 0

4

a C¿1x i2 ¢x, with ¢x = 50. [4.2]

i=1

Evaluate. [4.1] 12. 14.

11 + 2x2-2 dx

50 100 150 200 250 300 x Number of units produced

Approximate the total cost of producing 200 car stereos by computing the sum

a) ln x + C

dx

C′

300

b) - x-1 + C

L L

20x 4 dx a3t2 + 5t +

13.

L

13e x + 22 dx

1 b dt 1assume t 7 02 t

Find the area under the curve over the indicated interval. [4.3] c) - 11 + x22-1 + C

1 dx, x 7 0 Lx 2x L1 + x 1 Lx

2

2

d) -

dx

e) 2x1>2 + C

dx 2x

L 11 + x 222

1 11 + 2x2-1 + C 2

f) ln 11 + x22 + C

dx

15. y = 4 - x 2 ; 3- 2, 14

16. y = x 2 + 2x + 1; 30, 34 In each case, give an interpretation of the shaded region. [4.2, 4.3] 17.

18. Sales on the tth day

6.

Column B

L 2x

400

Keyboard speed (in words per minute)

5.

500

7

f1x2 dx + f1x2 dx = f1x2 dx. [4.4] L-1 L2 L-1 4. Every integral can be evaluated using integration by parts. [4.6]

1

600

t

Time (in minutes)

466

CHAPTER 4

•

Integration

t Time (in days)

Chapter 4 Review Exercises

Evaluate. [4.3, 4.4] b

19.

x 5 dx

La

20.

1

21.

22.

4

23.

L-2

L-1

1x 3 - x 42 dx

f1x2 dx , where f1x2 = b

2 dx L1 x

x + 2, 2 - 12 2x,

La

for x … 0, for x 7 0

25. y f

f a

b x

b x

a

f

a

x

50. 52.

27. Find the area of the region bounded by y = 3x 2 and y = 9x. [4.4]

28.

30.

L

x e dx ln 14x2

L

2x

29.

dx

31.

L

32. 34.

L L

3xe dx

33.

3x 2 ln x dx

35.

24t 5 L 4t 6 + 3 L

dt

2e -3x dx

3

L L

ln 2x dx 2

x 4e 3x dx

38.

1 L 49 - x 2

dx

x dx 7x + 1 L

49.

L 24 + 5x dx Lx + 9 L

1assume x 7 - 92 51.

ln 17x2 dx

53.

12t 2 L 4t 3 + 7

dt

5

L L L

5x 4e x dx t 71t 8 + 3211 dt x ln 18x2 dx

Evaluate. [4.5–4.7] 54. 56.

t 4 ln 1t 5 + 32

L

58. 60.

t + 3 5

dt

ln 2x dx L x L

ln ¢ 3

L

x - 3 ≤ dx x - 4

x 2x + 3 dx

55. 57.

dx Le + 2 x

L

x 91 ln x dx dx

59.

L x 1 ln x24

61.

x2 dx L 2x + 1

TECHNOLOGY CONNECTION

Evaluate using Table 1. [4.7] 36.

47.

x dx

Evaluate using integration by parts. Do not use Table 1. [4.6] 3x

x 3e 0.1x dx

SYNTHESIS

Evaluate using substitution. Do not use Table 1. [4.5] 3 x4

xe 8x dx

Integrate using any method. [4.3–4.6]

48. b

L

45. Business: total revenue. A company estimates that its revenue will grow continuously at a rate given by the function S¿1t2 = 3e 3t , where S¿1t2 is the rate at which revenue is increasing on the tth day. Find the accumulated revenue for the first 4 days. [4.3]

46. 26. y

41.

44. A particle starts out from the origin. Its velocity in mph after t hours is given by v1t2 = 3t 2 + 2t. Find the distance that the particle travels during the first 4 hr 1from t = 0 to t = 42. [4.3]

f1x2 dx is positive, negative, or zero. [4.3]

24. y

x 6 ln x dx

43. Find the average value of y = xe -x over 30, 24. [4.4]

b

Decide whether

L

42. Business: total cost. Refer to Exercise 11. Calculate the total cost of producing 200 car stereos. [4.4]

4

1e x + x2 dx

L0

40.

1

37. 39.

L

467

x 2e 5x dx dx

L 2x 2 - 36

62. Use a graphing calculator to approximate the area between the following curves: y = 2x 2 - 2x, y = 12x 2 - 12x 3 . [4.4]

CHAPTER 4 TEST 1. Approximate 5

125 - x 22 dx L0 by computing the area of each rectangle and adding. y

Evaluate using substitution. Assume u 7 0 when ln u appears. Do not use Table 1. 13. 15.

25 20

dx x + 12 L L

14.

L

e -0.5x dx

t 31t 4 + 329 dt

Evaluate using integration by parts. Do not use Table 1.

15

16.

10 5 0

2

1

3

4

18.

4.

L L

23x dx ae x +

3.

L

1000x 5 dx

1 + x 3>8 b dx 1assume x 7 02 x

Find the area under the curve over the indicated interval. 5. y = x - x 2 ; 30, 14

4 ; 31, 34 x

6. y =

Running speed (in miles per hour)

9.

L0

e

L0

19.

dx L x17 - x2

21. Find the area of the region in the first quadrant bounded by y = x and y = x 5 . 22. Business: cost from marginal cost. An air conditioning company determines that the marginal cost, in dollars, for the xth air conditioner is given by C¿1x2 = - 0.2x + 500, C102 = 0. Find the total cost of producing 100 air conditioners.

t Time (in hours)

e2

dx

10.

5

11.

L

2 x dx

where W(t) is the speed, in words per minute, at time t. How many words are translated during the second minute (from t = 1 to t = 2)?

12x + 3x 22 dx -2x

x 3 ln x 4 dx

W1t2 = - 6t 2 + 12t + 90, t in 30, 44,

2

1

L

23. Social science: learning curve. A translator’s speed over 4-min interval is given by

Evaluate. L-1

17.

20. Find the average value of y = 4t 3 + 2t over 3- 1, 24.

7. Give an interpretation of the shaded area.

8.

xe 5x dx

Evaluate using Table 1.

5 x

Evaluate. 2.

L

g1x2 dx , where g1x2 = b

dx Le x

x2 , 6 - x,

24. A robot leaving a spacecraft has velocity given by v1t2 = - 0.4t 2 + 2t, where v(t) is in kilometers per hour and t is the number of hours since the robot left the spacecraft. Find the total distance traveled during the first 3 hr.

for x … 2, for x 7 2

b

12. Decide whether

La

f1x2 dx is positive, negative, or zero.

y f b x

a

Integrate using any method. Assume u>0 when ln u appears. 25.

468

CHAPTER 4

•

Integration

6 dx L 5 + 7x

26.

L

x 5e x dx

Extended Technology Application

31.

L

28.

dx

30.

L 64 - x

2

L

L L

2x ln x dx

x + 3 34. ln ¢ ≤ dx x + 5 L

x 4e -0.1x dx

36.

x ln 113x2 dx

31 ln x23 - 41 ln x22 + 54 x

L

1x + 422

L

x2

dx

dx

39. Use a calculator to approximate the area between the following curves:

x 3 2x 2 + 4 dx

y = 3x - x 2 , y = 2x 3 - x 2 - 5x. dx

Extended Technology Application Business: Distribution of Wealth Lorenz Functions and the Gini Coefficient The distribution of wealth within a population is of great interest to many economists and sociologists. Let y = f1x2 represent the percentage of wealth owned by x percent of the population, with x and y expressed as decimals between 0 and 1. The assumptions are that 0% of the population owns 0% of the wealth and that 100% of the population owns 100% of the wealth. With these requirements in place, the Lorenz function is defined to be any continuous, increasing and concave upward function connecting the points 10, 02 and 11, 12, which represent the two extremes. The function is named for economist Max Otto Lorenz (1880–1962), who developed these concepts as a graduate student in 1905–1906. If the collective wealth of a society is equitably distributed among its population, we would observe that (1, 1) Line of equality

(0, 0)

x Percentage of population

Most equitable distribution

y Percentage of wealth

y Percentage of wealth

33.

37.

3 L2 5x - 4

TECHNOLOGY CONNECTION

Evaluate using any method. L

L 23x - 2

dx

8x3 + 10

38. Evaluate 1 5x dx without using Table 1. 1Hint: 5 = e ln 5 .2

SYNTHESIS

32.

x

35.

(0, 0)

(1, 1)

f(x) x Percentage of population

“x% of the population owns x% of the wealth,” and this is modeled by the function f1x2 = x, where 0 … x … 1. This is an example of a Lorenz function that is often called the line of equality. In many societies, the distribution of wealth is not equitable. For example, the Lorenz function f1x2 = x 3 would represent a society in which a large percentage of the population owns a small percentage of the wealth. For example, in this society, we observe that f10.72 = 0.73 = 0.343, meaning that 70% of the population owns just 34.3% of the wealth, with the implication that the other 30% owns the remaining 65.7% of the wealth. In the graphs below, we see the line of equality in the left-most graph, and increasingly inequitable distributions as we move to the right. y

(0, 0)

(1, 1)

f(x) x Percentage of population

y

(1, 1)

Percentage of wealth

29.

6

x 5e x dx

Percentage of wealth

27.

469

f(x)

(0, 0)

x Percentage of population

Less equitable distribution

CHAPTER 4

•

Integration

Note that the area between the line of equality and the graph of the Lorenz function f1x2 is small if the distribution of wealth is close to equitable and is large when the distribution is very unequitable. The Gini coefficient (named for the Italian statistician and demographer Corrado Gini, 1884–1965) is a measure of the difference between the actual distribution of wealth in a society and the ideal distribution represented by the line of equality. It is the ratio of the area between the line of equality and the graph of the Lorenz function to the area below the line of equality and above the x-axis. In the figure that follows, the Gini coefficient is represented by the formula

2. Suppose the Lorenz function for a country is given by f1x2 = x3.5 , 0 … x … 1. a) What percentage of the wealth is owned by 60% of the population? b) Calculate the Gini index. Regression for Determining Lorenz Functions If data exist on the distribution of wealth in a society, a Lorenz function can be determined using regression.

EXERCISES 3. The data in the table show the amount of wealth distributed within a population.

A . Gini coefficient = A + B (1, 1)

y Percentage of wealth

470

(0, 0)

Line of equality A

x

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

y

0.0178

0.06

0.122

0.201

0.297

0.409

0.536

0.677

0.833

Lorenz function, f(x)

B Percentage of population

x

The area A is found by calculating the area between 1 two curves, 10 1x - f1x22 dx, where x is the line of equality and f1x2 is the Lorenz function for a particular society. We observe that A + B is a triangle with area 1 1 2 112112 = 2 . Thus, the Gini coefficient can be written as an integral: 1

Gini coefficient =

A L0 = A + B

1x - f1x22 dx 1 2

¢ ≤ 1

= 2

L0

1x - f1x22 dx.

For the most equitable distribution of wealth, the Gini coefficient would be 0, since there would be no difference (area) between the graph of the Lorenz function and the line of equality; for the most inequitable distribution of wealth, the Gini coefficient would be 1. Often, the Gini coefficient is multiplied by 100 to give the Gini index: a Gini coefficient of 0.34 gives a Gini index of 34.

EXERCISES 1. Suppose the Lorenz function for a country is given by f1x2 = x2 , 0 … x … 1. a) What percentage of the wealth is owned by 60% of the population? b) Calculate the Gini index (the value will be between 0 and 100).

Use regression to determine a power function that best fits these data. (Note: Entering the point 10, 02 may cause an error message to appear. However, the point 11, 12 should be entered along with the rest of the data.) a) Express the Lorenz function in the form f1x2 = x n . The coefficient should be 1, so you may have to do some rounding. b) Determine the Gini coefficient and the Gini index. c) What percentage of the wealth is owned by the lowest 74% of this population?

Extended Technology Application

4. A fast-food chain has many hundreds of franchises nationwide. Ideally, each franchise would generate equal amounts of revenue for the chain, but in reality, some perform better than others. An internal audit reveals the following results: the lowest 30% of the franchises account for just 6% of the total revenue, the lowest 50% account for 20% of the total revenue, and the lowest 70% account for 43.5% of the total revenue. (Assume that 100% of the franchises account for 100% of the total revenue.)

EXERCISES 5. Verify your results for Exercises 3 and 4 using the n - 1 function G1n2 = . n + 1 6. The United States had a Gini index of 45.0 in 2007. (Source: Department of Labor Statistics.) Express this as a decimal: G = 0.45. a) Solve for n, and write the Lorenz function in the form f1x2 = xn . b) According to this model, what percentage of the wealth was owned by the least wealthy 55% of U.S. citizens in 2007? 7. Canada’s Gini index is usually 30.0. a) Determine the Lorenz function. b) What percentage of wealth is owned by the least wealthy 55% of the citizens in Canada? Sometimes, raw data may not fit “neatly” into the f1x2 = xn form, especially in cases where the distribution very heavily favors a small percentage of the population that holds most of the wealth. In these cases, an exponential function of the form f1x2 = a # b x , 0 … x … 1 may work better, as long as the value of a is extremely small.

a) Use regression to determine a power function that models these data, and write the Lorenz function in the form f1x2 = x n . The coefficient should be 1, so you may have to do some rounding. b) Determine the Gini coefficient and the Gini index. c) What percentage of total revenue is generated by the lowest 45% of the franchises? d) What percentage of total revenue is generated by the top 10% of the franchises? Gini Coefficient as a Function of n Functions of the form f1x2 = x n , 0 … x … 1, where n Ú 1, meet the criteria for Lorenz functions. We can develop a function G1n2 that will allow us to calculate the Gini coefficient directly, given a value of n. 1

G1n2 = 2

L0

1x - xn2 dx

1 1 1 = B 2 a x2 xn+1b d 2 n + 1 0

1 1 b 2 n + 1 2 = 1 n + 1 n - 1 = . n + 1 = 2a

471

472

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Integration

8. In 2004, the distribution of net worth within the United States was as given in the following table: PERCENTAGE OF POPULATION

0.4

0.6

0.8

0.9

0.95

0.99

1

PERCENTAGE OF WEALTH

0.002

0.04

0.153

0.287

0.41

0.656

1

(Source: Prof. E. N. Wolff, Levy Institute of Economics at Bard College, 2007.)

a) According to the table, what percentage of net worth was held by the top 1%? (Hint: What percentage did the other 99% hold?) b) Use regression to fit an exponential function g1x2 = a # b x to these data. c) Determine the area between the line of equality and the graph of g1x2 over the interval 30, 14. (Hint: Integrate a # b x using formula 11 from Table 1 in Section 4.7.) d) Determine the Gini coefficient and the Gini index. (Note: E. N. Wolff calculated the Gini coefficient as 0.829.) e) What percentage of the net worth was held by the lowest 50% of the population? f) What percentage of the net worth was held by the top 15% of the population?

Applications of Integration

5

Chapter Snapshot What You’ll Learn

Why It’s Important

5.1

In this chapter, we explore a wide variety of applications of integration to business and economics (consumer and producer surplus), environmental science (exponential growth and decay), and probability and statistics (expected value).We also see how to use integration to find volumes of solids and to solve differential equations.

5.2 5.3 5.4 5.5 5.6 5.7

An Economics Application: Consumer Surplus and Producer Surplus Applications of Integrating Growth and Decay Models Improper Integrals Probability Probability: Expected Value; The Normal Distribution Volume Differential Equations

Where It’s Used BUNGEE JUMPING Regina loves to go bungee jumping. The table shows the number of half-hours that Regina is willing to go bungee jumping at various prices. If Regina goes bungee jumping for 6 half-hours per month, what is her consumer surplus? At a price of $11.50 per halfhour, what is Regina’s consumer surplus? This problem appears as Exercise 21 in Exercise Set 5.1.

REGINA’S DEMAND DATA TIME SPENT (in half-hours per month)

PRICE (per half-hour)

8 7 6 5 4 3 2 1

$ 2.50 5.00 7.50 10.00 12.50 15.00 17.50 20.00

473

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5.1 OBJECTIVE • Given demand and supply functions, find the consumer surplus and the producer surplus at the equilibrium point.

An Economics Application: Consumer Surplus and Producer Surplus It has been convenient to think of demand and supply as quantities that are functions of price. For purposes of this section, we will find it convenient to think of them as prices that are functions of quantity: p = D1x2 and p = S1x2. Indeed, such an interpretation is common in economics. We can use integration to calculate quantities of interest to economists, such as consumer surplus and producer surplus. The consumer’s demand curve is the graph of p = D1x2, which shows the price per unit that the consumer is willing to pay for x units of a product. It is usually a decreasing function since the consumer expects to pay less per unit for large quantities of the product. The producer’s supply curve is the graph of p = S1x2, which shows the price per unit the producer is willing to accept for selling x units. It is usually an increasing function since a higher price per unit is an incentive for the producer to make more units available for sale. The equilibrium point 1x E, pE2 is the intersection of these two curves. p

TECHNOLOGY CONNECTION

$5

EXERCISE

S(x)

1. Graph the demand and supply

4

D1x2 = 1x - 522 and S1x2 = x 2 + x + 3 using the viewing window 30, 5, 0, 304, with Yscl = 5. Find the equilibrium point using the INTERSECT feature.

Price

functions

3 Equilibrium point, (xE, pE)

pE 1

D(x)

5

xE

15

20

25

x

Quantity

Price $10 9 8 7 6 5 0

Consumer surplus (1, $8.75) p  D(x) Total expenditure x 1 2 3 Units per month (average number of movies)

FIGURE 1

Utility is a function often considered in economics. When a consumer receives x units of a product, a certain amount of pleasure, or utility, U, is derived from them (see Exercise 27 in Exercise Set 1.3). For example, the number of movies that you see in a month gives you a certain utility. If you see four movies (unless they are not entertaining), you get more utility than if you see no movies. The same notion applies to having a meal in a restaurant or paying your heating bill to warm your home. To help to explain the concepts of consumer surplus and producer surplus, we will consider the utility of seeing movies over a fixed amount of time, say, 1 month. We are also going to make the assumption that the movies seen are of about the same quality. Samantha is a college student who likes movies. At a price of $10 per ticket, she will see no movies. At a price of $8.75 per ticket, she will see one movie per month, and at a price of $7.50 per ticket, she will see two movies per month. As the price per ticket decreases, Samantha tends to see more movies. As long as the number of movies 1x2 is small, Samantha’s demand function for movies can be modeled by p = 10 - 1.25x. We want to examine the utility she receives from going to the movies. At a ticket price of $8.75, Samantha sees one movie. Her total expenditure is (1) $8.75 = $8.75, as shown by the blue region in Fig. 1. However, the area under Samantha’s demand curve over the interval 30, 14 is $9.38 (rounded). This is what going to one movie per month is worth to Samantha—that is, what she is willing to pay. Since she spent $8.75, the difference in area, represented by the orange triangle, $9.38 - $8.75 = $0.63, can be interpreted as the pleasure Samantha gets, but does not have to pay for, from the one movie. Economists define this amount as the consumer surplus. It is the extra utility that consumers enjoy when prices decrease as more units are purchased.

5.1

Suppose Samantha goes to two movies per month at $7.50 per ticket. Her total expenditure is (2) $7.50 = $15.00, which is represented by the blue region in Fig. 2. The area under Samantha’s demand curve over the interval 30, 24 is $17.50. Therefore, Samantha’s consumer surplus is $2.50, which measures the pleasure Samantha received, but did not have to pay for, from the two movies. Suppose that the graph of a demand Price function is a curve, as shown at the right.

Price $10 9 8 7 6 5 0

475

• An Economics Application: Consumer Surplus and Producer Surplus

Consumer surplus (2, $7.50) p  D(x) Total expenditure

Consumer surplus

x 1 2 3 Units per month (average number of movies)

Total expenditure

FIGURE 2

P

0

D(x) Q

Units

If Samantha goes to Q movies when the price is P, then her total expenditure is QP. The total area under the curve is the total utility, or the total enjoyment received, and is Q

D1x2 dx.

L0

The consumer surplus is the total area under the curve minus the total expenditure. This surplus is the total utility minus the total cost and is given by Q

L0

D1x2 dx - QP.

DEFINITION Suppose that p = D1x2 describes the demand function for a commodity. Then the consumer surplus is defined for the point 1Q, P2 as Q

L0

D1x2 dx - QP.

■ EXAMPLE 1 Find the consumer surplus for the demand function given by D1x2 = 1x - 522 when x = 3.

Solution When x = 3, we have D132 = 13 - 522 = 1- 222 = 4. Then

Price $25 20

Consumer surplus =

Surplus ⫽ $27

15

Expenditure ⫽ 3($4) ⫽ $12

10

(3, 4)

5 0

3

D(x) ⫽ (x ⫺ 5)2

1

2 3 4 Number of units

5

3

=

1x 2 - 10x + 252 dx - 12

L0 3 x3 = c - 5x 2 + 25x d - 12 3 0

x

= ca

Quick Check 1 Find the consumer surplus for the demand function given by D1x2 = x 2 - 6x + 16 when x = 1.

L0

1x - 522 dx - 3 # 4

33 03 - 51322 + 25132b - ¢ - 51022 + 25102b d - 12 3 3

= 19 - 45 + 752 - 0 - 12 = $27.

O Quick Check 1

O

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• Applications of Integration

TECHNOLOGY CONNECTION

Exploratory

Graph D1x2 = 1x - 522, the demand function in Example 1, using the viewing window 30, 5, 0, 304, with Yscl = 5. To find the consumer surplus at x = 3, we first find D132. Then we graph y = D132. What is the point of intersection of y = D1x2 and y = D132? From the intersection, use DRAW to create a vertical line down to the x-axis. What does the area of the resulting rectangle represent? What does the area above the horizontal line and below the curve represent?

Let’s now look at a supply curve for a movie theater, as shown in Figs. 3 and 4. Suppose the movie theater will not sell tickets to a movie for any price at or below $4 (because this would not be enough to cover operating costs and return a profit), but will sell one ticket for one movie at $5.75 or two tickets for two movies at $7.50 each. For small numbers of movies 1x2, the theater’s supply curve is modeled by p = 4 + 1.75x. The price $5.75 is within what Samantha is willing to pay for one movie, and the theater will take in a revenue of (1) $5.75 = $5.75 for selling Samantha one ticket for one movie. The area of the yellow region in Fig. 3 represents the total perperson cost to the theater for showing one movie, which is $4.88 (rounded). Since the theater takes in $5.75 for selling one ticket, the difference, $5.75 - $4.88 = $0.87, represents the surplus over cost and is a contribution toward profit for the theater. Economists call this the producer surplus. It is the benefit a producer receives when supplying more units at a higher price than the price at which the producer expects to sell units. It is the extra revenue the producer receives as a result of not being forced to sell fewer units at a lower price. At a price of $7.50, the theater will show Samantha 2 movies and collect total receipts of 2($7.50), or $15. The area of the yellow region in Fig. 4 represents the total cost to the theater of showing Samantha 2 movies, which is $11.50. The area of the green triangle is $15.00 - $11.50 = $3.50 and is the producer’s surplus. It is a contribution to the theater’s profit. Price

Price $10 9 8 7 6 5 4 0

p  S(x)

Producer surplus (1, $5.75)

Total receipts 2 3 4 5 x Units per month (average number of movies) 1

FIGURE 3

$10 9 8 7 6 5 4 0

p  S(x)

(2, $7.50) Total receipts

2 3 4 5 x Units per month (average number of movies) 1

FIGURE 4

Suppose that the graph of the supply function is a curve, as shown at the right. If the theater shows Samantha Q movies when the price is P, the total receipts are QP. The producer surplus is the total receipts minus the area under the curve and is given by Q

QP -

Producer surplus

L0

S1x2 dx.

Price

Producer surplus S(x)

P

0

Q

Units

DEFINITION Suppose that p = S1x2 is the supply function for a commodity. Then the producer surplus is defined for the point 1Q, P2 as Q

QP -

L0

S1x2 dx.

5.1

• An Economics Application: Consumer Surplus and Producer Surplus

477

■ EXAMPLE 2 Find the producer surplus for S1x2 = x 2 + x + 3 when x = 3. Solution When x = 3, S132 = 32 + 3 + 3 = 15. Then

P

15

(3, 15)

Producer surplus ⫽ $22.50

10 5 0

1

= 45 - c

Total receipts ⫽ 3($15) ⫽ $45 2

3

Consumer surplus

O

0

FIGURE 5

S(x) Producer surplus D(x)

pE xE

3 x x2 + + 3x d 3 2 0

33 32 03 02 + + 3132b - a + + 3102b d 3 2 3 2

= 45 - a9 +

Find the producer surplus for S1x2 = 13 x 2 + 43 x + 4 when x = 1.

(xE, pE)

L0

Units

1x 2 + x + 32 dx

3

= 45 - c a

x

Quick Check 2

Price

3

Producer surplus = 3 # 15 -

S(x) ⫽ x2 ⫹ x ⫹ 3

$20

9 + 9 - 0b 2

= $22.50.

O Quick Check 2 The equilibrium point 1x E, pE2 in Fig. 5 is the point at which the supply and demand curves intersect. It is the point at which sellers and buyers come together and purchases and sales actually occur. Let’s reconsider the example involving Samantha and the movie theater. When the theater charged $5.75 for one ticket, Samantha saw one movie. Since seeing the movie was worth $9.38 to Samantha, she derived $9.38 - $5.75 = $3.63 in utility. To Samantha, this was a very good deal, since she paid much less than she was willing to pay. However, the theater lost potential revenue by “undercharging” Samantha. We see in Fig. 6 that at a price of Price $11 $7.50 per ticket, Samantha’s demand D(x) ⫽ 10 ⫺ 1.25x 10 curve and the theater’s supply curve 9 Consumer intersect. This point is advantageous 8 surplus (2, $7.50) for both Samantha and the theater, 7 since Samantha is willing to see two Producer 6 5 movies at a price of $7.50 per ticket, surplus S(x) ⫽ 4 ⫹ 1.75x 4 while the theater can increase its sur3 plus by selling the two tickets to 2 Samantha. In other words, if the price 1 per ticket is set too low, the theater 0 1 2 3 Units will certainly sell tickets but will lose revenue it could be receiving if the FIGURE 6 price were set slightly higher, since Samantha (and the general population) are willing to pay more according to the demand curve. On the other extreme, if the theater sets the price too high, it simply will not sell enough tickets to make a profit. The $7.50 ticket price is the best “middle ground” for producer (the theater) and consumer (Samantha) alike.

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■ EXAMPLE 3 Given

D1x2 = 1x - 522 and

S1x2 = x 2 + x + 3, Price $30 25

find each of the following. a) The equilibrium point b) The consumer surplus at the equilibrium point c) The producer surplus at the equilibrium point

S(x) = x 2 + x + 3

20 Consumer surplus 9 Producer surplus

D(x) = (x – 5)2

0

1

2

3 Units

4

5

Solution a) To find the equilibrium point, we set D1x2 = S1x2 and solve: 1x - 522 x 2 - 10x + 25 - 10x + 25 22 2

= x2 + x + 3 = = = =

x2 + x + 3 x + 3 11x x.

Thus, x E = 2. To find pE, we substitute x E into either D1x2 or S1x2. If we choose D1x2, we have pE = D1x E2 = = = =

D122 12 - 522 1- 322 $9.

Thus, the equilibrium point is 12, $92. b) The consumer surplus at the equilibrium point is xE

D1x2 dx - x EpE,

L0 2

or

L0

1x - 522 dx - 2 # 9 = B

1x - 523 3 12 - 52

2

R - 18 0

3

= B = =

3

1- 323 3

-

-

10 - 523

1- 523 3

44 L $14.67. 3

c) The producer surplus at the equilibrium point is xE

x EpE -

L0

S1x2 dx,

3

R - 18

- 18 = -

27 125 54 + 3 3 3

x

Exercise Set 5.1

479

or 2#9 -

Quick Check 3

O

Given D1x2 = x 2 - 6x + 16 and S1x2 = 13 x 2 + 43 x + 4, find each of the following. Assume x … 5. a) The equilibrium point b) The consumer surplus at the equilibrium point c) The producer surplus at the equilibrium point

2

L0

1x 2 + x + 32 dx = 2 # 9 - c = 18 - c a = 18 - a =

2 x2 x3 + + 3x d 3 2 0

23 22 03 02 + + 3 # 2b - a + + 3 # 0b d 3 2 3 2

8 + 2 + 6b 3

22 L $7.33. 3

O Quick Check 3

Section Summary • A demand curve is the graph of a function p = D1x2, which represents the unit price p a consumer is willing to pay for x items. It is usually a decreasing function. • A supply curve is the graph of a function p = S1x2, which represents the unit price p a producer is willing to accept for x items. It is usually an increasing function. • Consumer surplus at a point 1Q, P2 is defined as

• The equilibrium point 1x E, pE2 is the point at which the supply and demand curves intersect. The consumer surplus at the equilibrium point is xE

L0

D1x2 - x EpE.

The producer surplus at the equilibrium point is xE

Q

L0

D1x2 dx - QP.

x EpE -

• Producer surplus at a point 1Q, P2 is defined as

L0

S1x2 dx.

Q

QP -

L0

S1x2 dx.

EXERCISE SET

5.1 In each of Exercises 1–14, D1x2 is the price, in dollars per unit, that consumers are willing to pay for x units of an item, and S1x2 is the price, in dollars per unit, that producers are willing to accept for x units. Find (a) the equilibrium point, (b) the consumer surplus at the equilibrium point, and (c) the producer surplus at the equilibrium point. 1. D1x2 = - 56 x + 9, S1x2 = 12 x + 1 2. D1x2 = - 3x + 7, S1x2 = 2x + 2

3. D1x2 = 1x - 422, S1x2 = x 2 + 2x + 6 4. D1x2 = 1x - 322, S1x2 = x 2 + 2x + 1 5. D1x2 = 1x - 622, S1x2 = x 2

11. D1x2 =

100

12. D1x2 =

2x

, S1x2 = 2x

1800 2x + 1

, S1x2 = 22x + 1

13. D1x2 = 1x - 422, S1x2 = x 2 + 2x + 8 14. D1x2 = 13 - x, for 0 … x … 13; S1x2 = 2x + 17

SYNTHESIS For Exercises 15 and 16, follow the directions given for Exercises 1–14.

6. D1x2 = 1x - 822, S1x2 = x 2

15. D1x2 = e -x + 4.5, S1x2 = e x-5.5

7. D1x2 = 1000 - 10x, S1x2 = 250 + 5x

16. D1x2 = 256 - x, S1x2 = x

8. D1x2 = 8800 - 30x, S1x2 = 7000 + 15x

17. Explain why both consumers and producers feel good when consumer and producer surpluses exist.

9. D1x2 = 5 - x, for 0 … x … 5; S1x2 = 2x + 7 10. D1x2 = 7 - x, for 0 … x … 7; S1x2 = 22x + 1

18. Do some research on consumer and producer surpluses in an economics book. Write a brief description.

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•

Applications of Integration

TECHNOLOGY CONNECTION For Exercises 19 and 20, graph each pair of demand and supply functions. Then: a) Find the equilibrium point using the INTERSECT feature or another feature that will allow you to find this point of intersection. b) Graph y  D1x E2 and determine the regions of both consumer and producer surpluses. c) Find the consumer surplus. d) Find the producer surplus. 19. D1x2 =

Time Spent (in half-hours per month)

Price (per half-hour)

8 7 6 5 4 3 2 1

$ 2.50 5.00 7.50 10.00 12.50 15.00 17.50 20.00

x + 8 x2 + 4 , S1x2 = x + 1 20

20. D1x2 = 15 -

1 3 x,

3

S1x2 = 22x a) Make a scatterplot of the data, and determine the type of function that you think fits best. b) Fit that function to the data using REGRESSION. c) If Regina goes bungee jumping for 6 half-hours per month, what is her consumer surplus? d) At a price of $11.50 per half-hour, what is Regina’s consumer surplus?

21. Bungee jumping. Regina loves to go bungee jumping. The table shows the number of half-hours that Regina is willing to go bungee jumping at various prices.

Answers to Quick Checks 1. $2.33

5.2

2. $0.89 3. (a) 12, 82;

(b) $6.67; (c) $4.44

Applications of Integrating Growth and Decay Models

OBJECTIVES

Business and Economics Applications

• Find the future value of an investment.

We studied the exponential growth and decay models provided by the functions P1t2 = P0e kt and P1t2 = P0e - kt in Sections 3.3 and 3.4. Here we consider applications of the integrals of these functions. To ease our later work, let’s find formulas for evaluating these integrals. For the growth model, the formula is

• Find the accumulated future value of a continuous income stream. • Find the present value of an amount due in the future. • Find the accumulated present value of an income stream. • Calculate the total consumption of a natural resource.

T

L0

P0 # kt T e d k 0 P0 kT # = 1e - e k 02 k

P0e kt dt = c

=

Using the substitution u = e kt Evaluating the integral

P0 kT 1e - 12. k T

Similarly, for the decay model, the formula is have the following integration formulas.

L0

P0e - kt dt =

P0 11 - e - kT2. Thus, we k

T

Growth formula:

L0

P0e kt dt =

T

Decay formula:

481

• Applications of Integrating Growth and Decay Models

5.2

L0

P0e - kt dt =

P0 kT 1e - 12 k

(1)

P0 11 - e - kT2 k

(2)

Now let’s consider several applications of these formulas to business and economics.

Future Value Recall the basic model for the growth of an amount of money, presented in the following definition.

DEFINITION If P0 is invested for t years at interest rate k, compounded continuously (Section 3.3), then

(3)

P1t2 = P0e kt,

where P = P0 at t = 0. The value P is called the future value of P0 dollars invested at interest rate k, compounded continuously, for t years.

■ EXAMPLE 1 Business: Future Value of an Investment. Find the future value of $3650 invested for 3 yr at an interest rate of 5%, compounded continuously.

Solution Using equation (3) with P0 = 3650, k = 0.05, and t = 3, we get

Quick Check 1 Business: Future Value of an Investment. Find the future value of $10,000 invested for 3 yr at an interest rate of 6%, compounded continuously.

P132 = = L =

3650e 0.05132 3650e 0.15 365011.1618342 $4240.69.

The future value of $3650 after 3 yr will be about $4240.69.

O Quick Check 1

P Future value $4240.69

$3650 0

3

t

Accumulated Future Value of a Continuous Income Stream Let’s consider a situation involving the accumulation of future values. The owner of a parking space near a convention center receives a yearly profit of $3650 at the end of each of 4 years; this is called an income stream. The owner invests the $3650 at 5% interest compounded continuously. When $3650 is received at the end of the first year, it is invested for 4 - 1, or 3 yr. The future value is $4240.69, as we saw in Example 1. When $3650 is received at the end of the second year, it is invested for 4 - 2, or 2 yr. The future value of this investment is 3650e 0.0514 - 22, or $4,033.87. Note that this future value is less than $4240.69 because the time period is shorter. When $3650 is received at the end of the third year, it is invested for 4 - 3, or 1 yr. That future value is 3650e 0.0514 - 32, or $3837.14, smaller than each of the previous future values. When the last $3650 is received after the fourth year, it is reinvested for 4 - 4, or 0 yr. This

O

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• Applications of Integration

amount has no time to earn interest, so its future value is $3650. The accumulated, or total, future value of the income stream is the sum of the four future values: After 1st yr, t = 3: $3650 3650e 0.05132 $4,240.69 0.05122 After 2nd yr, t = 2: $3650 3650e $4,033.87 0.05112 After 3rd yr, t = 1: $3650 3650e $3,837.14 0.05102 3650e $3,650.00 After 4th yr, t = 0: $3650 Total future value of the income stream = $15,761.70 Next, let’s suppose that the owner of the parking space receives the profit at a rate of $3650 per year but in 365 payments of $10 per day. Each day, when the owner gets $10, it is invested at 5%, compounded continuously, and due at the end of the fourth year. The first day’s investment grows to 3650 0.0514 - 1>3652 e = 10e 0.0514 - 1>3652 L $12.2124, 365

since it will be invested for only 1 day, or 1> 365 yr, less than the full 4 yr. The value of the investment on the second day will grow to 3650 0.0514 - 2>3652 e = 10e 0.0514 - 2>3652 L $12.2107, 365 at the end of the fourth year, and so on, for every day in the 4-yr period. Assuming that all deposits are made into the same account, the total of the future values is 10e 0.0514 - 1>3652 + 10e 0.0514 - 2>3652 + Á + 10e 0.0512>3652 + 10e 0.0511>3652 + 10. Reversing the order of the terms in this sum, we have 10 + 10e 0.0511>3652 + 10e 0.0512>3652 + Á + 10e 0.0514 - 2>3652 + 10e 0.0514 - 1>3652.

1 1 If we express 10 as 3650 # 365 and let ¢t = 365 , then we have a Riemann sum, with t in years, which can be approximated by the definite integral 4

L0

(4)

3650e 0.05t dt.

Let’s further refine how the parking space owner receives profit. First, let’s review the notion of instantaneous rate of change. The speedometer on a car provides an instantaneous speed, or rate of change. If the speedometer reads 58 mph, this means that at that instant the car’s speed is 58 mph, and if the car continues at this speed for 1 hr, it will travel 58 mi. Instead of receiving an income stream at the rate of $10 a day for 4 yr, suppose the owner could receive the money continuously at a rate of $3650 per year for 4 yr. This means that over the course of 4 yr, $3650 in profit will be received at a constant rate of $3650 per year in what is called a continuous income stream, or flow. If at each instant the money is invested at 5%, compounded continuously, then the accumulated future value of the continuous income stream is approximated by the definite integral in equation (4). Let’s calculate that definite integral: 4

L0

3650e 0.05t dt = c

3650 0.05142 1e - 12 d 0.05 L $16,162.40.

Growth formula (1) Approximating using a calculator

Economists call $16,162.40 the accumulated future value of a continuous income stream.

5.2

• Applications of Integrating Growth and Decay Models

483

DEFINITION

Accumulated Future Value of a Continuous Income Stream Let R1t2 be a function that represents the rate, per year, of a continuous income stream, let k be the interest rate, compounded continuously, at which the continuous income stream is invested, and let T be the number of years for which the income stream is invested. Then the accumulated future value of the continuous income stream is given by T

A =

L0

R1t2e kt dt.

(5)

If R1t2 is a constant function, it can be factored out of the integral, and the formula becomes, after evaluating and simplifying, A =

R1t2 k

1e kT - 12.

(6)

If R1t2 is a nonconstant function, then equation (6) does not apply and the integral in equation (5) must be evaluated using some other technique such as integration by parts, tables, a graphing calculator, iPlot, or some other kind of software.

■ EXAMPLE 2 Business: Insurance Settlement. A cardiac surgeon, Sarah Maka-

Quick Check 2

O

Business: Insurance Settlement. Repeat Example 2 but assume that the insurance settlement is a continuous income stream of $125,000 per year for 25 yr and the money is invested at 5%, compounded continuously.

hone, earns an annual income of $450,000 per year but is involved in an automobile accident that injures her legs in such a way that she can no longer stand up to perform heart surgery. In a legal settlement with an insurance company, Sarah is granted a continuous income stream of $225,000 per year for 20 yr, half her normal yearly income since she can practice other kinds of medicine while seated. Sarah invests the money at 3.2%, compounded continuously, in the Halmos Global Equities Fund. Find the accumulated future value of the continuous income stream.

Solution This is an income stream flowing at a constant rate, so we can use equation (6), with R1t2 = $225,000, k = 0.032, and T = 20. We have A =

225,000 0.0321202 1e - 12 L $6,303,381.18. 0.032

O Quick Check 2

TECHNOLOGY CONNECTION

Present Value

Exploratory

We saw in Example 1 that the future value of $3650 invested for 3 yr at a continuously compounded interest rate of 5% is $4240.69. We call $3650 the present value of $4240.69 invested for 3 yr at interest rate 5%, compounded continuously. It answers the question, “What do we have to invest now at a certain interest to attain a certain future value?” (see Section 3.4). In general, the present value P0 of an amount P invested at interest rate k and due t years later is found by solving the growth equation for P0:

When using a calculator to do computations like those in Example 2, enter the equation as shown in the example and do the calculation all at once. If you use some other procedure, the last digits of your answer may differ from what is obtained in the example. Check with your instructor for guidance.

P0e kt = P P P0 = kt = Pe -kt. e

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DEFINITION P

The present value, P0, of an amount P due t years later, at interest rate k, compounded continuously, is given by P0 = Pe -kt.

$4240.69

Present value

Quick Check 3 Business: Finding the Present Value of a Trust. Mira Bell, following the birth of a grandchild, wants to set up a trust fund that will be worth $120,000 on the child’s 18th birthday. Mira can get an interest rate of 5.6%, compounded continuously, for the time period. What amount will Mira have to deposit in the trust fund to achieve her goal?

Future value

P(t) ⫽ 3650e 0.05t

P0 ⫽ $3650 0

3

t

■ EXAMPLE 3 Business: Finding the Present Value of a Trust. In 10 years, Sam Bixby is going to receive $250,000 under the terms of a trust established by his uncle. If the money in the trust fund is invested at 4.8% interest, compounded continuously, what is the present value of Sam’s legacy?

Solution Using the equation for present value given above, we have P0 = 250,000e -0.0481102 L $154,695.85.

O Quick Check 3 Accumulated Present Value of a Continuous Income Stream To find the accumulated present value of a continuous income stream, when R1t2 is constant, we can work backward from equation (6): A =

R1t2 k

1e kT - 12.

We are looking for the principal B, the amount of a one-time deposit, at the interest rate k, that will yield the same accumulated value as the income stream. We choose B such that Be kT = A in equation (6). Then we solve for B: 1e kT - 12 k R1t2 e kT - 1 = a b k e kT

Be kT = Be kT e kT

B = B =

R1t2

Dividing by e kT

R1t2 e kT 1 a kT - kT b k e e R1t2 k

11 - e - kT2.

Simplifying

O

5.2

485

• Applications of Integrating Growth and Decay Models

DEFINITION

Accumulated Present Value of a Continuous Income Stream Let R1t2 be a function that represents the rate, per year, of a continuous income stream, let k be the interest rate, compounded continuously, at which the continuous income stream is invested, and let T be the number of years over which the income stream is received. If R1t2 is a constant function, then B, the accumulated present value of the continuous income stream, is given by B =

R1t2 k

11 - e - kT2.

(7)

If R1t2 is a nonconstant function, the accumulated present value of the continuous income stream is given by the following integral: T

B =

L0

R1t2e -kt dt

(8)

Accumulated present value is a useful tool in business decision making when evaluating a purchase, an investment, or a contract. It brings alternatives and allows for comparisons.

■ EXAMPLE 4 Business: Determining the Value of a Franchise. Silver Spoon, Inc., operates frozen yogurt franchises. Chris Nelson, noting how much he enjoys the yogurt and yearning to be an entrepreneur, considers buying a franchise in his home town, Carmel, Indiana. As part of his decision to purchase, he wants to determine the accumulated present value of the income stream from the franchise over an 8-yr period. Silver Spoon tells Chris that he should expect a constant annual income stream given by R11t2 = $275,000,

which Chris knows he can invest at an interest rate of 5%, compounded continuously. However, Chris took a business calculus course like this one, and he does a linear regression on data from the annual reports of Silver Spoon, which indicates that there will be a nonconstant annual income stream of R21t2 = $80,000t.

a) Evaluate the accumulated future value of the income stream at rate R11t2. Then evaluate the accumulated present value of the income stream, and interpret the results. b) Evaluate the accumulated future value of the income stream at rate R21t2. Then evaluate the accumulated present value of the income stream, and interpret the results. Round all answers to the nearest ten dollars.

Solution a) Chris will have a constant income stream of $275,000 per year for 8 yr. Using equation (6), the accumulated future value is A =

R11t2 k

1e kT - 12 =

275,000 0.05182 1e - 12 L $2,705,040. 0.05

This gives Chris a sense of the value of the franchise over the 8-yr period. The accumulated present value is found by using equation (7): B =

R11t2 k

11 - e -kT2 =

275,000 11 - e -0.051822 L $1,813,240. 0.05

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The first result tells us that if Chris were to buy the franchise now and invest the predicted income stream at 5%, compounded continuously, he would have $2,705,040 in 8 yr. The second result tells us that the first amount is worth $1,813,240 at the present. b) With a nonconstant income stream, R21t2 = 80,000t per year, using equation (5), the accumulated future value is 8

L0

180,000t2e 0.05t = 80,000

8

L0

te 0.05t dt.

To evaluate this integral, we can use any of a variety of integration methods: integration by parts, tables, a graphing calculator, or iPlot. We use Formula 6, from Table 1 in Chapter 4 (p. 454), with a = 0.05 and x = t: L

xe ax dx =

1 # ax 1 # e 0.05t10.05t - 12 e 1ax - 12 + C = 2 0.0025 a = 400e 0.05t10.05t - 12 = 20te 0.05t - 400e 0.05 + C.

Then, te 0.05t dt = 80,000 C 120182e 0.05182 - 400e 0.051822 - 120102e 0.05102 - 400e 0.051022 D

8

80,000

L0

= 80,000 C 1160e 0.4 - 400e 0.42 - 1- 4002 D = 80,0003- 240e 0.4 + 4004 L $3,356,970,

and 8

Quick Check 4 Business: Determining the Value of a Franchise. Repeat Example 4, but with the following income streams: R11t2 = $265,000, R21t2 = 75,000t,

and an interest rate of 8%, compounded continuously.

L0

180,000t2e -0.05t = 80,000

8

L0

te -0.05t dt L $1,969,660.

The first result tells us that if Chris were to buy the franchise now and invest the predicted income stream at 5%, compounded continuously, he would have $3,356,970 in 8 yr. The second result tells us that the first amount is worth $1,969,660 at the present. Chris’s computations yield a higher accumulated present value than that claimed by Silver Spoon, which gives him an indication that he is dealing with a reputable company.

O Quick Check 4 ■ EXAMPLE 5 Business: Creating a College Trust. Emma and Jake Tuttle have a new grandchild, Erica. They want to create a college trust fund for her that will yield $100,000 by her 18th birthday. a) What lump sum would they have to deposit now, in the Hilbert Prime Money Market Fund, at 6% interest, compounded continuously, to yield $100,000? b) They discover that the required lump sum is more than they can afford at the time, so they decide to invest a constant stream of R1t2 dollars per year. Find R1t2 such that the accumulated future value of the continuous money stream is $100,000, assuming that the interest rate is 6%, compounded continuously.

Solution a) The lump sum is the present value of $100,000, at 6% interest, compounded continuously, for 18 yr: P0 = Pe -kt = 100,000e -0.061182 L $33,959.55.

O

• Applications of Integrating Growth and Decay Models

5.2

487

b) We want R1t2 such that 1e 0.061182 - 12 0.06 0.061100,0002 = R1t21e 1.08 - 12 6000 = R1t2 1.08 1e - 12 R1t2

100,000 =

R1t2 L $3085.34

Quick Check 5

O

Business: Creating a College Trust. Repeat Example 5 for a yield of $50,000 and an interest rate of 4%.

Using equation (4)

A continuous money stream of $3085.34 per year, invested at 6%, compounded continuously for 18 yr, will yield a future value of $100,000.

O Quick Check 5 ■ EXAMPLE 6 Business: Contract Buyout. A business executive is working under a contract that pays him $500,000 each year for 5 yr. After 2 yr, the company offers him a buyout of his contract. How much should the company offer him? Assume an annual percentage rate of 4.75%, compounded continuously.

Solution We can view the $500,000 as a continuous money stream. After 2 yr, the contract’s accumulated future value, A2, is A2 =

500,000 0.0475122 1e - 12 L $1,049,040.58. 0.0475

If the contract were allowed to run the full 5 yr, the accumulated future value, A5, would be A5 =

500,000 0.0475152 1e - 12 L $2,821,842.07. 0.0475

The difference is A5 - A2 = $2,821,842.07 - $1,049,040.58 = $1,772,801.49.

Quick Check 6 Business: Contract Buyout. Repeat Example 6 for a $400,000 contract and an interest rate of 3.2%.

Since the company is offering a lump sum payment to buy out the contract, the executive should expect an amount that, if allowed to grow at 4.75%, compounded continuously for the remaining 3 yr, would yield $1,772,801.49. That is, he should receive the present value of the difference, or P0 = 1,772,801.49e -0.0475132 = $1,537,351.39.

O Quick Check 6 Life and Physical Sciences: Consumption of Natural Resources Another application of the integration of models of exponential growth uses P1t2 = P0e kt as a model of the demand for natural resources. Suppose that P0 represents the annual amount of a natural resource (such as coal or oil) used at time t = 0 and that the growth rate for the use of this resource is k. Then, assuming exponential growth in demand (which is the case for the use of many resources), the amount used annually t years in the future is P1t2, given by P1t2 = P0e kt.

The total amount used during an interval 30, T4 is then given by T

L0

P1t2 dt =

T

L0

P0e kt dt = c

P0 kt T P0 kT e d = 1e - 12. k k 0

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Consumption of a Natural Resource Suppose that P1t2 is the annual consumption of a natural resource in year t. If consumption of the resource is growing exponentially at growth rate k, then the total consumption of the resource after T years is given by T

L0

P0e kt dt =

P0 kT 1e - 12, k

(9)

where P0 is the annual consumption at time t = 0.

P P(t) = P0e kt

P0 0

T

t

■ EXAMPLE 7 Physical Science: Gold Mining. In 2000 (t = 0), world gold production was 2547 metric tons, and it was growing exponentially at the rate of 0.6% per year. (Source: U.S. Geological Survey, U.S. Dept. of the Interior, 6/20/06.) If the growth continues at this rate, how many tons of gold will be produced from 2000 to 2013?

Solution Using equation (9), we have 13

L0

2547e 0.006t dt =

2547 0.0061132 1e - 12 0.006

= 424,5001e 0.078 - 12 L 424,50011.081122659 - 12 L 34,437. From 2000 to 2013, approximately 34,437 metric tons of gold will be produced. Quick Check 7 Life and Physical Science: Minerals from Avatar®. The movie Avatar is set in the year 2154 on the moon Pandora, of the planet Polyphemus in the star system of Alpha Centauri. The conflict in the movie is centered around a precious but scarce mineral, Unobtanium. a) In 2010, the universe’s production of Unobtanium was 6800 metric tons and it was being used at the rate of 0.8% per year. If Unobtanium continues to be used at this rate, how many tons of Unobtanium will be used between 2010 and 2024? b) In 2010, the universe’s reserve of Unobtanium was 86,000 metric tons. Assuming that the growth rate of 0.8% per year continues and that no new reserves are discovered, when will the universe reserves of Unobtanium be depleted?

■ EXAMPLE 8 Physical Science: Depletion of Gold Reserves. The world reserves of gold in 2000 were estimated to be 77,000 metric tons. (Source: U.S. Geological Survey, U.S. Dept. of the Interior, 6/20/06; data exclude China.) Assuming that the growth rate for production given in Example 7 continues and that no new reserves are discovered, when will the world reserves of gold be depleted?

Solution Using equation (9), we want to find T such that 77,000 =

2547 0.006T 1e - 12. 0.006

We solve for T as follows: 77,000 0.1814 1.1814 ln 1.1814 ln 1.1814 28

= L L L L L

424,5001e 0.006T - 12 e 0.006T - 1 Dividing both sides by 424,500 0.006T e Taking the natural logarithm of each side ln e 0.006T 0.006T Recall that ln e k = k. T. Dividing both sides by 0.006 and rounding

Thus, assuming that world production of gold continues to increase at 0.6% per year and no new reserves are found (and disregarding Chinese reserves, for which data are unavailable), the world reserves of gold will be depleted 28 yr from 2000, in 2028.

O Quick Check 7

O

Exercise Set 5.2

489

Section Summary • The future value of an investment is given by P = P0e kt, where P0 dollars are invested for t years at interest rate k, compounded continuously. • The accumulated future value of a continuous income stream is given by

• The present value is given by P0 = Pe -kt, where the amount P is due t years later and is invested at interest rate k, compounded continuously. • The accumulated present value of a continuous income stream is given by

T

A =

L0

R1t2e kt dt,

where R1t2 represents the rate of the continuous income stream, k is the interest rate, compounded continuously, at which the continuous income stream is invested, and T is the number of years for which the income stream is invested. • If R1t2 is a constant function, then A =

R1t2 k

1e kT - 12.

T

B =

L0

R1t2e -kt dt,

where R1t2 represents the rate of the continuous income stream, k is the interest rate, compounded continuously, at which the continuous income stream is invested, and T is the number of years over which the income stream is received. • If R1t2 is a constant function, then B =

R1t2 k

11 - e - kT2.

EXERCISE SET

5.2 For all the exercises in this exercise set, use a graphing calculator. Find the future value P of each amount P0 invested for time period t at interest rate k, compounded continuously. 1. P0 = $100,000, t = 6 yr, k = 3% 2. P0 = $55,000, t = 8 yr, k = 4% 3. P0 = $140,000, t = 9 yr, k = 5.8% 4. P0 = $88,000, t = 13 yr, k = 4.7% Find the present value P0 of each amount P due t years in the future and invested at interest rate k, compounded continuously. 5. P = $100,000, t = 6 yr, k = 3% 6. P = $100,000, t = 8 yr, k = 4%

Find the accumulated present value of each continuous income stream at rate R(t), for the given time T and interest rate k, compounded continuously. 13. R 1t2 = $250,000, T = 18 yr, k = 4% 14. R 1t2 = $425,000, T = 15 yr, k = 7% 15. R 1t2 = $800,000, T = 20 yr, k = 8% 16. R 1t2 = $520,000, T = 25 yr, k = 6% 17. R 1t2 = $5200t, T = 18 yr, k = 7% 18. R 1t2 = $6400t, T = 20 yr, k = 4%

19. R 1t2 = $2000t + 7, T = 30 yr, k = 8% 20. R 1t2 = t 2, T = 40 yr, k = 7%

7. P = $1,000,000, t = 25 yr, k = 7%

APPLICATIONS

8. P = $2,000,000, t = 20 yr, k = 9%

Business and Economics

Find the accumulated future value of each continuous income stream at rate R(t), for the given time T and interest rate k, compounded continuously. Round to the nearest $10. 9. R 1t2 = $50,000, T = 22 yr, k = 7%

10. R 1t2 = $125,000, T = 20 yr, k = 6% 11. R 1t2 = $400,000, T = 20 yr, k = 8% 12. R 1t2 = $50,000, T = 22 yr, k = 7%

21. Present value of a trust. In 18 yr, Maggie Oaks is to receive $200,000 under the terms of a trust established by her grandparents. Assuming an interest rate of 5.8%, compounded continuously, what is the present value of Maggie’s legacy? 22. Present value of a trust. In 16 yr, Claire Beasley is to receive $180,000 under the terms of a trust established by her aunt. Assuming an interest rate of 6.2%, compounded continuously, what is the present value of Claire’s legacy?

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23. Salary value. At age 35, Rochelle earns her MBA and accepts a position as vice president of an asphalt company. Assume that she will retire at the age of 65, having received an annual salary of $95,000, and that the interest rate is 6%, compounded continuously. a) What is the accumulated present value of her position? b) What is the accumulated future value of her position? 24. Salary value. At age 25, Del earns his CPA and accepts a position in an accounting firm. Del plans to retire at the age of 65, having received an annual salary of $125,000. Assume an interest rate of 7%, compounded continuously. a) What is the accumulated present value of his position? b) What is the accumulated future value of his position? 25. Future value of an inheritance. Upon the death of his uncle, David receives an inheritance of $50,000, which he invests for 16 yr at 7.3%, compounded continuously. What is the future value of the inheritance? 26. Future value of an inheritance. Upon the death of his aunt, Burt receives an inheritance of $80,000, which he invests for 20 yr at 8.2%, compounded continuously. What is the future value of the inheritance? 27. Decision making. A group of entrepreneurs is considering the purchase of a fast-food franchise. Franchise A predicts that it will bring in a constant revenue stream of $80,000 per year for 10 yr. Franchise B predicts that it will bring in a constant revenue stream of $95,000 per year for 8 yr. Based on a comparison of accumulated present values, which franchise is the better buy, assuming the going interest rate is 6.1%, compounded continuously, and both franchises have the same purchase price? 28. Decision making. A group of entrepreneurs is considering the purchase of a fast-food franchise. Franchise A predicts that it will bring in a constant revenue stream of $120,000 per year for 10 yr. Franchise B predicts that it will bring in a constant revenue stream of $112,000 per year for 8 yr. Based on a comparison of accumulated present values, which franchise is the better buy, assuming the going interest rate is 7.4%, compounded continuously, and both franchises have the same purchase price? 29. Decision making. An athlete attains free agency and is looking for a new team. The Bronco Crunchers offer a salary of 100,000t for 8 yr. The Doppler Radars offer a salary of 83,000t for 9 yr. a) Based on the accumulated present values of the salaries, which team has the better offer, assuming the going interest rate is 6%, compounded continuously? b) What signing bonus should the team with the lower offer give to equalize the offers? 30. Capital outlay. A company determines that the rate of revenue coming in from a new machine is R11t2 = 8000 - 100t,

in dollars per year, for 8 yr, after which the machine will have to be replaced. The company also determines that a different brand of the machine will yield revenue at a rate of R21t2 = 7600 - 85t.

a) Find the accumulated present value of the income stream from each machine at an interest rate of 16%, compounded continuously. b) Find the difference in the accumulated present values. 31. Trust fund. Bob and Ann MacKenzie have a new grandchild, Brenda. They want to create a trust fund for her that will yield $250,000 on her 24th birthday, when she might want to start her own business. a) What lump sum would they have to deposit now at 5.8%, compounded continuously, to achieve $250,000? b) The amount in part (a) is more than they can afford, so they decide to invest a constant money stream of R1t2 dollars per year. Find R1t2 such that the accumulated future value of the continuous money stream is $250,000, assuming an interest rate of 5.8%, compounded continuously. 32. Trust fund. Ted and Edith Markey have a new grandchild, Kurt. They want to create a trust fund for him that will yield $1,000,000 on his 22nd birthday so that he can start his own business when he is out of college. a) What lump sum would they have to deposit now at 6.2%, compounded continuously, to achieve $1,000,000? b) The amount in part (a) is more than they can afford, so they decide to invest a constant money stream of R1t2 dollars per year. Find R1t2 such that the accumulated future value of the continuous money stream is $1,000,000, assuming an interest rate of 6.2%, compounded continuously. 33. Early retirement. Lauren Johnson signs a 10-yr contract as a loan officer for a bank, at a salary of $84,000 per year. After 7 yr, the bank offers her early retirement. What is the least amount the bank should offer Lauren, given that the going interest rate is 7.4%, compounded continuously? 34. Early sports retirement. Tory Johnson signs a 10-yr contract to play for a football team at a salary of $5,000,000 per year. After 6 yr, his skills deteriorate, and the team offers to buy out the rest of his contract so they can drop his name from the roster. What is the least amount Tory should accept for the buyout, given that the going interest rate is 8.2%, compounded continuously? 35. Disability insurance settlement. A movie stuntman receives an annual salary of $180,000 per year, but becomes a quadriplegic after jumping from a cliff into water that is too shallow. He can never work again as a stuntman. Through a legal settlement with an insurance company, he is granted a continuous income stream of $120,000 per year for 20 yr. The stuntman invests the money at 8.2%, compounded continuously.

Exercise Set 5.2

a) Find the accumulated future value of the continuous income stream. Round your answer to the nearest $10. b) Thinking that he might not live 20 yr, the stuntman negotiates a flat sum payment from the insurance company, which is the accumulated present value of the continuous income steam. What is that amount? Round your answer to the nearest $10. 36. Disability insurance settlement. Dale is a furnace maintenance employee who receives an annual salary of $70,000 per year. He becomes partially paralyzed after falling through a ceiling while working on an attic air conditioner. Through a legal settlement with his employer’s insurance company, he is granted a continuous income stream of $40,000 per year for 25 yr. Dale invests the money at 8%, compounded continuously. a) Find the accumulated future value of the continuous income stream. Round your answer to the nearest $10. b) Thinking that he might not live 25 yr more, Dale negotiates a flat sum payment from the insurance company, which is the accumulated present value of the continuous stream plus $100,000. What is that amount? Round your answer to the nearest $10. 37. Lottery winnings and risk analysis. Lucky Larry wins $1,000,000 in a state lottery. The standard way in which a state pays such lottery winnings is at a constant rate of $50,000 per year for 20 yr. a) If Lucky invests each payment from the state at 7%, compounded continuously, what is the accumulated future value of the income stream? Round your answer to the nearest $10. b) What is the accumulated present value of the income stream at 7%, compounded continuously? This amount represents what the state has to invest at the start of its lottery payments, assuming the 7% interest rate holds. c) The risk for Lucky is that he doesn’t know how long he will live or what the future interest rate will be; it might drop or rise, or it could vary considerably over 20 yr. This is the risk he assumes in accepting payments of $50,000 a year over 20 yr. Lucky has taken a course in business calculus so he is aware of the formulas for accumulated future value and present value. He calculates the accumulated present value of the income stream for interest rates of 4%, 6%, 8%, and 10%. What values does he obtain? d) Lucky thinks “a bird in the hand (present value) is worth two in the bush (future value)” and decides to negotiate with the state for immediate payment of his lottery winnings. He asks the state for $600,000. They offer $400,000. Discuss the pros and cons of each amount. Lucky finally accepts $500,000. Is this a good decision? 38. Negotiating a sports contract. Gusto Stick is an excellent professional baseball player who has just become a free agent. His attorney begins negotiations with an interested

491

team by asking for a contract that provides Gusto with an income stream given by R11t2 = 800,000 + 340,000t, over 10 yr, where t is in years. (Round all answers to the nearest $100.) a) What is the accumulated future value of the offer, assuming an interest rate of 8%, compounded continuously? b) What is the accumulated present value of the offer, assuming an interest rate of 8%, compounded continuously? c) The team counters by offering an income stream given by R21t2 = 600,000 + 210,000t. What is the accumulated present value of this counteroffer? d) Gusto comes back with a demand for an income stream given by R31t2 = 1,000,000 + 250,000t. What is the accumulated present value of this income stream? e) Gusto signs a contract for the income stream in part (d) but decides to live on $500,000 each year, investing the rest at 8%, compounded continuously. What is the accumulated future value of the remaining income, assuming an interest rate of 8%, compounded continuously?

Life and Physical Sciences 39. Demand for natural gas. In 2006 1t = 02, the world consumption of natural gas was approximately 101.4 trillion cubic feet and was growing exponentially at about 2.6% per year. (Source: International Energy Outlook 2005, U.S. Energy Information Administration, U.S. Department of Energy.) If the demand continues to grow at this rate, how many cubic feet of natural gas will the world use from 2006 to 2020? 40. Demand for aluminum ore (bauxite). In 2005 1t = 02, bauxite production was approximately 153 million metric tons, and the demand was growing exponentially at a rate of 2.5% per year. (Source: U.S. Energy Information Administration.) If the demand continues to grow at this rate, how many tons of bauxite will the world use from 2005 to 2030? 41. Depletion of natural gas. The world reserves of natural gas were approximately 6112 trillion cubic feet in 2006. (Source: Oil and Gas Journal, Jan. 1, 2006.) Assuming the growth described in Exercise 39 continues and that no new reserves are found, when will the world reserves of natural gas be depleted? 42. Depletion of aluminum ore (bauxite). In 2005, the world reserves of bauxite were about 23 billion metric tons. (Source: U.S. Geological Survey summaries, Jan. 2005.) Assuming that the growth described in Exercise 40 continues and that no new reserves are discovered, when will the world reserves of bauxite be depleted? 43. Demand for and depletion of oil. Between 2006 and 2010, the annual world demand for oil was projected to increase from approximately 30.8 billion barrels to 34.5 billion barrels. (Source: U.S. Department of Energy and Oil and Gas Journal, Jan. 1, 2006.)

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SYNTHESIS Capitalized cost. The capitalized cost, c, of an asset over its lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed by the formula c  c0 

L

m1t2e rt dt,

L0 where c0 is the initial cost of the asset, L is the lifetime (in years), r is the interest rate (compounded continuously), and m(t) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions. 46. c0 = $500,000, r = 5%, m 1t2 = $20,000, L = 20

47. c0 = $400,000, r = 5.5%, m 1t2 = $10,000, L = 25

a) Assuming an exponential growth model, compute the growth rate of demand. b) Predict the demand in 2015. c) The world reserves of crude oil in 2006 were estimated at 1293 billion barrels. Assuming that no new oil is found, when will the reserves be depleted?

48. c0 = $600,000, r = 4%, m 1t2 = $40,000 + $1000e 0.01t, L = 40

49. c0 = $300,000, r = 5%, m 1t2 = $30,000 + $500t, L = 20

The model T

 kt

50. Describe the idea of present value to a friend who is not a business major. Then describe accumulated present value.

P dt  11  e  kT2 k

Pe L0 can be applied to calculate the buildup of a radioactive material that is being released into the atmosphere at a constant annual rate. Some of the material decays, but more continues to be released. The amount present at time T is given by the integral above, where P is the amount released per year and k is the half-life.

51. Look up some data on rate of use and current world reserves of a natural resource not considered in this section. Predict when the world reserves for that resource will be depleted.

Answers to Quick Checks

44. Radioactive buildup. Plutonium-239 has a decay rate of approximately 0.003% per year. Suppose that plutonium239 is released into the atmosphere for 20 yr at a constant rate of 1 lb per year. How much plutonium-239 will be present in the atmosphere after 20 yr?

1. $11,972.17 2. $6,225,857.39 3. $43,793.78 4. With R1, accumulated future value is $2,969,590, and accumulated present value is $1,565,840. With R2, accumulated future value is $3,717,970, and accumulated present value is $1,584,850. 5. $24,337.61, $1896.75 6. $1,219,822.71 7. (a) 100,736 metric tons; (b) 12 yr, or by 2022

45. Radioactive buildup. Cesium-137 has a decay rate of 2.3% per year. Suppose cesium-137 is released into the atmosphere for 20 yr at at rate of 1 lb per year. How much cesium-137 will be present in the atmosphere after 20 yr?

5.3 OBJECTIVES • Determine whether an improper integral is convergent or divergent. • Solve applied problems involving improper integrals.

Improper Integrals Let’s try to find the area of the region under the graph of y = 1>x 2 over the interval 31, q 2. y

y

y = 12 x

1

x

1

b

x

Note that this region is of infinite extent. We have not yet considered how to find the area of such a region. Let’s find the area under the curve over the interval from

• Improper Integrals

5.3

TECHNOLOGY CONNECTION

EXPLORATORY EXERCISES

1 to b, and then see what happens as b gets very large. The area under the graph over 31, b4 is b

dx 1 b = c d 2 x 1 L1 x 1 1 = a- b - a- b b 1

1. Using a graphing calculator or

iPlot, find 10

dx

L1 x 3 100

L1

,

dx x

1000

3

x3

L1

.

2. Predict the value of q

dx

L1 x

3

.

1 + 1 b 1 = 1 - . b = -

, and

dx

493

Then lim 1area from 1 to b2 = lim a1 -

b: q

b: q

1 b = 1. b

We define the area from 1 to infinity to be this limit. Here we have an example of an infinitely long region with a finite area. Such areas may not always be finite. Let’s try to find the area of the region under the graph of y = 1>x over the interval 31, q 2. y

y

y = 1x

A x

1

1

b

x

By definition, the area A from 1 to infinity is the limit as b approaches q of the area from 1 to b, so b

dx = lim 3ln x4b1 b: q L1 x = lim 1ln b - ln 12

A = lim

b: q b: q

= lim ln b. b: q

In Section 3.2, we graphed y = ln x and saw that the function is always increasing. Therefore, the limit lim ln b does not exist and we have an infinitely long region b: q with an infinite area. Note that the graphs of y = 1>x 2 and y = 1>x have similar shapes, but the region under one of them has a finite area and the other does not. An integral such as q

La

f1x2 dx,

with an upper limit of infinity, is an example of an improper integral. Its value is defined to be the following limit.

DEFINITION q

La

b

f1x2 dx = lim

b: q

La

f1x2 dx

494

CHAPTER 5

• Applications of Integration

If the limit exists, then we say that the improper integral converges, or is convergent. If the limit does not exist, then we say that the improper integral diverges, or is divergent. Thus, q

dx = 1 converges, and 2 L1 x

q

dx diverges. L1 x

■ EXAMPLE 1 Determine whether the following integral is convergent or divergent, and calculate its value if it is convergent: q

4e -2x dx.

L0

Solution We have q

b

4e -2x dx = lim

b: q L 0

L0

= lim c b: q

4e -2x dx

4 -2x b e d -2 0

= lim 3- 2e -2x4b0 b: q

#

= lim 3- 2e -2b - 1- 2e -2 024 b: q

= lim 1- 2e -2b + 22 b: q

= lim a2 b: q

2 b. e 2b

As b approaches q , we know that e 2b approaches q (see the graphs of y = a x in Chapter 3), so 2 : 0 and e 2b

Quick Check 1 Determine whether the following integral is convergent or divergent, and calculate its value if it is convergent: q

2 dx. 3 L2 x

q

Thus,

a2 -

2 b : 2. e 2b

4e -2x dx = lim a2 b: q

L0

2 b = 2. e 2b

The integral is convergent.

O Quick Check 1 Following are definitions of two other types of improper integrals.

DEFINITIONS b

1.

L- q

b

f1x2 dx =

q

2.

L- q

lim

a: - q

La

f1x2 dx q

c

f1x2 dx =

L- q

f1x2 dx +

Lc

f1x2 dx,

where c can be any real number. q

In order for 1- q f1x2 dx to converge, both integrals on the right in the second part of the definition must converge.

O

• Improper Integrals

5.3

495

Applications of Improper Integrals In Section 5.2, we learned that the accumulated present value of a continuous money flow (income stream) of P dollars per year, at a constant rate, from now until T years in the future can be found by integration: T

Pe -kt dt =

L0

P 11 - e -kT2, k

where k is the interest rate and interest is compounded continuously. Suppose that the money flow is to continue perpetually (forever). Under this assumption, the accumulated present value of the money flow is q

L0

Pe -kt dt = lim

T

T: q L 0

Pe -kt dt

P 11 - e -kT2 k P 1 P = lim a1 - kT b = . T: q k k e

= lim

T: q

THEOREM 1 The accumulated present value of a continuous money flow into an investment at the constant rate of P dollars per year perpetually is given by q

L0

Pe -kt dt =

P , k

where k is the interest rate and interest is compounded continuously. Quick Check 2

■ EXAMPLE 2 Business: Accumulated Present Value. Find the accumulated present value of an investment for which there is a perpetual continuous money flow of $2000 per year. Assume that the interest rate is 8%, compounded continuously.

Solution The accumulated present value is 2000> 0.08, or $25,000.

O Quick Check 2

TECHNOLOGY CONNECTION

Exploratory

When an amount P of radioactive material is being released into the atmosphere annually, the total amount that has been released at time T is given by

We can explore the situation of Example 2 with a calculator. q To evaluate 10 2000e -0.08x dx, we first consider t

2000 2000e -0.08x dx = 11 - e -0.08t2. 0.08 L0 Then we examine what happens as t gets large. Graph f 1x2 =

2000 11 - e -0.08x2 0.08

using the window 30, 10, 0, 300004, with Xscl = 1 and Yscl = 5000. On the same set of axes, graph y = 25,000. Then change the viewing window to 30, 50, 0, 300004, with Xscl = 10 and Yscl = 5000, and finally to 30, 100, 0, 300004. What happens as x gets larger? What is the significance of 25,000?

T

L0

Pe -kt dt =

P 11 - e -kT2. k

As T approaches q (the radioactive material is released forever), the buildup of radioactive material approaches a limiting value P> k. It is no wonder that scientists and environmentalists are so concerned about radioactive waste. The radioactivity is “here to stay.” y ∞ 0

Pe–kt dt = lim

⎣

⎣

O

Find the accumulated present value of an investment for which there is a perpetual continuous money flow of $10,000 per year. Assume that the interest rate is 6%, compounded continuously.

P (1 – e–kT) = P –– –– ⎣ k

T→∞ ⎣ k

T

496

CHAPTER 5

• Applications of Integration

Section Summary • An improper integral has infinity as one or both of its bounds and is evaluated using the limit: q

b

f1x2 dx = lim

b: q L a

La L- q

q

f1x2 dx,

b

L0

b

f1x2 dx =

q

lim

• The accumulated present value of a continuous money flow into an investment at the rate of P dollars per year perpetually is given by

a: - q L a

f1x2 dx,

Pe -kt dt =

P , k

where k is the interest rate compounded continuously. q

c

f1x2 dx =

f1x2 dx + f1x2 dx, L- q L- q Lc where c is any real number. and

EXERCISE SET

5.3 Determine whether each improper integral is convergent or divergent, and calculate its value if it is convergent. q

1.

q

dx

L2 x

2.

2

q

4.

7.

q

3e q

-3x

dx

6.

8.

L1 x 3

dx 9. 2 + x L0

10.

4 dx 3 + x L0

-2

dx

12.

e dx

14.

L0

x 2 dx

16.

L0

xe x dx

18.

me -mx dx, m 7 0

20.

dt

Lp t L- q

P¿1x2 = 200e -0.032x. Suppose that it were possible for the firm to make infinitely many units of this item. What would its total profit be?

ln x dx

L1 q

1.001

Qe -kt dt, k 7 0

L0

22. 24.

2t

L1 t + 1 2

q

t dt

32. Total profit from marginal profit. Find the total profit in Exercise 31 if P¿1x2 = 200x -1.032, where x Ú 1.

q

q

23.

x 4 dx

L5 q

q

21.

31. Total profit from marginal profit. A firm is able to determine that its marginal profit, in dollars, from producing x units of an item is given by

q

q

19.

dx

e 2x dx

L0

q

17.

-2

q x

L3

7x

L2

q

15.

30. Accumulated present value. Find the accumulated present value of an investment for which there is a perpetual continuous money flow of $3500 per year at an interest rate of 6%, compounded continuously.

q

4x

L0

29. Accumulated present value. Find the accumulated present value of an investment for which there is a perpetual continuous money flow of $3600 per year at an interest rate of 7%, compounded continuously.

q

q

13.

Business and Economics dx

L1 x 4

q

L2

APPLICATIONS

4e -4x dx

L0 q

dx

q

11.

28. Find the area, if it is finite, of the region bounded by y = 1> 213x - 223, x = 6, and 36, q 2.

dx L4 x

q

L0

27. Find the area, if it is finite, of the region bounded by 2 y = 2xe -x , x = 0, and 30, q 2.

L4 x 2 q

dx 3. L3 x 5.

dx

26. Find the area, if it is finite, of the region under the graph of y = 1>x over the interval 32, q 2.

3x 2

33. Total cost from marginal cost. A company determines that its marginal cost, in dollars, for producing x units of a product is given by

dt

L1 1x 3 + 122

dx

25. Find the area, if it is finite, of the region under the graph of y = 1>x 2 over the interval 32, q 2.

C¿1x2 = 3600x -1.8, where x Ú 1. Suppose that it were possible for the company to make infinitely many units of this product. What would the total cost be?

Exercise Set 5.3

34. Total production. A firm determines that it can produce tires at a rate of r1t2 = 2000e -0.42t, where t is the time, in years. Assuming that the firm endures forever (it never gets tired), how many tires can it make? 35. Accumulated present value. Find the accumulated present value of an investment for which there is a perpetual continuous money flow of $5000 per year, assuming continuously compounded interest at a rate of 8%. 36. Accumulated present value. Find the accumulated present value of an investment for which there is a perpetual continuous money flow of $2000e -0.01t per year, assuming continuously compounded interest at a rate of 7%. Capitalized cost. The capitalized cost, c, of an asset for an unlimited lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed by the formula ˆ

c  c0 

L0

m1t2e

rt

42. Suppose that the treatment uses palladium-103, which has a half-life of 16.99 days. a) Find the decay rate, k, of palladium-103. b) How much energy (measured in rems) is transmitted in the first month if the initial rate of transmission is 10 rems per year? c) What is the total amount of energy that the implant will transmit to the body?

SYNTHESIS Determine whether each improper integral is convergent or divergent, and calculate its value if it is convergent. q

43. 45.

L0 x

L0 1x + 123>2 xe -x dx

48.

Radioactive implant treatments. In the treatment of prostate cancer, radioactive implants are often used. The implants are left in the patient and never removed. The amount of energy that is transmitted to the body from the implant is measured in rem units and is given by E

a

L0

P0e kt dt,

L- q

e 2x dx xe -x dx 2

L- q

Life science: drug dosage. Suppose that an oral dose of a drug is taken. Over time, the drug is assimilated in the body and excreted through the urine. The total amount of the drug that has passed through the body in time T is given by T

38. c0 = $700,000, r = 5%, m1t2 = $30,000

40. Radioactive buildup. Cesium-137 has a decay rate of 2.3% per year. Suppose that a nuclear accident causes cesium-137 to be released into the atmosphere perpetually at the rate of 1 lb each year. What is the limiting value of the radioactive buildup?

L1 2x

q

2

L0

dx

0

dx

46.

37. c0 = $500,000, r = 5%, m1t2 = $20,000

39. Radioactive buildup. Plutonium has a decay rate of 0.003% per year. Suppose that a nuclear accident causes plutonium to be released into the atmosphere perpetually at the rate of 1 lb each year. What is the limiting value of the radioactive buildup?

44.

2>3

q

dt,

Life and Physical Sciences

q

dx

q

47.

where c0 is the initial cost of the asset, r is the interest rate (compounded continuously), and m(t) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions.

497

L0

E1t2 dt,

where E(t) is the rate of excretion of the drug. A typical rate-ofexcretion function is E1t2  te kt, where k>0 and t is the time, in hours. Use this information for Exercises 49 and 50. q

49. Find 10 E1t2dt, and interpret the answer. That is, what does the integral represent? 50. A physician prescribes a dosage of 100 mg. Find k. 51. Consider the functions y =

1 x2

and y =

1 . x

Suppose that you go to a paint store to buy paint to cover the region under each graph over the interval 31, q 2. Discuss whether you could be successful and why or why not. 52. Suppose that you are the owner of a building that yields a continuous series of rental payments and you decide to sell the building. Explain how you would use the concept of the accumulated present value of a perpetual continuous money flow to determine a fair selling price.

where k is the decay constant for the radioactive material, a is the number of years since the implant, and P0 is the initial rate at which energy is transmitted. Use this information for Exercises 41 and 42.

TECHNOLOGY CONNECTION

41. Suppose that the treatment uses iodine-125, which has a half-life of 60.1 days. a) Find the decay rate, k, of iodine-125. b) How much energy (measured in rems) is transmitted in the first month if the initial rate of transmission is 10 rems per year? c) What is the total amount of energy that the implant will transmit to the body?

Approximate each integral.

53. Graph the function E and shade the area under the curve for each situation in Exercises 49 and 50. q

54.

4

L1 1 + x

q

55.

dx 2

Answers to Quick Checks 1. Convergent; 14

2. $166,666.67

6 x dx L1 5 + e

498

CHAPTER 5

• Applications of Integration

Probability

5.4 OBJECTIVES

A number between 0 and 1 that represents the likelihood that an event will occur is referred to as the event’s probability. A probability of 0 means that the event is impossible (will not occur), and a probability of 1 means that the event is certain to occur. In this section, we will see that integration is a useful tool for calculating probabilities.

• Verify certain properties of probability density functions.

Experimental and Theoretical Probability There are two types of probability, experimental and theoretical. If we toss a coin a great number of times—say, 1000—and count the number of times we get heads, we can determine the probability of the coin landing heads up. If it lands heads up 503 times, we calculate the probability of it landing heads up to be

• Solve applied problems involving probability density functions.

503 , or 0.503. 1000 This is an experimental determination of probability. Such a determination of probability is discovered by the observation and study of data and is quite common and very useful. Here, for example, are some probabilities that have been determined experimentally: 1. If a person has a heart attack, the probability he or she will die is 13. 2. If you kiss someone who has a cold, the probability of catching the cold is 0.07. 3. A person who has just been released from prison has an 80% probability of returning to prison. If we consider tossing a coin and reason that we are just as likely to get heads as tails, we would calculate the probability of it landing heads up to be 12, or 0.5. This is a theoretical determination of probability. Here, for example, are some probabilities that have been determined theoretically, using mathematics: 1. If there are 30 people in a room, the probability that two of them have the same birthday (excluding year) is 0.706. (See Exercise 42.) 2. While on a trip, you meet someone and, after a period of conversation, discover that you have a common acquaintance. “It’s a small world!” is actually not a very appropriate reaction in this situation, because the probability of such an occurrence is surprisingly high—just over 22%. (This is called the “small world” problem.) In summary, experimental probabilities are determined by making observations and gathering data. Theoretical probabilities are determined by reasoning mathematically. Examples of experimental and theoretical probabilities like those above lead us to see the value of a study of probability. You might ask, “What is true probability?” In fact, there is none. Experimentally, we can determine probabilities within certain limits. These may or may not agree with the probabilities that we obtain theoretically. There are situations in which it is much easier to determine one type of probability than the other. For example, it would be quite difficult to determine the theoretical probability of catching a cold. In the discussion that follows, we will consider primarily theoretical probability. Eventually, calculus will come to bear on our considerations.

A desire to calculate odds in games of chance gave rise to the theory of probability.

Quick Check 1 What is the probability of drawing each of the following from a well-shuffled deck of cards? a) a three b) a heart c) the three of hearts

■ EXAMPLE 1 What is the probability of drawing an ace from a well-shuffled deck of cards?

Solution Since there are 52 possible outcomes, and each card has the same chance

4 1 of being drawn, and since there are 4 aces, the probability of drawing an ace is 52 or 13 , or about 7.7%.

O

Quick Check 1

O

5.4

• Probability

499

■ EXAMPLE 2 A jar contains 7 black balls,

Assume that a jar has the same starting assortment of colored balls as in Example 2 and you remove 1 ball without looking. a) What is the probability that it is black or yellow? b) What is the probability that it is not green?

Solution a) There are 20 balls altogether and of these 3 are 3 red, so the probability of drawing a red ball is 20 . 0 b) There are no white balls, so the probability of drawing a white one is 20 , or 0.

O Quick Check 2 Below is a table of probabilities for the situation in Example 2. Note that the sum of these probabilities is 1. We are certain that we will draw either a black, yellow, green, or red ball. The probability of that event is 1. Let’s arrange these data from the table into what is called a relative frequency graph, or histogram, which shows the proportion of times that each event occurs (the probability of each event). If we assign a width of 1 to each rectangle in this graph, then the sum of the areas of the rectangles is 1. That is, it is certain that you will draw a ball of one of these colors. Color

Probability

Black (B)

7 20

Yellow (Y)

6 20

Green (G)

4 20

Red (R)

3 20

Probabilities

Quick Check 2

6 yellow balls, 4 green balls, and 3 red balls, all the same size and weight. The jar is shaken well, and you remove 1 ball without looking. a) What is the probability that the ball is red? b) What is the probability that it is white?

7 ––– 20 6 ––– 20 5 ––– 20 4 ––– 20 3 ––– 20 2 ––– 20 1 ––– 20

B

Y

G

R

Continuous Random Variables Suppose that we throw a dart at a number line in such a way that it always lands in the interval 31, 34, Let x be the number that the dart hits. There is an infinite number of possibilities for x. Note that x is a quantity that can be observed (or measured) repeatedly and whose possible values comprise an interval of real numbers. Such a variable is called a continuous random variable.

x

0

1

2

3

4

5

Suppose that we throw the dart a large number of times and it lands 43% of the time in the subinterval 31.6, 2.84 of the main interval 31, 34. The probability, then, that the dart lands in the interval 31.6, 2.84 is 0.43. Let’s consider some other examples of continuous random variables.

O

500

CHAPTER 5

• Applications of Integration

■ EXAMPLE 3 Suppose that buses traveling from Philadelphia to New York City require at least 2 hr and at most 5 hr for the trip. If x is the number of hours a bus takes to make the trip, then x is a continuous random variable distributed over the interval 32, 54. [2, 5] 0

1

2

3

4

5

6

■ EXAMPLE 4 Suppose that x is the corn acreage of any farm in the United States or

Canada. The interval is 30, a4, where a is the highest acreage. Not knowing what the highest acreage is, we could use 30, q 2 to allow for all possibilities. Note: It might be argued that there is a value in 30, a4 or 30, q 2 for which no farm has that acreage, but for practical convenience, all values are included in our consideration.

0

[0, a]

a

[0, ∞)

0

(a)

(b)

Then x is a continuous random variable distributed over the interval 30, a4 or 30, q 2. Considering Example 3 on the travel times of buses, suppose that we want to know the probability that a bus will take between 4 hr and 5 hr, as represented by P134, 542, or P14 … x … 52. There may be a function y = f1x2 such that the area under the graph over a subinterval of 32, 54 gives the probability that a particular trip time appears in the subinterval. For example, suppose that we have a constant function f1x2 = 13 that gives us these probabilities. Look at its graph. y 1 –1– 3

1

2

The area under the curve is 3 # 13, or 1. The probability that a trip takes between 4 hr and 5 hr is the area that lies over the interval 34, 54. That is, P134, 542 =

1 3

3

4

x

6

y 1

Area = –13– of 1

–1– 3

= 33 13%.

The probability that a trip takes between 2 hr and 4.5 hr is 56, or 83 13%. This is the area of the rectangle over 32, 4.54. Note that when f1x2 = 13, any interval between the numbers 2 and 5 of width 1 has probability 13. This does not happen for all functions. Suppose instead that 3 2 f1x2 = 117 x .

5

1

2

5

6

x

Area =

–5– 6

3

4 4.5 5

6

x

3

4

y 1

of 1

–1– 3

1

2

5.4

As shown in the graph, the area under the graph of f from 4 to 5 is given by the definite integral over the interval 34, 54 and yields the probability that a trip takes between 4 hr and 5 hr. We have

501

• Probability

y

f

P([3, 4]) = 0.32

5

P134, 542 =

L4

TECHNOLOGY CONNECTION

5

Exploratory

=

3 2 x dx L4 117

=

3 x3 5 c d 117 3 4

Graph the function f 1x2 =

3 2 117 x

using a viewing window of 30, 5, 0, 14. Then successively evaluate each of the following integrals, shading the appropriate area if possible: 3

L2

1

2

3 4 5 6 P([2, 3]) = 0.16

x

5 1 cx 3 d 117 4 1 61 = 153 - 4 32 = L 0.52. 117 117

=

Thus, according to this model, there is a probability of 0.52 that a bus trip takes between 4 hr and 5 hr. The function f is called a probability density function. Its integral over any subinterval gives the probability that x “lands” in that subinterval. Similar calculations for the bus trip example are shown in the following table.

3 2 117 x dx, 4 3 2 117 x dx,

L3

P([4, 5]) = 0.52

f1x2 dx

5

and

L4

3 2 117 x dx.

Trip Time

Add your results, and explain the meaning of the total.

Probability That a Trip Time Occurs during the Interval 3

2 hr to 3 hr

P132, 342 =

3 hr to 4 hr

P133, 442 =

4 hr to 5 hr

P134, 542 =

2 hr to 5 hr

P132, 542 =

L2

3 2 117 x dx

L 0.16

4

L3

3 2 117 x dx

L 0.32

5

y

L4

3 2 117 x dx

L 0.52

5

f P([a, b]) = 1

L2

3 2 117 x dx

= 1.00

The results in the table lead us to the following definition of a probability density function. a

x

b

DEFINITION

FIGURE 1

Let x be a continuous random variable. A function f is said to be a probability density function for x if:

y

f

1. For all x in the domain of f, we have 0 … f1x2. 2. The area under the graph of f is 1 (see Fig. 1). 3. For any subinterval 3c, d4 in the domain of f (see Fig.2), the probability that x will be in that subinterval is given by

P([c, d])

d

a

FIGURE 2

c

d

b

x

P13c, d42 =

Lc

f1x2 dx.

502

CHAPTER 5

• Applications of Integration

■ EXAMPLE 5 Verify that Property 2 of the definition of a probability density function holds for f1x2 =

3 2 x , for 2 … x … 5. 117

Solution 5

3 2 3 1 3 5 x dx = c x d 117 3 L2 117 2

Quick Check 3 Assume that x is a continuous random variable. Verify that 3 g1x2 = 14 2x, for 1 … x … 4, is a probability density function.

=

5 1 cx 3 d 117 2

=

1 153 - 2 32 117

=

117 = 1 117

O Quick Check 3 ■ EXAMPLE 6 Business: Life of a Product. A company that produces compact fluorescent bulbs determines that the life t of a bulb is from 3 to 6 yr and that the probability density function for t is given by f1t2 =

24 , for 3 … t … 6. t3

a) Verify Property 2 of the definition of a probability density function. b) Find the probability that a bulb will last no more than 4 yr. c) Find the probability that a bulb will last at least 4 yr and at most 5 yr.

Solution 6

a) We want to show that

f1t2 dt = 1. We have

L3

24 t -2 6 -3 dt = 24 t dt = 24c d 3 -2 3 L3 t L3 6

6

= - 12c

1 6 1 1 d = - 12a 2 - 2 b 2 t 3 6 3

= - 12a

1 1 3 - b = - 12a - b = 1. 36 9 36

b) The probability that a bulb will last no more than 4 yr is 4

P13 … t … 42 =

4

24 dt = 24 t -3 dt 3 L3 t L3

= 24c

t -2 4 1 4 d = - 12c 2 d -2 3 t 3

= - 12a

1 1 1 1 - 2 b = - 12a - b 2 16 9 4 3

= - 12a -

7 7 b = L 0.58. 144 12

O

5.4

Quick Check 4

O

The time between arrivals of subway trains at a station is modeled by the probability density function h1x2 =

Probability

503

c) The probability that a bulb will last at least 4 yr and at most 5 yr is 5

P14 … t … 52 =

24 dt 3 L4 t 5

t -3dt L4 t -2 5 1 5 = 24c d = - 12c 2 d -2 4 t 4

= 24

10 , for 5 … x … 10, x2

where x is in minutes. Find the probability that: a) the time between trains is between 5 and 7 minutes; b) the time between trains is between 8 and 10 minutes.

•

= - 12a

1 1 1 1 b - 2 b = - 12a 2 25 16 5 4

= - 12a -

27 9 b = = 0.27. 400 100

O Quick Check 4 Constructing Probability Density Functions Suppose that you have an arbitrary nonnegative function f1x2 whose definite integral over some interval 3a, b4 is K. Then b

f1x2 dx = K.

La

Multiplying on both sides by 1> K gives us b

1 1 f1x2 dx = # K = 1, or K La K

b

1# f1x2 dx = 1. La K

Thus, when we multiply the function f1x2 by 1> K, we have a function whose area over the given interval is 1. Such a function satisfies the definition of a probability density function.

■ EXAMPLE 7 Find k such that f1x2 = kx 2

is a probability density function over the interval 31, 44. Then write the probability density function.

Solution We have 4

L1

x 2 dx = c =

x3 4 d 3 1

43 13 64 1 63 = = = 21. 3 3 3 3 3 4

Quick Check 5 k is x a probability density function over the interval 32, 74. Then write the probability density function. Find k such that g1x2 =

Thus, for f to be a probability density function, we must have 11 kx 2 dx = 1, or 4 k 11 x 2 dx = 1, and 1 k = 21 . The probability density function is f1x2 =

O Quick Check 5

1 2 21 x ,

for 1 … x … 4.

O

504

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• Applications of Integration

Uniform Distributions Suppose that the probability density function of a continuous random variable is constant. How is it described? Consider the graph shown below. y 1 –1– 3

1 y

1 f(x) = ––—– b–a

1 ––—– b–a

a

b

x

3

2

4

5

6

x

The length of the shaded rectangle is the length of the interval 32, 54, which is 3. In order for the shaded area to be 1, the height of the rectangle must be 31. Thus, f1x2 = 13. The length of the shaded rectangle shown at the left is the length of the interval 3a, b4, which is b - a. In order for the shaded area to be 1, the height of the rectangle must be 1>1b - a2. Thus, f1x2 = 1>1b - a2.

DEFINITION A continuous random variable x is said to be uniformly distributed over an interval 3a, b4 if it has a probability density function f given by f1x2 =

1 , for a … x … b. b - a

■ EXAMPLE 8 A number x is selected at random from the interval 340, 504. The probability density function for x is given by f1x2 =

1 10 ,

for 40 … x … 50.

Find the probability that a number selected is in the subinterval 342, 484.

Solution The probability is 48

P142 … x … 482 = =

L42

1 10

1 10 148

dx =

1 48 10 3x442

- 422 =

6 10

= 0.6.

■ EXAMPLE 9 Business: Quality Control. A company produces sirens used for Quick Check 6 Business: Quality Control. The probability density function for the weight x, in pounds, of bags of feed sold at a feedstore is f1x2 = 18, for 45 … x … 53. A bag is selected at random. Find the probability the bag weighs between 47.5 and 50.25 lb.

tornado warnings. The maximum loudness, L, of the sirens ranges from 70 to 100 decibels. The probability density function for L is f1L2 =

1 30 ,

for 70 … L … 100.

A siren is selected at random off the assembly line. Find the probability that its maximum loudness is from 70 to 92 decibels.

Solution The probability is 92

P170 … L … 922 = =

L70

1 30 dL

1 30 192

=

1 92 30 3L470

- 702 =

22 30

=

11 15

L 0.73.

O Quick Check 6 Exponential Distributions The duration of a phone call, the distance between successive cars on a highway, and the amount of time required to learn a task are all examples of exponentially distributed random variables. That is, their probability density functions are exponential.

O

5.4

• Probability

505

DEFINITION A continuous random variable is exponentially distributed if it has a probability density function of the form f1x2 = ke -kx, over the interval 30, q 2.

To see that f1x2 = 2e -2x is such a probability density function, note that q

L0

2e -2x dx = lim

b

b: q L 0

2e -2x dx = lim 3- e -2x4b0 = lim a b: q

b: q

-1 - 1- 12b = 1. e 2b

The general case, q

L0

y

ke -kx dx = 1,

k

can be verified in a similar way. f(x) = ke – kx Why is it reasonable to assume that the distance between cars is exponentially distributed? Part of the reason is x that there are many more cases in which distances are small, though we can find other distributions that are “skewed” in this manner. The same argument holds for the duration of a phone call. That is, there are more short calls than long ones. The rest of the reason might lie in an analysis of the data involving such distances or phone calls.

■ EXAMPLE 10 Business: Transportation Planning. The distance x, in feet, between successive cars on a certain stretch of highway has a probability density function f1x2 = ke -kx, for 0 … x 6 q , where k = 1>a and a is the average distance between successive cars over some period of time. A transportation planner determines that the average distance between cars on a certain stretch of highway is 166 ft. What is the probability that the distance between two successive cars, chosen at random, is 50 ft or less?

Solution

A transportation planner can determine the probabilities that cars are certain distances apart.

Quick Check 7

1 k = 166 L 0.006024.

The probability density function for x is f1x2 = 0.006024e -0.006024x, for 0 … x 6 q . The probability that the distance between the cars is 50 ft or less is

The response time for a paramedic unit has the probability density function f1t2 = 0.05256e for 1 … t 6 q ,

We first determine k:

50

P10 … x … 502 =

,

where t is in minutes. Find the probability of each response time: a) between 5 and 10 min; b) between 8 and 20 min; c) between 1 and 45 min; d) more than 60 min.

L0

= c

-0.05t

0.006024e -0.006024x dx

0.006024 -0.006024x 50 e d - 0.006024 0

= C - e -0.006024x D 0

50

#

#

= 1- e -0.006024 502 - 1- e -0.006024 02

= - e -0.301200 + 1 = 1 - e -0.301200 = 1 - 0.739930 L 0.260.

O Quick Check 7

O

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• Applications of Integration

Section Summary • The probability of an event is a number between 0 and 1, with 0 meaning that the event is impossible and 1 meaning that the event is certain. • Probabilities may be determined experimentally (by conducting trials) or theoretically (by reasoning). • A continuous random variable is a quantity that can be observed (or measured) and whose possible values comprise an interval of real numbers. • If x is a continuous random variable, then function f is a probability density function for x if it meets the following criteria: (1) For all x in 3a, b4, 0 … f1x2.

(2) The area under the graph of f over 3a, b4 is 1; that is, b 1a f1x2 dx = 1. (3) The probability that x is within the subinterval 3c, d4 d is given by P13c, d42 = 1c f1x2 dx. • A continuous random variable x is uniformly distributed over an interval 3a, b4 if its probability density function 1 has the form f1x2 = over the interval. b - a • A continuous random variable x is exponentially distributed over an interval 30, q 2 if its probability density function has the form f1x2 = ke -kx over the interval.

EXERCISE SET

5.4 In Exercises 1–12, verify Property 2 of the definition of a probability density function over the given interval. 1. f 1x2 = 14 x, 31, 34

2. f 1x2 = 2x, 30, 14

3. f 1x2 = 3,

4. f 1x2 = 15, 33, 84

5. f 1x2 =

C 0, 13 D

3 2 64 x ,

30, 44

6. f 1x2 =

3 2 26 x ,

31, 34

1 , 31, e4 x 1 8. f 1x2 = e x, 30, 14 e - 1 7. f 1x2 =

12. f 1x2 = 4e -4x, 30, q 2

Find k such that each function is a probability density function over the given interval. Then write the probability density function. 13. f 1x2 = kx, 32, 54

14. f 1x2 = kx, 31, 44

15. f 1x2 = kx 2, 3- 1, 14

16. f 1x2 = kx 2, 3- 2, 24

17. f 1x2 = k, 31, 74

18. f 1x2 = k, 33, 94

19. f 1x2 = k12 - x2, 30, 24

23. f 1x2 = ke x, 30, 34

for 0 … x … 10.

3 2 125 x ,

for 0 … x … 5.

a) Find P11 … x … 42, the probability that the dart lands in 31, 44. b) Interpret your answer to part (a). 27. A number x is selected at random from the interval 34, 204. The probability density function for x is given by f 1x2 =

1 16 ,

for 4 … x … 20.

Find the probability that a number selected is in the subinterval 39, 204. 28. A number x is selected at random from the interval 35, 294. The probability density function for x is given by f 1x2 =

1 24 ,

for 5 … x … 29.

Find the probability that a number selected is in the subinterval 314, 294.

20. f 1x2 = k14 - x2, 30, 44 k , 31, 34 x

1 50 x,

a) Find P12 … x … 62, the probability that the dart lands in 32, 64. b) Interpret your answer to part (a).

f 1x2 =

10. f 1x2 = 13 x 2, 3- 2, 14

21. f 1x2 =

f 1x2 =

26. In Exercise 25, suppose that the dart always lands in the interval 30, 54, and that the probability density function for x is given by

9. f 1x2 = 32 x 2, 3- 1, 14 11. f 1x2 = 3e -3x, 30, q 2

25. A dart is thrown at a number line in such a way that it always lands in the interval 30, 104. Let x represent the number that the dart hits. Suppose that the probability density function for x is given by

22. f 1x2 =

k , 31, 24 x

24. f 1x2 = ke x, 30, 24

APPLICATIONS Business and Economics 29. Transportation planning. Refer to Example 10. A transportation planner determines that the average distance between cars on a certain highway is 100 ft. What is the

Exercise Set 5.4

30. Transportation planning. Refer to Example 10. A transportation planner determines that the average distance between cars on a certain highway is 200 ft. What is the probability that the distance between two successive cars, chosen at random, is 10 ft or less? 31. Duration of a phone call. A telephone company determines that the duration t, in minutes, of a phone call is an exponentially distributed random variable with a probability density function f 1t2 = 2e -2t, 0 … t 6 q . Find the probability that a phone call will last no more than 5 min. 32. Duration of a phone call. Referring to Exercise 31, find the probability that a phone call will last no more than 2 min. 33. Time to failure. The time to failure, t, in hours, of a machine is often exponentially distributed with a probability density function f1t2 = ke -kt, 0 … t 6 q , where k = 1>a and a is the average amount of time that will pass before a failure occurs. Suppose that the average amount of time that will pass before a failure occurs is 100 hr. What is the probability that a failure will occur in 50 hr or less? 34. Reliability of a machine. The reliability of the machine (the probability that it will work) in Exercise 33 is defined as T

R1T2 = 1 -

L0

0.01e -0.01t dt,

where R 1T2 is the reliability at time T. Write R 1T2 without using an integral.

a) Find a value k to make the function kf 1x2 a probability density function. b) Use your answer from part (a) to approximate the probability that a female aged 1 through 85 who died in 2003 was between 25 and 40 years old. 36. Mortality rate. For every 100,000 males in the United States of any age x between 1 and 85, the number that died in 2003 can be approximated by the function f 1x2 = 17.359e 0.067x.

(Source: Centers for Disease Control and Prevention.) f(x) = 17.359 e 0.067x

y 3500

Number of deaths (per 100,000 males)

probability that the distance between two successive cars, chosen at random, is 40 ft or less?

507

3000 2500 2000 1500 1000 500 0

20

40

60

80

x

Age

a) Find a value k to make the function kf 1x2 a probability density function. b) Use your answer from part (a) to approximate the probability that a male aged 1 through 85 who died in 2003 was between 20 and 30 years old.

Social Sciences 37. Time in a maze. In a psychology experiment, the time t, in seconds, that it takes a rat to learn its way through a maze is an exponentially distributed random variable with the probability density function f 1t2 = 0.02e -0.02t, 0 … t 6 q . Find the probability that a rat will learn its way through a maze in 150 sec or less.

Life and Physical Sciences 35. Mortality rate. For every 100,000 females in the United States of any age x between 1 and 85, the number that died in 2003 can be approximated by the function f 1x2 = 8.1305e 0.074x.

(Source: Centers for Disease Control and Prevention.) f(x) = 8.1305 e 0.074x

y Number of deaths (per 100,000 females)

3500 3000 2500 2000 1500

The time that it takes a rat to learn its way through a maze is an exponentially distributed random variable.

1000 500 0

20

40

60

Age

80

x

38. Time in a maze. Using the situation and the equation in Exercise 37, find the probability that a rat will learn its way through the maze in 50 sec or less.

508

CHAPTER 5

•

Applications of Integration

SYNTHESIS

a) Verify the claim made at the start of this section that the probability that two people in a group of 30 have the same birthday is about 70%. b) How many people are required to make the probability that two of them share a birthday greater than 50%? c) How many people are in your calculus class? What is the probability of at least one shared birthday among you and your classmates? Test your calculation experimentally.

39. The function f 1x2 = x 3 is a probability density function over 30, b4. What is b? 40. The function f 1x2 = 12x 2 is a probability density function over 3- a, a4. What is a?

41. Use your answer to Exercise 37 to find the probability that a rat requires more than 150 sec to learn its way through the maze.

TECHNOLOGY CONNECTION

42. The birthday problem. Assume that birthdays are uniformly distributed throughout the year and that February 29 is omitted from consideration. The probability of at least one shared birthday (month and day only) among n randomly chosen people is 364 # 363 # 362 # Á # 1366 - n2 P1n2 = 1 - a b. 365n - 1

43–54. Verify Property 2 of the definition of a probability density function for each of the functions in Exercises 1–12.

Answers to Quick Checks 13 4 1 1. (a) 52 or 13 ; (b) 52 or 14; 4

5.5

3 14 2x dx

3.

For example, in a group of 10 people the probability of at least one shared birthday is 364 # 363 # 362 # Á # 356 P1102 = 1 - a b = 0.117. 3659

=

C

D =

3 2 3>2 4 14 3 x 1

L1 g1x2 Ú 0 on 31, 44 4. (a) 0.57; (b) 0.25

13 2. (a) 20 ;

1 (c) 52

A

3 16 14 3

-

2 3

B = 1;

4 (b) 16 20 or 5

0.7982 , for 2 … x … 7 x 6. 0.34375 7. (a) 0.181; (b) 0.318; (c) 0.889; (d) 0.053 5. k = 0.7982; g1x2 =

Probability: Expected Value; The Normal Distribution Expected Value

OBJECTIVES

Let’s again consider throwing a dart at a number line in such a way that it always lands in the interval 31, 34. This time we assume a uniform distribution so that it is equally likely that the dart will land anywhere in the interval.

2

• Find E(x), E(x ), the mean, the variance, and the standard deviation. • Evaluate normal distribution probabilities using a table. • Calculate percentiles for a normal distribution.

0

1

2

3

4

5

Suppose that we throw the dart at the line 100 times and keep track of the numbers it hits.Then we calculate the arithmetic mean (or average) x of all these numbers: 100

y

1 1 2

a xi 100 x 1 + x 2 + x 3 + Á + x 100 1 i=1 = = a xi # . 100 100 100 i=1

x =

2

f( x) =

1

1 , 2

Assuming that the x i’s are uniformly distributed over 31, 34, as shown to the left,

1≤x≤3

2

3

#1 a x i n , or i=1 n

4

x

is analogous to 3

L1

x # f1x2 dx,

#1 2 a ax i 2 b n , i=1 n

Note that the interval width is 2.

5.5

•

Probability: Expected Value; The Normal Distribution

509

where f1x2 = 1>2 is a probability density function for x. Because the width of 31, 34 is 2, we can regard 2>n as ¢x. The probability density function gives a “weight” to x. We add all the values of ax i #

1 2 ba b n 2

n 1 2 when we find a ax i # b a b. Similarly, we add all the values of n 2 i=1

1x # f1x221¢x2

when we find 11 x # f1x2 dx : 3

3

L1

y 0.9 0.8 0.7 0.5

1 1 x2 3 1 dx = c d = 332 - 124 = 2. 2 2 2 1 4

This result, representing the average value of the distribution, is not surprising, since the distribution is uniform and centered around 2. Although the average of 100 dart throws may not be 2, as n : q , we will have x : 2. Suppose that we use the probability density function f1x2 = 14 x over the interval 31, 34. As we can see from the graph, this function gives more “weight” to the right side of the interval than to the left. Perhaps more points are awarded if a dart lands on the right. Then we have

1.0

0.6

x#

3

f(x) = 1 –x 4

L1

0.4

x # f1x2 dx =

L1

x#

1 x dx 4

3

0.3 0.2

=

1 x 2 dx 4 L1

=

1 x3 3 c d 4 3 1

=

1 26 133 - 132 = L 2.17. 12 12

0.1 –1

3

1

2

3

4

x

2.17

Suppose that we continue to throw the dart and compute averages. The more times we throw the dart, the closer we expect the averages to come to 2.17.

DEFINITION

Let x be a continuous random variable over the interval 3a, b4 with probability density function f. The expected value of x is defined by b

E1x2 =

La

x # f1x2 dx.

The concept of the expected value of a random variable can be generalized to functions of the random variable. Suppose that y = g1x2 is a function of the random variable x. Then we have the following.

510

CHAPTER 5

•

Applications of Integration

DEFINITION The expected value of g1x2 is defined by b

E1g1x22 =

La

g1x2 # f1x2 dx,

where f is a probability density function for x.

■ EXAMPLE 1 Given the probability density function f1x2 =

1 2

x, over 30, 24,

find E1x2 and E1x 22.

Solution 2

2

x#

1 1 2 x dx = x dx 2 L0 L0 2 2 1 x3 2 1 = c d = cx 3 d 2 3 0 6 0

E1x2 =

=

1 3 12 - 032 6

=

1 # 4 8 = ; 6 3

E1x 2 =

2

2

Quick Check 1 Given the probability density function f1x2 =

1 2

-

1 8

x, over 30, 44,

find E1x2 and E 1x 22.

L0

x

2

# 1 x dx = 2

2

1 3 1 x4 2 x dx = c d 2 4 0 L0 2

=

1 4 2 1 cx d = 12 4 - 042 8 8 0

=

1# 16 = 2 8

O Quick Check 1 DEFINITION The mean, M , of a continuous random variable x is defined to be E1x2. That is, b

m = E1x2 =

La

xf1x2 dx,

where f is a probability density function for x defined over 3a, b4. (The symbol m is the lowercase Greek letter mu.)

O

5.5

y ⫽ f(x)

a

•

Probability: Expected Value; The Normal Distribution

We can get a physical idea of the mean of a random variable by pasting the graph of the probability density function on cardboard and cutting out the area under the curve over the interval 3a, b4. Then we try to find a balance point on the x-axis. That balance point is the mean, m.

y ⫽ f(x)

μ

b

x

a

x

b

μ

511

Variance and Standard Deviation Because two very different distributions can have the same mean, it is useful to have a second statistic that serves as a measure of how the data in a distribution are spread out. Statistics that provide such a measure are the variance and (especially) the standard deviation of a distribution. A full derivation of how these statistics were developed is beyond the scope of this book.

DEFINITION

The variance, S2, of a continuous random variable x, defined on 3a, b4, with probability density function f, is s2 = E1x 22 - m2 = E1x 22 - 3E1x242 b

=

La

x 2f1x2 dx - c

b

La

2

xf1x2 dx d .

The standard deviation, S, of a continuous random variable is defined as s = 2variance. (The symbol s is the lowercase Greek letter sigma.)

■ EXAMPLE 2 Given the probability density function f1x2 =

1 2

x, over 30, 24,

find the mean, the variance, and the standard deviation.

Solution From Example 1, we have E1x2 = 43 and E1x 22 = 2. Thus, mean = m = E1x2 = 43;

variance = s2 = E1x 22 - 3E1x242 Quick Check 2

= 2 -

Given the probability density function

=

f1x2 =

1 2

-

1 8

x, over 30, 44,

find the mean, the variance and the standard deviation.

18 9

-

standard deviation = s = =

O Quick Check 2

1 3

A 43 B 2 = 2 16 9

= 29;

29 2

22 L 0.47.

16 9

O

512

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•

Applications of Integration

Loosely speaking, the standard deviation is a measure of how closely bunched the graph of f is, that is, how far the points on f are, on average, from the line x = m, as indicated below. y

y

σ, small σ, large μ

a

b x

μ

a

b x

The Normal Distribution Suppose that the average score on a test is 70. Usually there are about as many scores above the average as there are below the average; and the farther away from the average a particular score is, the fewer people there are who get that score. On this test, it is probable that more people scored in the 80s than in the 90s, and more people scored in the 60s than in the 50s. Test scores, heights of human beings, and weights of human beings are all examples of random variables that are often normally distributed. Consider the function g1x2 = e -x >2, over the interval 1- q , q 2. 2

TECHNOLOGY CONNECTION

Exploratory Use a graphing calculator or iPlot to confirm that q

e -x >2 dx = 22p, 2

L- q

letting y1 = e -x >2. 2

1. Set the limits of integration at

- 2 and 2.

2. Repeat, setting the limits at

This function has the entire set of real numbers as its domain. Its graph is the bellshaped curve shown below. We can find function values by using a calculator: x

g1x2

0 1 2 3 -1 -2 -3

1 0.6 0.1 0.01 0.6 0.1 0.01

y 1

g(x) = e – x

2 /2

0.5

–3

–2

–1

1

2

3

x

- 3 and 3.

3. Compare your results with

22p L 2.507.

This function has an antiderivative, but that antiderivative has no basic integration formula. Nevertheless, it can be shown that the improper integral converges over the interval 1- q , q 2 to a number given by q

e -x >2 dx = 22p. 2

L- q

That is, although an elementary expression for the antiderivative cannot be found, there is a numerical value for the improper integral evaluated over the set of real numbers. Note that since the area is not 1, the function g is not a probability density function, but the following function is: f1x2 =

1 22p

e -x >2. 2

•

5.5

513

Probability: Expected Value; The Normal Distribution

DEFINITION A continuous random variable x has a standard normal distribution if its probability density function is f1x2 =

1

e -x >2, over 1- q , q 2. 2

22p

This standard normal distribution has a mean of 0 and a standard deviation of 1. Its graph follows. y 1

TECHNOLOGY CONNECTION 1 ≈ 0.4 √2π

Exploratory Use a graphing calculator or iPlot to approximate b

1

σ =1 2 f(x) = 1 e –x / 2 √2π

e -x >2 dx 2

L-b 22p

–3

–2

–1

1

2

3

x

μ=0

for b = 10, 100, and 1000. What does this suggest about q

1

The general case is defined as follows.

e -x >2 dx? 2

L- q 22p

DEFINITION

This is a way to verify part of the assertion that f 1x2 =

1 22p

A continuous random variable x is normally distributed with mean m and standard deviation s if its probability density function is given by

e -x >2 2

is a probability density function. Use a similar approximation procedure to show that the mean is 0 and the standard deviation is 1.

f1x2 =

1 s22p

e -11>2231x - m2>s4 , over 1- q , q 2. 2

The graph of any normal distribution is a transformation of the graph of the standard normal distribution. This can be shown by translating the graph of a normal distribution along the x-axis and adjusting how tightly clustered the graph is about the mean. Some examples follow. μ = –2 μ = 0 μ = 2 σ =1 σ =1 σ =1

μ =0 σ = 0.5 μ =0 σ =1

–2

2 1 Normal distributions with same standard deviations but different means –1

0

–2

–1

0

1

μ =0 σ =2 2

Normal distributions with same means but different standard deviations

514

CHAPTER 5

•

Applications of Integration

The normal distribution is extremely important in statistics; it underlies much of the research in the behavioral and social sciences. Because of this, tables of approximate values of the definite integral of the standard normal distribution have been prepared using numerical approximation methods like the Trapezoidal Rule given in Exercise Set 4.1. Table A at the back of the book (p. 621) is such a table. It contains values of z

P10 … x … z2 =

1

L0 22p

e -x >2 dx. 2

The symmetry of the graph of this function about the mean allows many types of probabilities to be computed from the table. Some involve addition or subtraction of areas.

■ EXAMPLE 3 Let x be a continuous random variable with a standard normal distribution. Using Table A at the back of the book, find each of the following. a) P10 … x … 1.682 c) P1- 2.43 … x … 1.012 e) P1- 2.98 … x … - 0.422

b) P1- 0.97 … x … 02 d) P11.90 … x … 2.742 f) P1x Ú 0.612

Solution a) P10 … x … 1.682 is the area bounded by the standard normal curve and the lines x = 0 and x = 1.68. We look this up in Table A by going down the left column to 1.6, then moving to the right to the column headed 0.08. There we read 0.4535. Thus,

y

0

x

1.68

P10 … x … 1.682 = 0.4535. y

b) Because of the symmetry of the graph, P1- 0.97 … x … 02 = P10 … x … 0.972 = 0.3340.

– 0.97 0

c) P1- 2.43 … x … 1.012 = P1- 2.43 … x … 02 + P10 … x … 1.012 = P10 … x … 2.432 + P10 … x … 1.012 = 0.4925 + 0.3438 = 0.8363

0.97

x

1.01

x

y

– 2.43

0

d) P11.90 … x … 2.742 = P10 … x … 2.742 - P10 … x … 1.902 = 0.4969 - 0.4713 = 0.0256

y

0

e) P1- 2.98 … x … - 0.422 = P10.42 … x … 2.982 = P10 … x … 2.982 - P10 … x … 0.422 = 0.4986 - 0.1628 = 0.3358

1.90 2.74 x

y

−2.98

−0.42 0.42

2.98

x

5.5

Quick Check 3

O

Let x be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. a) P A - 1 … x … 12 B b) P1x … - 0.772 c) P A x Ú 25 B

•

Probability: Expected Value; The Normal Distribution

f) P1x Ú 0.612 = P1x Ú 02 - P10 … x … 0.612 = 0.5000 - 0.2291 Because of the symmetry about the line x = 0, half the area is on each side of the line, and since the entire area is 1, we have P 1x Ú 02 = 0.5000.

= 0.2709

515

y

0 0.61

x

O Quick Check 3 For most normal distributions, m Z 0 and s Z 1. It would be a hopeless task to make tables for all values of the mean m and the standard deviation s. For any normal distribution, the transformation z =

x - m s

standardizes the distribution, since 1x - m2>s is a measure of how many standard deviations x is from m. Subtracting m from all x-values and then dividing by s preserves the order of the x-values while permitting the use of Table A. Such converted values are called z-scores, or z-values. P1a … x … b2 = Pa μ = 150

160

180

w

a - m b - m b, … z … s s

and this last probability can be found using Table A.

■ EXAMPLE 4 The weights, w, of the students in a calculus class are normally distributed with a mean, m, of 150 lb and a standard deviation, s, of 25 lb. Find the probability that the weight of a student in the class is between 160 lb and 180 lb.

Solution We first standardize the weights: b - m 180 - 150 = = 1.2; s 25 a - m 160 - 150 160 is standardized to = = 0.4. s 25

180 is standardized to μ=0

0.4

1.2

z

Quick Check 4 Referring to Example 4, find the probability of each of the following. a) The weight of a student is below 165 lb. b) The weight of a student is between 135 lb and 155 lb. c) The weight of a student is above 175 lb.

These z-values measure the distance of w from m in terms of s.

Then we have P1160 … w … 1802 = = = =

P10.4 … z … 1.22 ⎫ ⎢ Now we can P10 … z … 1.22 - P10 … z … 0.42 ⎬ use Table A. ⎢ 0.3849 - 0.1554 ⎭ 0.2295.

Thus, the probability that the weight of a student in that class is 160 lb to 180 lb is 0.2295. That is, about 23% of the students weigh between 160 lb and 180 lb.

O Quick Check 4

O

516

CHAPTER 5

•

Applications of Integration

TECHNOLOGY CONNECTION

Statistics on a Calculator It is possible to use a TI-83/84 Plus to make an approximation of the probability in Example 4 without performing a standard conversion, using Table A, or entering the normal probability density function. We first select an appropriate window, 30, 300, - 0.002, 0.024, with Xscl = 50 and Yscl = 0.01. Next, we use the ShadeNorm command, which we find by pressing F DISTR  [. We enter the values as shown and press [. (If necessary, the ClearDraw option on the DRAW menu can be used to clear the graph.)

know the probability that a student weighs less than 145 lb, we can use 0 as the left endpoint: normalcdf(0,145,150,25) 0.4207403112

If we want to know the probability that a student weighs more than 160 lb, we can use 300 as the right endpoint: normalcdf(160,300,150,25) 0.3445783019

Left endpoint of interval Right endpoint of interval

For cases involving an open-ended (infinite) bound, a rule of thumb is to set that bound at a value approximately five standard deviations below or above the mean. This will give areas that are accurate to over six decimal places.

ShadeNorm(160,180,150,25)

Area = .229509 low = 160

up = 180

EXERCISES

Mean Standard deviation

The area is shaded and given as 0.229509, or about 23%. Alternatively, the probability can be calculated directly by pressing F and DISTR and selecting normalcdf. The values are entered as shown: Mean Standard deviation normalcdf(160,180,150,25) 0.2295085713

1. The weights of the students in a calculus class are nor-

mally distributed with mean m = 150 lb and standard deviation s = 25 lb. a) What is the probability that a student’s weight is from 125 lb to 170 lb? b) What is the probability that a student’s weight is greater than 200 lb?

2. SAT scores. In a recent year, combined SAT reading

Left endpoint Right endpoint

Open-ended intervals can be handled by selecting a “distant” value for the open end. For example, if we want to

and math scores were normally distributed with mean m = 1020 and standard deviation s = 140. a) What percentage of the scores were between 400 and 900? b) What percentage of the scores were above 840?

Percentiles Suppose you take an exam and score better than 85% of all the students taking that exam. We say that your score is in the 85th percentile. This leads us to the definition of percentile:

DEFINITION If x is a continuous random variable and f is a probability density function over an interval 3a, b4, the pth percentile is a value c, with a 6 c 6 b, such that c p = f1x2 dx. 100 La

Note that p>100 is an area; that is, we can interpret percentiles as areas. The factor 100 converts the value of the area into a value between 0 and 1.00. For the standard normal distribution, the value c is the standard deviation, which is denoted by the variable z.

5.5

•

Probability: Expected Value; The Normal Distribution

517

■ EXAMPLE 5 For the standard normal distribution, with m = 0 and s = 1, determine the percentile corresponding to each of the following z-values. a) z = 0

b) z = - 1.75

c) z = 2.25

Solution a) The percentile corresponding to z = 0 is the area under the curve shown at the right from - q to 0 (i.e., to the left of 0). This is half of the total area of the standard normal distribution. Thus, a z-value of 0 corresponds to the 50th percentile: a score exactly at the mean is higher than 50% of all other scores. b) We use Table A to determine the area from 0 to 1.75, which is 0.4599. By symmetry, the area between - 1.75 and 0 is also 0.4599. Since the area from - q to 0 is 0.5, to get the area from - q to - 1.75, we subtract: 0.5 - 0.4599 = 0.0401. Therefore, a z-value of - 1.75 corresponds to the 4th percentile (rounded). Quick Check 5

O

For the standard normal distribution, with m = 0 and s = 1, determine the percentile corresponding to each of the following z-values. a) z = - 1 b) z = 0.25 c) z = 2.8

y

Area  0.5

–3

–2

–1

0  z0

1

2

3

x

1

2

3

x

y

Area  0.04 –3

–2

–1

0 

z  1.75

c) Table A shows that the area from 0 to 2.25 is 0.4878. We add this to 0.5, so the total area from - q to 2.25 is 0.9878. A z-value of 2.25 corresponds to the 98.78th percentile.

y

Area  0.9878

–3

–2

–1

0 

1

2

z  2.25

O Quick Check 5

TECHNOLOGY CONNECTION

Percentiles on a Calculator On the T1-83 Plus and TI-84 Plus calculators, z-values can be determined for percentiles of the standard normal distribution 1m = 0 and s = 12. Press F and DISTR and select InvNorm. Enter the percentile as a decimal between 0 and 1, and press [. The result is the z-value that corresponds to the given percentile. For example, the z-value that corresponds to the 70th percentile is 0.524.

tion is s = 12, then the x-value that corresponds to the 70th percentile is 96.293. Standard Percentile Mean deviation InvNorm(0.7,90,12) 96.29280612

EXERCISES

Percentile

Use InvNorm to find the x-values for the given percentiles. InvNorm(0.7) 0.5244005101

1. 50th percentile, m = 0, s = 1 2. 25th percentile, m = 0, s = 1

For any mean and standard deviation, the x-value can be found directly by entering the percentile, followed by the given mean and standard deviation, separated by commas. For example, if the mean is m = 90 and the standard devia-

3. 95th percentile, m = 5, s = 3 4. 3rd percentile, m = 10, s = 4 5. 58.45th percentile, m = 100, s = 15

3

x

518

CHAPTER 5

•

Applications of Integration

■ EXAMPLE 6 A large class takes an exam, and the students’ scores are normally

distributed; the mean score is m = 72, and the standard deviation is s = 4.5. The professor curves the grading scale so that anyone who scored in the top 10% receives an A. What is the minimum score needed to get an A?

Solution The top 10% corresponds to the 90th percentile. We need to determine the z-value that corresponds to an area of 0.9. From Table A, we see that an area of 0.4 is achieved when z = 1.28 (with rounding). Therefore, when the area of 0.5 of the left half of the distribution is included, we see that a z-value of 1.28 corresponds to the 90th percentile. y

Area  0.5

–3

–1

90th percentile

Quick Check 6 Speeds along a stretch of highway are normally distributed and have a mean of m = 59 mph with a standard deviation of s = 8. A policeman will issue a speeding citation to any driver whose speed is in the top 2% of this distribution. What is the minimum speed that will get a driver a citation?

–2

Area  0.4

0 

2

1

3

x

z  1.28 “Top 10%”

We use the transformation formula to determine the x-value (test score) that corresponds to z = 1.28: x - 72 4.5 5.76 = x - 72 x = 77.76. 1.28 =

Multiplying both sides by 4.5

Therefore, a score of 78 (rounded) is the minimum score needed to get an A.

O Quick Check 6 Curving the grading scale is a common technique to base grades on the collective performance of a group (the entire class, in Example 6). It is considered to be a competitive grading system, in which each student competes against fellow students for a good grade, as opposed to a noncompetitive system, in which a grade is based on a raw percentage. Many standardized exams such as the SAT and the GRE use competitive grading: your standing is based on your percentile.

■ EXAMPLE 7 Business: Quality Control. Bottles of cola are to contain a volume with a mean of 591 mL, but some variation is expected. Any bottle at or below the 20th percentile of the volume distribution is rejected. Suppose we know that a bottle that contains 593 mL of cola is in the 65th percentile. What is the smallest volume that will be accepted? Assume that the volumes are normally distributed.

Solution We are not given the standard deviation, but we can determine it from the given information. Since we know that a bottle with 593 mL is in the 65th percentile, we need to determine a z-value that corresponds to the 65th percentile. Table A shows that the area from 0 to 0.385 (interpolated) is 0.15, which we add to the 0.5 from the

O

5.5

•

Probability: Expected Value; The Normal Distribution

519

left half of the distribution. Therefore, z = 0.385. We use the transformation formula to solve for s, with x = 593 and m = 591: 0.385 =

593 - 591 s

0.385 =

2 s

s =

2 = 5.19. 0.385

Solving for s

We now determine the z-value that corresponds to the 20th percentile. Table A shows that the area from 0 to 0.84 is 0.3, so, by symmetry, the area from - 0.84 to 0 is also 0.3. Therefore, the area to the left of - 0.84 is 0.2. The 20th percentile corresponds to a z-value of - 0.84. We now solve for the x-value that corresponds to this z-value:

Quick Check 7

O

High school students take a state-wide exam, and anyone scoring in the top 15% is awarded a scholarship for college. The scores are normally distributed, with a mean of m = 79. Your friend scored 87, which was in the 80th percentile, and did not get a scholarship. Your score was 91. Did you get a scholarship?

0.84 =

x - 591 5.19

- 4.46 = x - 591 x = 586.6.

Multiplying both sides by 5.19 Adding 591

Rounding, we conclude that any bottle containing 586 mL of cola or less is rejected and any bottle containing 587 mL or greater is accepted.

O Quick Check 7

Section Summary Assume that x is a continuous random variable and f is a probability density function for x on an interval 3a, b4. • The expected value of x is b

E1x2 =

La

x # f1x2 dx.

• If g is a function of x on the interval 3a, b4, then the expected value of g1x2 is b

E1g1x22 =

La

g1x2 # f1x2 dx.

b

La

x # f1x2 dx.

• The variance 1s22 of x is

x 2 f1x2 dx - c

f1x2 =

1 22p

e -x >2, 2

1 s22p

e -11>2231x - m2>s4 , 2

over the interval 1- q , q 2.

b

La

It is used to describe the “spread” of the data. • The standard normal distribution of x is defined by the probability density function

f1x2 =

s2 = E1x 22 - m2 =

s = 2variance.

over the interval 1- q , q 2, where the mean m = 0 and the standard deviation s = 1. • The general case of a normally distributed random variable x with mean m and standard deviation s has the probability density function

• The mean 1m2 is the expected value: m = E1x2 =

• The standard deviation is the square root of the variance:

b

La

2

xf1x2 dx d .

• For a normal distribution, data values x are converted into x - m z-values by the transformation formula: z = . s • The pth percentile is a value c, with a 6 c 6 b, such that c p = f1x2 dx. 100 La

520

CHAPTER 5

•

Applications of Integration

EXERCISE SET

5.5 For each probability density function, over the given interval, find E1x2, E1x 22, the mean, the variance, and the standard deviation. 1. f 1x2 = 14, 33, 74

2. f 1x2 = 15, 33, 84

5. f 1x2 =

6. f 1x2 =

3. f 1x2 = 18 x, 30, 44 7. f 1x2 =

1 4 x, 3 2 2x ,

31, 34

3 - 1, 14

9. f 1x2 =

1 #1 , 31.5, 7.54 ln 5 x

10. f 1x2 =

1 #1 , 30.8, 3.24 ln 4 x

4. f 1x2 = 29 x, 30, 34 8. f 1x2 =

2 3 x,

31, 24

1 2 3x ,

3- 2, 14

Let x be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. 11. P10 … x … 2.132

12. P10 … x … 0.362

13. P1 - 1.37 … x … 02

14. P1- 2.01 … x … 02

15. P1- 1.89 … x … 0.452

16. P1- 2.94 … x … 2.002

17. P11.35 … x … 1.452

18. P10.76 … x … 1.452

19. P1 - 1.27 … x … - 0.582 20. P1 - 2.45 … x … - 1.242 21. P1x Ú 3.012

22. P1x Ú 1.012

23. a) P1 - 1 … x … 12 b) What percentage of the area is from - 1 to 1? 24. a) P1 - 2 … x … 22 b) What percentage of the area is from - 2 to 2? Let x be a continuous random variable that is normally distributed with mean M  22 and standard deviation S  5. Using Table A, find each of the following. 25. P124 … x … 302

26. P122 … x … 272

27. P119 … x … 252

28. P118 … x … 262

29–46. Use a graphing calculator to do Exercises 11–28. 47. Find the following percentiles for a standard normal distribution. a) 30th percentile b) 50th percentile c) 95th percentile 48. In a normal distribution with m = 60 and s = 7, find the x-value that corresponds to the b) 75th percentile a) 35th percentile 49. In a normal distribution with m = - 15 and s = 0.4, find the x-value that corresponds to the b) 92nd percentile a) 46th percentile

50. In a normal distribution with m = 0 and s = 4, find the x-value that corresponds to the b) 84th percentile a) 50th percentile

APPLICATIONS Business and Economics 51. Mail orders. The number of orders, N, received daily by an online vendor of used CDs is normally distributed with mean 250 and standard deviation 20. The company has to hire extra help or pay overtime on those days when the number of orders received is 300 or higher. What percentage of days will the company have to hire extra help or pay overtime? 52. Bread baking. The number of loaves of bread, N, baked each day by Fireside Bakers is normally distributed with mean 1000 and standard deviation 50. The bakery pays bonuses to its employees on those days when at least 1100 loaves are baked. What percentage of days will the bakery have to pay a bonus? Manufacturing. In an automotive body-welding line, delays encountered during the manufacturing process can be modeled by various probability distributions. (Source: R. R. Inman, “Empirical Evaluation of Exponential and Independence Assumptions in Queueing Models of Manufacturing Systems,” Production and Operations Management, Vol. 8, 409–432 (1999).) 53. The processing time for the robogate has a normal distribution with mean 38.6 sec and standard deviation 1.729 sec. Find the probability that the next operation of the robogate will take 40 sec or less. 54. The processing time for the automatic piercing station has a normal distribution with mean 36.2 sec and standard deviation 2.108 sec. Find the probability that the next operation of the piercing station will take between 35 and 40 sec. 55. Test score distribution. In 2004, combined SAT reading and math scores were normally distributed with mean 1026 and standard deviation 113. Find the SAT scores that correspond to these percentiles. (Source: www.collegeboard.com.) a) 35th percentile b) 60th percentile c) 92nd percentile

General Interest 56. Test score distribution. The scores on a biology test are normally distributed with mean 65 and standard deviation 20. A score from 80 to 89 is a B. What is the probability of getting a B?

Exercise Set 5.5

57. Test score distribution. In a large class, students’ test scores had a mean of m = 76 and a standard deviation s = 7. a) The top 12% of students got an A. Find the minimum score needed to get an A (round to the appropriate integer). b) The top 30% (excluding those who got an A) got a B. Find the minimum score needed to get a B (round to the appropriate integer). 58. Average temperature. Las Vegas, Nevada, has an average daily high temperature of 104 degrees in July, with a standard deviation of 6 degrees. (Source: www.wunderground. com.) a) In what percentile is a temperature of 112 degrees? b) What temperature would be at the 67th percentile? c) What temperature would be in the top 0.5% of all July temperatures for this location?

521

Median. Let x be a continuous random variable over 7a, b8 with probability density function f. Then the median of the x-values is that number m for which m

1 f 1x2 dx  . 2 La Find the median.

63. f 1x2 = 12 x, 30, 24

65. f 1x2 = ke -kx, 30, q 2

64. f 1x2 = 32 x 2, 3- 1, 14

66. Business: coffee production. Suppose that the amount of coffee beans loaded into a vacuum-packed bag has a mean weight of m ounces, which can be adjusted on the filling machine. Suppose that the amount dispensed is normally distributed with s = 0.2 oz. What should m be set at to ensure that only 1 bag in 50 will have less than 16 oz?

59. Heights of basketball players. Players in the National Basketball Association have a mean height of 79 in. (6 ft 7 in.).(Source: www.apbr.org.) If a basketball player who is 7 ft 2 in. tall is in the top 1% of players by height, in what percentile is a 6 ft 11 in. player?

67. Business: does thy cup overflow? Suppose that the mean amount of cappuccino, m, dispensed by a vending machine can be set. If a cup holds 8.5 oz and the amount dispensed is normally distributed with s = 0.3 oz, what should m be set at to ensure that only 1 cup in 100 will overflow?

60. Bowling scores. At the time this book was written, the bowling scores, S, of author Marv Bittinger (shown below) were normally distributed with mean 201 and standard deviation 23. a) Find the probability that a score is from 185 to 215, and interpret your results. b) Find the probability that a score is from 160 to 175, and interpret your results. c) Find the probability that a score is greater than 200, and interpret your results.

68. Explain the uses of integration in the study of probability. 69. You are told2 that “the antiderivative of the function f 1x2 = e -x >2 has no basic integration formula.” Make some guesses of functions that might seem reasonable to you as antiderivatives and show why they are not.

TECHNOLOGY CONNECTION 70. Approximate the integral q

e -x dx. 2

L- q

Answers to Quick Checks 1. E1x2 = 43, E1x 22 =

8 3

2. m = 43, s2 = 89, s =

28 L 0.943 3

3. (a) 0.533; (b) 0.221; (c) 0.345 4. (a) 0.726; (b) 0.305; (c) 0.159 5. (a) 15.9th percentile; (b) About 59.9th percentile; (c) 99.7th percentile 6. About 75.4 mph 7. Yes, you were in the 89.7th percentile.

SYNTHESIS For each probability density function, over the given interval, find E1x2, E1x 22, the mean, the variance, and the standard deviation. 61. f 1x2 =

1 , b - a

62. f 1x2 =

3a 3 x4

over 3a, b4

, over 3a, q 2

522

CHAPTER 5

• Applications of Integration

Volume

5.6

Consider the graph of y = f1x2 in Fig. 1. If the upper half-plane is rotated about the x-axis, then each point on the graph has a circular path, and the whole graph sweeps out a certain surface, called a surface of revolution. The plane region bounded by the graph, the x-axis, x = a, and x = b sweeps out a solid of revolution. To calculate the volume of this solid, we first approximate it as a finite sum of thin right circular cylinders, or disks (Fig. 2). We divide the interval 3a, b4 into equal subintervals, each of length ¢x. Thus, the height h of each disk is ¢x (Fig. 3). The radius of each disk is f1x i2, where x i is the right-hand endpoint of the subinterval that determines that disk. If f1x i2 is negative, we can use ƒ f1x i2 ƒ .

OBJECTIVE • Find the volume of a solid of revolution.

y

y

y

y  f(x)

y  f(x)

y  f(x)

h  Δx r  f(x i ), f(x i )  0

a

a

b

b

FIGURE 1

x

x

x

FIGURE 3

FIGURE 2

Since the volume of a right circular cylinder is given by V = pr 2h, or

Volume = area of the base # height,

each of the approximating disks has volume p ƒ f1xi2 ƒ 2 ¢x = p3f1xi242 ¢x.

Squaring makes use of the absolute value unnecessary.

The volume of the solid of revolution is approximated by the sum of the volumes of all the disks: n

V L a p3f1xi242 ¢x. i=1

The actual volume is the limit as the thickness of the disks approaches zero, or the number of disks approaches infinity: b

n

V = lim a p3f1xi242 ¢x = n: q i=1

La

p3f1x242 dx.

(See Section 4.2.) That is, the volume is the value of the definite integral of the function y = p3f1x242 from a to b.

THEOREM 2

For a continuous function f defined on 3a, b4, the volume, V, of the solid of revolution obtained by rotating the area under the graph of f from a to b about the x-axis is given by b

V =

La

p3f1x242 dx.

• Volume

5.6

523

■ EXAMPLE 1 Find the volume of the solid of revolution generated by rotating the region under the graph of y = 2x from x = 0 to x = 1 about the x-axis.

Solution y

1

V =

p3f1x242 dx

L0

y  √x

1

p C 2x D 2 dx

1

=

L0

1

= Explain how this could be interpreted as a solid of revolution.

= pc

Quick Check 1

O

Find the volume of the solid of revolution generated by rotating the region under the graph of y = x 3 from x = 0 to x = 2 about the x-axis.

px dx

L0

1 x

2 1

x d 2 0

=

p 2 1 cx d 2 0

=

p 2 p 11 - 022 = 2 2

O Quick Check 1 ■ EXAMPLE 2 Find the volume of the solid of revolution generated by rotating the region under the graph of y = ex from x = - 1 to x = 2 about the x-axis.

y

y  ex

Solution 2

V =

L-1 2

=

L-1

p3f1x242 dx p3e x42 dx

L-1

= c

Quick Check 2 Find the volume of the solid of revolution generated by rotating the region under the 1 graph of y = from x = 1 to x x = 3 about the x-axis.

pe 2x dx

p 2x 2 e d 2 -1

p 2x 2 ce d 2 -1 p 2#2 = 1e - e 21-122 2 p 4 = 1e - e -22 L 85.55 2 =

O Quick Check 2

1 2 x

2

=

1

O

524

CHAPTER 5

• Applications of Integration

■ EXAMPLE 3 Business: Water Storage. A city’s water storage tank is in the shape of the solid of revolution generated by rotating the region under the graph of f1x2 = 50

1 -

B

x2 402

from x = - 40 ft to x = 40 ft about the x-axis. What is the volume of this tank?

Solution The tank’s shape is called an oblate spheroid: its vertical diameter (80 ft) is less than its horizontal diameter (100 ft). The graph of f is shown at left. Rotating the graph of f about the x-axis gives the shape of the tank, but it is standing on end. The actual tank has this shape turned on its side. If we rotate the portion of the graph in the first quadrant, that is, from x = 0 to x = 40, we will get half of the solid. This has the advantage of using 0 as a bound of integration. We then multiply the result by 2 to determine the whole volume. The volume for 0 … x … 40 is

y 50 40

f

30 20 10 – 40 –30 –20 –10

10

20

30

40

x

40

V =

p B 50

L0

1 -

B

40

=

L0

pc2500a1 40

=

L0

pc2500 -

= pc2500x =

Quick Check 3

O

A tepee is a cone with a height of 15 ft at its center and a circular base with a radius of 8 ft. Determine the volume contained within this tepee. (Hint: Rotate the line 8 y = 15 x for x = 0 to x = 15 about the x-axis.)

x2 2 R dx 402 x2 b d dx 1600

25 2 x d dx 16

25 3 40 x d 48 0

502 = 2500, and the radical disappears due to squaring. Multiplying through by 2500 Antidifferentiating

200,000 p. 3

Multiplying this result by 2 gives the tank’s entire volume: 400,000 p L 418,879 ft3. 3 Since 1 ft3 holds 7.48 gal, this tank holds over 3.13 million gallons of water.

O Quick Check 3

Section Summary

• If a function f is continuous over an interval 3a, b4, then the volume of the solid formed by rotating the area under the graph of f from a to b about the x-axis is given by b

V =

La

p3f1x242 dx.

Exercise Set 5.6

525

EXERCISE SET

5.6 Find the volume generated by rotating about the x-axis the regions bounded by the graphs of each set of equations. 1. y = x, x = 0, x = 1 y yx

y = 50 1 +

B

x2 , for - 250 … x … 150, 22,500

where x and y are in feet, describes the shape of such a tower (laying on its side). Determine the volume of the tower by rotating the region bounded by the graph of y about the x-axis. (Hint: See Example 3.)

0 1 x

2. y = x, x = 0, x = 2 3. y = 2x, x = 1, x = 4 4. y = 2x, x = 1, x = 3 5. y = e x, x = - 2, x = 5 6. y = e x, x = - 3, x = 2 7. y =

1 , x = 1, x = 3 x

8. y =

1 , x = 1, x = 4 x

9. y = 10. y =

2 2x 1 2x

, x = 4, x = 9 , x = 1, x = 4

11. y = 4, x = 1, x = 3 12. y = 5, x = 1, x = 3 13. y = x , x = 0, x = 2 2

20. Volume of a football. A regulation football used in the National Football League is 11 in. from tip to tip and 7 in. in diameter at its thickest (the regulations allow for slight variation in these dimensions). (Source: NFL.) The shape of a football can be modeled by the function f 1x2 = - 0.116x 2 + 3.5, for - 5.5 … x … 5.5,

where x is in inches. Find the volume of the football by rotating the region bounded by the graph of f about the x-axis.

SYNTHESIS 21. Graph y = 24 - x 2 and y = 2r 2 - x 2, with r 7 0, and explain how the results can be used to calculate the volume of a common shape. (See Exercises 17 and 18.) 22. Prove that the volume of a right-circular cone of height h and radius r is V = 13 pr 2h. (Hint: Rotate a line starting at the origin and ending at the point 1h, r2 about the x-axis.)

14. y = x + 1, x = - 1, x = 2 15. y = 21 + x, x = 2, x = 10 16. y = 22x, x = 1, x = 2 17. y = 24 - x 2, x = - 2, x = 2

h

18. y = 2r 2 - x 2, x = - r, x = r (assume r 7 0) r

APPLICATIONS 19. Cooling tower volume. Cooling towers at nuclear power plants have a “pinched” chimney shape (which promotes cooling within the tower) formed by rotating a hyperbola around an axis. The function

Find the volume generated by rotating about the x-axis the regions bounded by the graphs of each set of equations. 23. y = 2ln x, x = e, x = e 3 24. y = 2xe -x, x = 1, x = 2

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25. Consider the function y = 1>x over the interval 31, q 2. We showed in Section 5.3 that the area under the curve does not exist; that is, q

1 dx L1 x diverges. Find the volume of the solid of revolution formed by rotating the region under the graph of y = 1>x over the interval 31, q 2 about the x-axis. That is, find q

TECHNOLOGY CONNECTION 26. Paradox of Gabriel’s horn or the infinite paint can. Though we cannot prove it here, the surface area of Gabriel’s horn (see Exercise 25) is given by q

S =

Show that the surface area of Gabriel’s horn does not exist. The paradox is that the volume of the horn exists, but the surface area does not. This is like a can of paint that has a finite volume but, when full, does not hold enough paint to paint the outside of the can.

2

1 pc d dx. x L1 y y

2p 1 1 + 4 dx. x L1 x A

1 x

Answers to Quick Checks

0 1

1.

128p 7

2.

2p 3

3. 320p ft3

x

This solid is sometimes referred to as Gabriel’s horn.

5.7 OBJECTIVES • Solve differential equations. • Verify that a given function is a solution of a differential equation. • Solve differential equations using separation of variables. • Solve applications involving differential equations.

Differential Equations A differential equation is an equation that involves derivatives, or differentials. In Chapter 3, we studied one very important differential equation, dP = kP, or P¿1t2 = k # P1t2, dt where P, or P1t2, is the population at time t. This equation is a model of uninhibited population growth. Its solution is the function P1t2 = P0e kt, where the constant P0 is the size of the population at t = 0. As this example illustrates, differential equations are rich in applications and have solutions that are functions.

Solving Certain Differential Equations In this section, we will frequently use the notation y¿ for a derivative—mainly because it is simple. Thus, if y = f1x2, then y¿ =

dy dx

= f ¿1x2.

We actually find solutions of certain differential equations when we find their antiderivatives or indefinite integrals. The differential equation dy dx

= g1x2, or y¿ = g1x2,

has the solution y = The motion of waves can be represented by differential equations.

L

g1x2 dx.

5.7

•

Differential Equations

527

■ EXAMPLE 1 Solve: y¿ = 2x. Solution y =

TECHNOLOGY CONNECTION

Exploratory For y¿ = 3x 2, write the general solution. Then use a graphing calculator, iPlot, or Graphicus to graph the particular solutions for C = - 2, C = 0, and C = 1.

L

2x dx = x 2 + C

Look again at the solution of Example 1. Note the constant of integration, C. This solution is called a general solution because taking all values of C gives all the solutions. Taking specific values of C gives particular solutions. For example, the following are particular solutions of y¿ = 2x: y y = x 2 + 3, y = x 2, y = x 2 - 3.

9 8 7

The graph shows the curves of these few particular solutions. The general solution can be regarded as the set of all particular solutions, a family of curves. Knowing the value of a function at a particular point may allow us to select a particular solution from the general solution.

6

y = x 2 + 3 (C = 3)

5

y = x 2 (C = 0)

4

y = x 2 – 3 (C = –3)

3 2 1 –3 –2 –1

1

2

3

x

–1 –2 –3

■ EXAMPLE 2 Solve f ¿1x2 = e x + 5x - x 1>2, given that f102 = 8.

Solution We first find the general solution: 1e x + 5x - x 1>22 dx L = e x + 52 x 2 - 23 x 3>2 + C.

f1x2 =

L

f ¿1x2 dx =

Next, since f102 = 8, we substitute to find C: 8 = e 0 + 52 # 02 8 = 1 + C 7 = C.

2 3

# 03>2 + C

Thus the particular solution is f1x2 = e x + 52 x 2 - 23 x 3>2 + 7.

Verifying Solutions To verify that a function is a solution to a differential equation, we find the necessary derivatives and substitute.

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■ EXAMPLE 3 Show that y = 4e x + 5e 3x is a solution of y– - 4y¿ + 3y = 0. Solution We first find y¿ and y–: d 14e x + 5e 3x2 = 4e x + 15e 3x; dx d y– = y¿ = 4e x + 45e 3x. dx y¿ =

Then we substitute in the differential equation, as follows: ?

y– - 4y¿ + 3y = 0 x 3x x 3x 14e + 45e 2 - 414e + 15e 2 + 314e x + 5e 3x2 0 4e x + 45e 3x - 16e x - 60e 3x + 12e x + 15e 3x 0 0 0 Quick Check 1

O

Show that y = x 2 + x is a 1 solution of y¿ + y = 3x + 2. x

TRUE

Since a true equation, 0 = 0, results, we know that y = 4e x + 5e 3x is a solution of the differential equation.

O Quick Check 1 Separation of Variables Consider the differential equation dy dx

(1)

= 2xy.

We treat dy>dx as a quotient, as we did in Sections 2.6 and 4.5. Multiplying equation (1) by dx and then by 1>y, we get dy = 2x dx, y Z 0. y

(2)

We have now separated the variables, meaning that all the expressions involving y are on one side and all those involving x are on the other. We then integrate both sides of equation (2): At the 1968 Olympic Games in Mexico City, Bob Beamon made a miraculous long jump of 29 ft, 221 in. Many believed that the jump’s recond length was due to the altitude, which was 7400 ft. Using differential equations for analysis, M. N. Bearley refuted the altitude theory in “The Long Jump Miracle of Mexico City” (Mathematics Magazine, Vol. 45, 241–246 (November 1972)). Bearley argues that the world-record jump was a result of Beamon’s exceptional speed (9.5 sec in the 100-yd dash) and the fact that he hit the take-off board in perfect position.

dy = 2x dx L Ly ln ƒ y ƒ = x 2 + C. We use only one constant because the two antiderivatives differ by, at most, a constant. Recall that the definition of logarithms says that if log a b = t, then b = a t. We have ln ƒ y ƒ = log e ƒ y ƒ = x 2 + C, so 2

ƒ y ƒ = ex

+C

, or y = ;e x # e C. 2

Thus, the solution to differential equation (1) is 2

y = C1e x , where C1 = ;e C. In fact, C1 is still an arbitrary constant.

5.7

•

Differential Equations

■ EXAMPLE 4 Solve: 3y 2

dy

+ x = 0, where y = 5 when x = 0.

dx

Solution We first separate the variables as follows: dy

= -x dx 3y 2 dy = - x dx.

3y 2

Adding - x to both sides Multiplying both sides by dx

We then integrate both sides: L

3y 2 dy =

- x dx L x2 + C y3 = 2 x2 y3 = C 2 y =

3

C -

B

x2 . 2

Taking the cube root of both sides

Since y = 5 when x = 0, we substitute to find C: 3

5 =

02 2

C -

B

Substituting 5 for y and 0 for x

3

5 = 2C 125 = C. Quick Check 2

The particular solution is

Solve: dy

- 3x = 0, dx where y = 9 when x = 0. 22y

Cubing both sides

y =

3

125 -

B

x2 . 2

O Quick Check 2 ■ EXAMPLE 5 Solve: dy dx

=

x . y

Solution We first separate the variables: dy

= x dx y dy = x dx.

y

We then integrate both sides: L

y dy =

x dx L y2 x2 = + C 2 2 y 2 = x 2 + 2C y 2 = x 2 + C1,

529

O

530

CHAPTER 5

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Applications of Integration

Quick Check 3

O

Solve: y¿ = x 2y.

where C1 = 2C. We make this substitution in order to simplify the equation. We then obtain the solutions y = 2x 2 + C1 and y = - 2x 2 + C1.

O Quick Check 3 ■ EXAMPLE 6 Solve: y¿ = x - xy. Solution Before we separate the variables, we replace y¿ with dy>dx: dy dx

= x - xy.

Then we separate the variables: dy = 1x - xy2 dx dy = x11 - y2 dx dy = x dx. 1 - y

Next, we integrate both sides: dy

TECHNOLOGY CONNECTION

Exploratory Solve y¿ = 2x + xy. Graph the particular solutions for C1 = - 2, C1 = 0, and C1 = 1.

=

x dx L x2 - ln ƒ 1 - y ƒ = + C 2 L1 - y

ln ƒ 1 - y ƒ = -

x2 - C 2

ƒ 1 - y ƒ = e -x /2 - C 2 1 - y = ;e -x /2 - C 2 1 - y = ;e -x /2e -C 2

Using base e to remove ln on the left side Remember: a x + y = a xa y.

Since C is an arbitrary constant, ;e -C is an arbitrary constant. Thus, we can replace ;e -C with C1: 1 - y = C1e -x /2 2 - y = C1e -x /2 - 1 2 y = 1 - C1e -x /2. 2

Multiplying both sides by - 1

Or, if we choose to replace - C1 with C2, we have y = 1 + C2e -x /2. 2

Addition is generally regarded as “simpler” than subtraction.

An Application to Economics: Elasticity ■ EXAMPLE 7 Suppose that for a certain product, the elasticity of demand is 1 for

all prices x 7 0. That is, E1x2 = 1 for x 7 0. Find the demand function q = D1x2. (See Section 3.6. Note that this use of the symbol E is unrelated to our earlier work on expected value.)

Solution Since E1x2 = 1 for all x 7 0, 1 = E1x2 = -

xD¿1x2

D1x2 x # dq = . q dx

Substituting

5.7

•

Differential Equations

531

Then -

q dq = . x dx

Separating the variables, we get dq dx = - . x q Now we integrate both sides: dq dx = x L Lq ln x = - ln q + C.

Note that both the price, x, and the quantity, q, can be assumed to be positive.

Then ln x + ln q = C ln 1xq2 = C xq = e C.

Using a property of logarithms Writing an equivalent exponential equation

We let C1 = e = xq. Then C

Quick Check 4 Find the demand function q = D1x2 if the elasticity of demand is E1x2 = x.

q =

C1 C1 , or x = . x q

This result characterizes those demand functions for which the elasticity is always 1.

O Quick Check 4 An Application to Psychology: Reaction to a Stimulus THE WEBER–FECHNER LAW In psychology, one model of stimulus–response asserts that the rate of change dR>dS of the reaction R with respect to a stimulus S is inversely proportional to the intensity of the stimulus. That is, dR k = , dS S where k is some positive constant.

To solve this equation, we first separate the variables: dR = k #

dS . S

We then integrate both sides: k#

dS S L L R = k ln S + C. dR =

We assume S 7 0.

(3)

Now suppose that we let S0 be the lowest level of the stimulus that can be detected. This is the threshold value, or the detection threshold. For example, the lowest level of sound that can be consistently detected is the tick of a watch from 20 ft away, under

O

532

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•

Applications of Integration

very quiet conditions. If S0 is the lowest level of stimulus that can be detected, it seems reasonable that R1S02 = 0. Substituting this condition into equation (3), we get 0 = k ln S0 + C, or

- k ln S0 = C.

Replacing C in equation (3) with - k ln S0 gives us

–dR –– dS

R = k ln S - k ln S0

k –dR –– = –– dS S

As a check, note that

= k1ln S - ln S02.

dR k = . dS S

Using a property of logarithms, we have S0

R

S

S R = k ln ––– S0

S0

S

R = k # ln

S . S0

Look at the graphs of dR>dS and R on the left. Note that as the stimulus gets larger, the rate of change decreases; that is, the change in reaction becomes smaller as the stimulation received becomes stronger. For example, suppose that a lamp has a 50-watt bulb in it. If the bulb were suddenly changed to 100 watts, you would probably be very aware of the difference. That is, your reaction would be strong. If the bulb were then changed to 150 watts, your reaction would not be as great as it was to the change from 50 to 100 watts. A change from a 150- to a 200-watt bulb would cause even less reaction, and so on. For your interest, here are some other detection thresholds. Stimulus Light Taste Smell Touch

Detection Threshold The flame of a candle 30 miles away on a dark night Water diluted with sugar in the ratio of 1 teaspoon to 2 gallons One drop of perfume diffused into the volume of three average-size rooms The wing of a bee dropped on your cheek at a distance of 1 centimeter (about 38 of an inch)

Section Summary • A differential equation is an equation that involves derivatives, or differentials. • The solution to a differential equation is a function, which can be a general solution of the form y = f1x2 + C or, if an initial condition is known, a particular solution in which a value of C is specified.

• Separation of variables is a method for solving some differential equations by writing all the expressions involving y on one side of the equation and all those involving x on the other.

Exercise Set 5.7

533

EXERCISE SET

5.7 Find the general solution and three particular solutions. 1. y¿ = 5x 4 3. y¿ = e

2x

2. y¿ = 6x 5 + x

4. y¿ = e

8 5. y¿ = - x2 + x5 x

4x

- x + 2

3 6. y¿ = + x2 - x4 x

Find the particular solution determined by the given condition. 7. y¿ = x 2 + 2x - 3; y = 4 when x = 0 8. y¿ = 3x 2 - x + 5; y = 6 when x = 0 9. f ¿1x2 = x 2/3 - x; f 112 = - 6 10. f ¿1x2 = x

2/5

+ x; f 112 = - 7

11. Show that y = x ln x + 3x - 2 is a solution of 1 = 0. x

y– -

1 = 0. x

13. Show that y = e x + 3xe x is a solution of y– - 2y¿ + y = 0. 14. Show that y = - 2e x + xe x is a solution of y– - 2y¿ + y = 0. Solve. 15.

dy dx

17. 3y 2

19.

21.

dy dx dy dx

= 4x 3y dy dx

= 8x

16.

dy dx

18. 3y 2

=

2x y

20.

=

6 y

22.

dy dx dy dx

23. y¿ = 3x + xy;

y = 5 when x = 0

24. y¿ = 2x - xy;

y = 9 when x = 0

= 5x 4y dy dx = =

= 5x

x 2y 7 y2

26. y¿ = 7y -2; y = 3 when x = 1

29.

= 3y

28.

dP = 2P dt

30.

dx

given that f 112 =

23 3.

APPLICATIONS Business and Economics 32. Total revenue from marginal revenue. The marginal revenue for a certain product is given by R¿1x2 = 300 - 2x. Find the total-revenue function, R1x2, assuming that R102 = 0. 33. Total cost from marginal cost. The marginal cost for a certain product is given by C¿1x2 = 2.6 - 0.02x. Find the total-cost function, C1x2, and the average cost, A1x2, assuming that fixed costs are $120; that is, C102 = $120.

dy dx

dI = hkI, dt where I is the investment, h is the investment productivity (constant), k is the marginal productivity to the consumer (constant), and t is the time. a) Use separation of variables to solve the differential equation. b) Rewrite the solution in terms of the condition I0 = I102.

dP - 200 = . dC 1C + 323/2 a) Find the profit function, P1C2, if P = $10 when C = $61. b) At what cost will the firm break even 1P = 02? 36. Stock growth. The growth rate of a certain stock, in dollars, can be modeled by dV = k1L - V2, dt where V is the value of the stock, per share, after t months; k is a constant; L = $24.81, the limiting value of the stock; and V102 = 20. Find the solution of the differential equation in terms of t and k.

25. y¿ = 5y ; y = 3 when x = 2

dy

1 - 4x + 2x, x

35. Total profit from marginal profit. A firm’s marginal profit, P, as a function of its total cost, C, is given by

-2

27.

f ¿1x2 =

34. Capital expansion. Domar’s capital expansion model is

12. Show that y = x ln x - 5x + 7 is a solution of y– -

31. Solve

= 4y

dP = 4P dt

37. Utility. The reaction R in pleasure units by a consumer receiving S units of a product can be modeled by the differential equation k dR = , dS S + 1 where k is a positive constant.

534

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•

Applications of Integration

a) Use separation of variables to solve the differential equation. b) Rewrite the solution in terms of the initial condition R102 = 0. c) Explain why the condition R102 = 0 is reasonable. Elasticity. Find the demand function q  D1x2, given each set of elasticity conditions. 38. E1x2 =

4 ; q = e when x = 4 x

39. E1x2 =

x ; q = 190 when x = 10 200 - x

40. E1x2 = 2, for all x 7 0 41. E1x2 = n, for some constant n and all x 7 0

Life and Physical Sciences 42. Exponential growth. a) Use separation of variables to solve the differentialequation model of uninhibited growth, dP = kP. dt b) Rewrite the solution of part (a) in terms of the condition P0 = P102.

Social Sciences 43. The Brentano–Stevens Law. The validity of the Weber– Fechner Law has been the subject of great debate among psychologists. An alternative model, R dR = k# , dS S where k is a positive constant, has been proposed. Find the general solution of this equation. (This model has also been referred to as the Power Law of Stimulus–Response.)

SYNTHESIS Solve. 44.

dy dx

= 5x 4y 2 + x 3y 2

45. e -1/x #

dy dx

= x -2 # y 2

46. Discuss as many applications as you can of the use of integration in this chapter 47. In Example 6 in this section, it is stated that “Since C is an arbitrary constant, ;e -C is an arbitrary constant.” Explain why the ; is necessary. First-order linear differential equations. A differential equation of the form y¿  M1x2y  N1x2 has the general solution y

1 P1x2N1x2 dx + C , where P1x2  e 1 M1x2 dx. P1x2

The method of solution is broken down into three steps: (1) determine P1x2; (2) determine 1 P1x2N1x2 dx; and (3) divide the result of step 2 by P1x2. Use this method to solve the differential equations in Exercises 48–50. 48. y¿ + xy = x 49. y¿ +

y x

(the same as the equation in Example 6)

= x

50. y¿ + x 2y = x 2

TECHNOLOGY CONNECTION 51. Solve dy>dx = 5>y. Graph the particular solutions for C1 = 5, C1 = - 200, and C1 = 100.

Answers to Quick Checks 1. y¿ = 2x + 1; therefore, 1 12x + 12 + 1x 2 + x2 = 2x + 1 + x + 1 = 3x + 2 x 2. y =

A 98 x 2 + 27 B 2>3 3. y = Ce x >3 4. q = Ce -x 3

CHAPTER 5 SUMMARY KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 5.1 If p = D1x2 is a demand function, then the consumer surplus at a point 1Q, P2 is

Price

Q

L0

p D(x) Total expenditure

If p = S1x2 is a supply function, then the producer surplus at a point 1Q, P2 is Q

L0

p S(x)

(Q, P)

(Q, P)

D1x2 dx - QP.

QP -

Producer surplus

Price

Consumer surplus

Total receipts

Units

S1x2 dx.

The equilibrium point 1x E, pE2 is the point at which the supply and demand curves intersect.

Units

Price Consumer surplus

S(x)

P D(x)

Producer surplus Q

Units

Let p = 12 - 1.5x be a demand function and p = 4 + 0.5x be a supply function. The two curves intersect at (4, 6), the equilibrium point. At this point, the consumer surplus is 4

L0

112 - 1.5x2 dx - 142162 = 36 - 24 = $12,

and the producer surplus is 142162 -

4

L0

14 + 0.5x2 dx = 24 - 20 = $4.

y Consumer surplus ⫽ $12

$12 11 10 9 8 7 6 5 4 0

Producer surplus ⫽ $4 p ⫽ 4 ⫹ 0.5x (x E, p E) ⫽ (4, 6) p ⫽ 12 ⫺ 1.5x 1

2

3

4

5

x

(continued)

Chapter 5 Summary

535

536

CHAPTER 5

•

Applications of Integration

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 5.2 The future value of P0 dollars invested at an interest rate k for t years, compounded continuously, is given by P = P0e kt.

The future value of $6000 invested at 6.75%, compounded continuously, for 5 yr is

The amount P0 is called the present value. If the future value P is known, then P0 = Pe - kt.

Sue wants to have $15,000 in 4 yr to make a down payment on a house. She opens a savings account that offers 4.5% interest, compounded continuously. The present value is the amount she needs to deposit now to have $15,000 in 4 yr:

P = 6000e 0.0675152 = $8408.64.

P = 15,000e -0.045142 = $12,529.05 The accumulated future value of a continuous income stream is given by T

A =

R 1t2e kt dt,

L0

where R(t) is the rate of the continuous income stream, k is the interest rate, and T is the number of years. If R1t2 is a constant function, then A =

R 1t2 k

T

L0

R 1t2e

R 1t2 k

1,150,000e 0.05t dt

L0 1,150,000 0.05162 1e - 12 = $8,046,752.57. = 0.05

-kt

The accumulated present value of the contract is 6

1,150,000e -0.05t dt L0 1,150,000 = 11 - e -0.051622 = $5,961,180.92. 0.05

B = dt.

If R(t) is a constant function, then B =

6

A =

1e kT - 12.

The accumulated present value of a continuous income stream is given by B =

A baseball pitcher signs an $8,000,000 6-year contract and will be paid $1,150,000 per year. The money will be invested at 5%, compounded continuously, for the 6-yr term. The accumulated future value is

11 - e -kT2.

If R(t) is not a constant function, the definite integral must be solved by an appropriate integration technique. Consumption of a natural resource can be modeled by T

L0

P0e kt dt =

P0 kt 1e - 12, k

where P 1t2 = P0e kt is the annual consumption of the natural resource in year t and consumption is growing exponentially at growth rate k.

Canada’s diamond mines produce diamonds according to the model P1t2 = 2.5e 0.272t, where t = 0 is 2000 and P1t2 is in millions of carats. (Source: USGS Mineral Commodities Summaries.) Using this model, we can forecast the total production of diamonds between 2000 and 2012: 12

L0

2.5e 0.272t dt =

2.5 1e 0.2721122 - 12 = 231.2 million carats. 0.272

Chapter 5 Summary

KEY TERMS AND CONCEPTS

537

EXAMPLES

SECTION 5.3 An integral with infinity as a bound is called an improper integral. All improper integrals are evaluated as limits: q

f 1x2 dx = lim

b: q

La b

L- q

f 1x2 dx =

q

L1 x

dx = lim 3

b: q L 1

f 1x2 dx,

b: q

lim

a: - q L a

f 1x2 dx.

b: q

=

If the limit exists, the improper integral is convergent. Otherwise, it is divergent.

dx d 2

2x 1

= lim c -

b

1 x3 1

= lim c -

b

La

b

1

b 1

21b22

- a-

1 21122

bd

1 2

y 1.5

y⫽

1 x3

1 Area ⫽

0.5 0

If both bounds are infinity, then the improper integral can be written as the sum of two integrals, where c is any real number: q

L- q

f 1x2 dx =

c

L- q

f 1x2 dx +

q

Lc

1

2

0

L- q

1 2

3

4

5

x

0

e 4x dx = lim

a: -q L a

e 4x dx

0 1 = lim c e 4x d a: -q 4 a 1 4102 - e 41a22 d = lim c 1e a: -q 4 1 = 4

f 1x2 dx.

y 2

y ⫽ e4x

1.5

Area ⫽

–2

The accumulated present value of a continuous money flow into an investment at the rate of P dollars per year perpetually is q

Pe -kt dt =

–1.5

–1

1 4

–0.5

1

0.5

x

An investment of $5000 per year perpetually at 7%, compounded continu5000 = 71,428.57. ously, has a present value of 0.07

P , k

L0 where k is the continuously compounded interest rate. (continued)

538

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•

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KEY TERMS AND CONCEPTS SECTION 5.4 In probability, a continuous random variable is a quantity that can be observed (or measured) repeatedly and whose possible values comprise an interval of real numbers. A function f is a probability density function for a continuous random variable x if it meets the following conditions:

• For all x in its domain, 0 … f 1x2. • The area under the graph of f is 1. • For any subinterval 3c, d4 in the domain of f, the probability that x will be in that d subinterval is P13c, d42 = 1c f 1x2 dx. A probability density function is always stated with its domain.

EXAMPLES The function f 1x2 = 29 x, for 0 … x … 3, is a probability density function since • f 1x2 Ú 0 for all x in 30, 34. 3

•

3 2 1 32 - 0 = 1. x dx = c x 2 d = 9 9 0 L0 9

y

2 f(x) ⫽ ⫺x 9

0.6 0.4

Area ⫽ 1

0.2 0

1

2

3

x

The probability that x is between 1.5 and 2.3 is 2.3

L1.5

2.3 12.322 11.522 2 1 x dx = c x 2 d = L 0.338. 9 9 9 9 1.5

y

2 f(x) ⫽ ⫺x 9

0.6 0.4

Area ⫽ 0.338

0.2 0

1

2 1.5

A continuous random variable is uniformly distributed over an interval 3a, b4 if it has a probability density function f given by f 1x2 =

1 , for a … x … b. b - a

3

x

2.3

Helicopter tours over Hoover Dam last from 45 to 55 min, with the times uniformly distributed. If x = time that a tour lasts, the probability density function f is given by f 1x2 =

1 , for 45 … x … 55. 10

The probability that a flight lasts between 48 and 53.5 min is 53.5

L48 A continuous random variable is exponentially distributed if it has a probability density function f of the form

f 1x2 = ke -kx, over the interval 30, q 2.

53.5 1 1 1 dx = c x d = 153.5 - 482 = 0.55. 10 10 48 10

The time x (in minutes) between shoppers entering a store is modeled by the probability density function f 1x2 = 3e -3x, for 0 … x 6 q .

The probability that the time between shoppers is 2 min or less is 2

L0

3e -3x dx = 3- e -3x420 = 1- e -6 - 1- 122 = 0.9975.

539

Chapter 5 Summary

KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 5.5

Consider the probability density function f 1x2 = 29 x over the interval 30, 34. Its mean is

Assume x is a continuous random variable over the interval 3a, b4 with probability density function f.

3

Then the mean 1m2 is the expected value of x: m = E 1x2 =

b

La

b

=

La

2

x 2 # f 1x2 dx - c

b

La

s2 = c

x # f 1x2 dx d . 2

The standard deviation 1s2 is the square root of the variance:

A continuous random variable x has a standard normal distribution if it has a probability density function f given by 1 22p

3

L0

x2 #

s = 20.5 L 0.71.

Weights of packages of ground coffee are normally distributed with mean m = 3 oz and standard deviation s = 0.5. The probability a packet of ground coffee has a weight between 2.75 oz and 3.15 oz is

e -x >2, over 1- q , q 2, 2

Pa

with m = 0 and s = 1.

3.15 - 3 2.75 - 3 … x … b = P1- 0.5 … z … 0.32 0.5 0.5 = 0.309 = 30.9% y

Tables or calculators are used to determine areas within the standard normal distribution.

Area ⫽ 0.309

To convert an x-value into a z-value for use with the standard normal distribution, we use the transformation formula z =

Let f 1x2 =

For the standard normal distribution, c is the standard deviation, which is denoted by z.

2 9

3.15

x over the interval 30, 34. The area to the left of z = 2 is

2

2 x dx = 0.44 Á . L0 9

c

p = f 1x2 dx. 100 La

x

3 2.75

x - m . s

For a probability density function f over an interval 3a, b4, the pth percentile is a value c, with a 6 c 6 b, such that

2 x dx d - m2 = 4.5 - 4 = 0.5. 9

Its standard deviation is

s = 2variance.

f 1x2 =

3

2 2 2 x dx = x dx = 2. 9 L0 9

Its variance is

s = E1x 2 - m 2

L0

x#

x # f 1x2 dx.

The variance 1s22 of x is 2

m =

Therefore, p>100 L 0.44, and z = 2 is the 44th percentile.

0.4

y

2 f(x) ⫽ ⫺x 9

y 0.6

Area  0.933

Area ⫽ 0.44

0.2

x

0 z  1.5

0

1

2

3

x

In the standard normal distribution, the area to the left of z = 1.5 is 0.933. Therefore, a z-value of 1.5 corresponds to the 93.3rd percentile. (continued)

540

CHAPTER 5

•

Applications of Integration

KEY TERMS AND CONCEPTS SECTION 5.6

If f is continuous over an interval 3a, b4, the volume of the solid of rotation formed by rotating the area under the graph of f from a to b about the x-axis is given by b

V =

La

EXAMPLES The volume of the solid formed by rotating the graph of f 1x2 = x = - 1 to x = 3 about the x-axis is 3

V =

p3 f 1x242 dx.

=

1 4

x 2 from

3

2 1 1 p x5 3 pa x 2 b dx = p a x 4 b dx = c d 4 16 16 5 -1 L-1 L-1

1 244 61 p 243 - a- bd = p = p. B 16 5 5 80 20

SECTION 5.7 A differential equation is an equation involving derivatives, or differentials.

The equation y¿ + 3y = 3x 2 + 2x is a differential equation since it involves the derivative y¿. The function y = x 2 is a solution of the differential equation y¿ + 3y = 3x 2 + 2x because 12x2 + 31x 22 = 3x 2 + 2x.

Separation of variables is a common method of solving differential equations.

The differential equation y¿ =

x2 can be solved by separating the variables: y

x2 y dx y dy = x 2 dx dy

The general solution of a differential equation is a function of the form y = f 1x2 + C.

L

=

y dy = y2

L

x 2 dx

x3 + C1 2 3 2 y 2 = x 3 + C, where C = 2C1. 3 =

Therefore, the general solution is y = ; 223 x 3 + C. If an initial condition is known, then a particular solution may be determined by solving for C.

If 10, 22 is an initial condition, we can solve the general solution for C. We choose the positive root since the output, y = 2, is positive. 2 =

23 102 2

2 = 2C 4 = C.

2

+ C

Therefore, the particular solution that passes through the point 10, 22 is y =

23 x 2

3

+ 4.

CHAPTER 5 REVIEW EXERCISES These review exercises are for test preparation. They can also be used as a practice test. Answers are at the back of the book. The blue bracketed section references tell you what part(s) of the chapter to restudy if your answer is incorrect.

Column A 6. Solid of revolution [5.6]

Column B f) y k f(x) = ke− kx

CONCEPT REINFORCEMENT Match each term in column A with the most appropriate graph in column B. Column A 1. Consumer surplus [5.1]

x

Column B a)

Classify each statement as either true or false.

y

7. The accumulated present value of an investment is the value of the investment as a tax-deductible present to a nonprofit charity. [5.2] 8. If an integral has - q or q as one of the limits of integration, it is an improper integral. [5.3] 9. If f is a probability density function over 3a, b4, then f 1x2 Ú 0 for all x in 3a, b4. [5.4]

x

2. Producer surplus [5.1]

y

b)

2 f(x) = 1 e−x /2 2π

x

1

3. Exponential distribution [5.4]

c)

11. To find the volume of the solid of revolution obtained by rotating the graph of y = f 1x2 about the x-axis, we must have f 1x2 Ú 0. [5.6] 12. If y = f 1x2 is a solution of y– + y¿ = 5, then y = f 1x2 + C is also a solution.[5.7]

y

1 b−a

REVIEW EXERCISES a

4. Standard normal distribution [5.5]

10. If f is a probability density function over 3a, b4 and x is a continuous random variable over 3a, b4, then the mean value of f 1x2 is 1b - a2>2. [5.5]

b

x

d) Price Supply

pE

Demand

Let D1x2  1x  622 be the price, in dollars per unit, that consumers are willing to pay for x units of an item, and S1x2  x 2  12 be the price, in dollars per unit, that producers are willing to accept for x units. 13. Find the equilibrium point. [5.1] 14. Find the consumer surplus at the equilibrium point. [5.1] 15. Find the producer surplus at the equilibrium point. [5.1]

xE

5. Uniform distribution [5.4]

Units

e) Price

16. Business: future value. Find the future value of $5000, at an annual percentage rate of 5.2%, compounded continuously, for 7 yr. [5.2]

Supply

pE

Demand xE

Units

17. Business: present value. Find the present value of $10,000 due in 5 yr, at an interest rate of 8.3%, compounded continuously. [5.2] 18. Business: future value of a continuous income stream. Find the accumulated future value of $2500 per year, at 6.25% compounded continuously, for 8 yr. [5.2]

Chapter 5 Review Exercises

541

542

CHAPTER 5

•

Applications of Integration

19. Business: present accumulated value of a trust. The DeMars family welcomes a new baby, and the parents want to have $250,000 in 18 yr for their child’s college education. Find the continuous money stream, at R 1t2 dollars per year, that they need to invest at 5.75% compounded continuously, to generate $250,000. [5.2] 20. Business: early retirement. Cal Earl signs a 7-yr contract as a session drummer for a major recording company. His contract gives him a salary of $150,000 per year. After 3 yr, the company offers to buy out the remainder of his contract. What is the least amount Cal should accept, if the going interest rate is 6.15%, compounded continuously? [5.2] 21. Physical science: iron ore consumption. In 2005 1t = 02, the world production of iron ore was estimated at 1.23 billion metric tons, and production was growing exponentially at the rate of 3% per year. (Source: U.S. Energy Information Administration.) If the production continues to grow at this rate, how much iron ore will be produced from 2005 to 2016? [5.2] 22. Physical science: depletion of iron ore. The world reserves of iron ore in 2005 were estimated to be 160 billion metric tons. (Source: U.S. Geological Survey.) Assuming that the growth rate in Exercise 21 continues and no new reserves are discovered, when will the world reserves of iron ore be depleted? [5.2] Determine whether each improper integral is convergent or divergent, and calculate its value if it is convergent. [5.3] q

23.

1

L1 x 2

q

24.

dx

q

25.

L0

L1

38. Business: pizza sales. The number of pizzas sold daily at Benito’s Pizzeria is normally distributed with mean m = 90 and standard deviation s = 20. What is the probability that at least 100 pizzas are sold during a day? [5.5] 39. Business: distribution of revenue. Benito’s Pizzeria has daily mean revenues that are normally distributed, with m = $5500 and s = $425. What is the lowest amount in the top 5% of daily revenues? [5.5] Find the volume generated by rotating about the x-axis the region bounded by the graphs of the given equations. [5.6] 40. y = x 3, x = 1, x = 2

Solve each differential equation. [5.7] 42.

44. 45.

= 11x 10y

dx dy

43.

dy dx

=

2 y

= 4y; y = 5 when x = 0

dx

dv = 5v -2; v = 4 when t = 3 dt 3x y

47. y¿ = 8x - xy

48. Economics: elasticity. Find the demand function q = D1x2, given the elasticity condition E 1x2 =

x ; q = 70 when x = 30. [5.7] 100 - x

49. Business: stock growth. The growth rate of a stock, in dollars per month, can be modeled by

26. Find k such that f 1x2 = k>x 3 is a probability density function over the interval 31, 24. Then write the probability density function. [5.4] 27. Business: waiting time. A person arrives at a random time at a doctor’s office where the waiting time t to see a doctor is no more than 25 min. The probability density 1 function for t is f 1t2 = 25 , for 0 … t … 25. Find the probability that a person will have to wait no more than 15 min to see a doctor. [5.4] Given the probability density function f 1x2  6x 11  x2

dy

46. y¿ =

e 4x dx

e -2x dx

1 , x = 0, x = 1 x + 2

41. y =

over 70, 18

find each of the following. [5.5] 28. E 1x 22

29. E 1x2

30. The mean

31. The variance

dV = k1L - V2, dt where V is the value of a share, in dollars, after t months; k is a constant; L = $36.37, the limiting value of the stock; and V 102 = 30. Find the solution of the differential equation in terms of t and k. [5.7]

SYNTHESIS

50. The function f 1x2 = x 8 is a probability density function over the interval 3- c, c4. Find c. [5.4]

Determine whether each improper integral is convergent or divergent, and calculate its value if it is convergent. [5.3] q

0

51.

x 4e -x dx 5

L- q

52.

32. The standard deviation

TECHNOLOGY CONNECTION

33. The percentile corresponding to x = 0.7

53. Approximate the integral

Let x be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. [5.5] 34. P10 … x … 1.852

35. P1 - 1.74 … x … 1.432

36. P1- 2.08 … x … - 1.182

37. P1x Ú 02

q

L1

ln x x2

dx. [5.3]

dx

L0 1x + 124>3

CHAPTER 5 TEST Let D1x2  1x  722 be the price, in dollars per unit, that consumers are willing to pay for x units of an item, and let S1x2  x 2  x  4 be the price, in dollars per unit, that producers are willing to accept for x units. Find: 1. The equilibrium point 2. The consumer surplus at the equilibrium point 3. The producer surplus at the equilibrium point 4. Business: future value. Find the future value of $12,000 invested for 10 yr at an annual percentage rate of 4.1%, compounded continuously. 5. Business: future value of a continuous income stream. Find the accumulated future value of $8000 per year, at an interest rate of 4.88%, compounded continuously, for 6 yr. 6. Physical science: demand for potash. In 2004 1t = 02, the world production of potash was approximately 49.9 million metric tons, and demand was increasing at the rate of 9.9% a year. (Source: U.S. Energy Information Administration.) If the demand continues to grow at this rate, how much potash will be produced from 2004 to 2016? 7. Physical science: depletion of potash. See Exercise 6. The world reserves of potash in 2004 were approximately 8300 million metric tons. (Source: U.S. Geological Survey.) Assuming the demand for potash continues to grow at the rate of 9.9% per year and no new reserves are discovered, when will the world reserves be depleted? 8. Business: accumulated present value of a continuous income stream. Bruce Kent wants to have $25,000 in 5 yr for a down payment on a house. Find the amount he needs to save, at R 1t2 dollars per year, at 6.125%, compounded continuously, to achieve the desired future value. 9. Business: contract buyout. Guy Laplace signs a 6-yr contract to play professional hockey at a salary of $475,000 per year. After 2 yr, his team offers to buy out the remainder of his contract. What is the least amount Guy should accept, if the going interest rate is 7.1%, compounded continuously? 10. Business: future value of a noncontinuous income stream. Stan signs a contract that will pay him an income given by R 1t2 = 100,000 + 10,000t, where t is in years and 0 … t … 8. If he invests this money at 5%, compounded continuously, what is the future value of the income stream? Determine whether each improper integral is convergent or divergent, and calculate its value if it is convergent. q

11.

L1

dx x

5

q

12.

4 dx L0 1 + 3x

13. Find k such that f 1x2 = kx 3 is a probability density function over the interval 30, 24. Then write the probability density function. 14. Business: times of telephone calls. A telephone company determines that the length of a phone call, t, in minutes, is an exponentially distributed random variable with probability density function f 1t2 = 2e -2t, 0 … t 6 q . Find the probability that a phone call will last no more than 3 min. Given the probability density function f 1x2  14 x over 71, 38, find each of the following. 15. E 1x2

16. E 1x 22

17. The mean

18. The variance

19. The standard deviation 20. The percentile corresponding to x = 2 Let x be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. 21. P10 … x … 1.32

22. P1- 2.31 … x … - 1.052

23. P1- 1.61 … x … 1.762 24. The price per pound p of wild salmon at various stores in a certain city is normally distributed with mean m = $12 and standard deviation s = $2.50. What is the probability that the price at a randomly selected store is at least $13.25 per pound? 25. Business: price distribution. If the price per pound p of wild salmon is normally distributed with mean m = $12 and standard deviation s = $2.50, what is the lowest price in the top 15% of salmon prices? Find the volume generated by rotating about the x-axis the regions bounded by the following. 26. y =

1 2x

, x = 1, x = 5

27. y = 22 + x, x = 0, x = 1 Solve each differential equation. 28. 30.

dy dx dy dt

= 8x 7y

29.

dy dx

=

9 y

= 6y; y = 11 when t = 0

31. y¿ = 5x 2 - x 2y

32.

dv = 2v -3 dt

33. y¿ = 4y + xy

Chapter 5 Test

543

544

CHAPTER 5

•

Applications of Integration

34. Economics: elasticity. Find the demand function q = D1x2, given the elasticity condition

SYNTHESIS

35. Business: stock growth. The growth rate of Fabric Industries stock, in dollars per month, can be modeled by

37. Determine whether the following improper integral is convergent or divergent, and calculate its value if it is convergent:

E 1x2 = 4 for all x 7 0.

dV = k1L - V2, dt where V is the value of a share, in dollars, after t months; L = $36, the limiting value of the stock; k is a constant; and V 102 = 0. a) Write the solution V 1t2 in terms of L and k. b) If V 162 = 18, determine k to the nearest hundredth. c) Rewrite V 1t2 in terms of t and k using the value of k found in part (b). d) Use the equation in part (c) to find V 1122, the value of the stock after 12 months. e) In how many months will the value be $30?

36. The function f 1x2 = x 3 is a probability density function over the interval 30, b4. What is b?

0

x 3e -x dx. 4

L- q

TECHNOLOGY CONNECTION 38. Approximate the integral q

1

L- q 1 + x 2

dx.

Extended Technology Application Curve Fitting and Volumes of Containers Consider the urn or vase shown at the right. How could we estimate the volume? One way would be to simply fill the container with a liquid and then pour the liquid into a measuring device. Another way, using calculus and the curvefitting or REGRESSION feature of a graphing calculator, would be to turn the urn on its side, as shown below, take a series of vertical measurements from the center to the top, use REGRESSION, and then integrate (either by hand or with the aid of the calculator).

20

–10

50

–20

Xscl=10, Yscl=4

The following table is a table of values for the red curve. x (in centimeters)

y (in centimeters)

3

4

5

10

10

17

15

16

20

10

25

5

28

3.5

34

7

EXERCISES 1. Using REGRESSION, fit a cubic polynomial function to the data. 2. Using the function found in Exercise 1, integrate over the interval 33, 344 to find the volume of the urn. (Hint: If the function in Exercise 1 is Y1, find the volume by using the O key to enter pY1^2 as Y2. Then use the CALC option to integrate.)

Extended Technology Application

545

546

CHAPTER 5

•

Applications of Integration

Now consider the bottle shown at the right. To find the bottle’s volume in a similar manner, we turn it on its side, use a measuring device to take vertical measurements, and proceed as we did with the urn.

y 2 1.275 1.275 1.125 1.125 1.275 1.250 1.000 0.875 0.750 1

1

The table of measurements is as follows. x (in inches)

y (in inches)

1

1.125

2

1.275

3

1.250

4

1.275

5

1.275

6

1.125

7

1.000

8

0.875

9

0.750

10

0.500

11

0.500

2

3

4

5

6

7

8

9

0.500 0.500 10

11 x

EXERCISES 3. Using REGRESSION, fit a quartic polynomial function to the data. 4. Using the function found in Exercise 3, integrate to find the volume of the bottle. Your answer will be in cubic inches. Convert it to fluid ounces using the fact that 1 in3 = 0.55424 fluid ounce. 5. The bottle in question holds 20 oz. How good was our curve-fitting procedure for making the volume estimate? 6. Find a curve that gives a better estimate of the volume. What is the curve and what is the estimated volume?

Functions of Several Variables

6

Chapter Snapshot What You’ll Learn

Why It’s Important

6.1 6.2 6.3 6.4 6.5 6.6

Functions that have more than one input are called functions of several variables. We introduce these functions in this chapter and learn to differentiate them to find partial derivatives. Then we use such functions and their partial derivatives to find regression lines and solve maximum–minimum problems. Finally, we consider the integration of functions of several variables.

Functions of Several Variables Partial Derivatives Maximum–Minimum Problems An Application: The Least-Squares Technique Constrained Optimization Double Integrals

Where It’s Used PREDICTING THE MINIMUM WAGE

The minimum hourly wage in the United States has grown over the years, as shown in the table. Find the regression line, and use it to predict the minimum hourly wage in 2015 and 2020. This problem appears as Exercise 5 in Section 6.4.

INCREASE IN THE MINIMUM WAGE NUMBER OF YEARS, x, SINCE 1990

MINIMUM HOURLY WAGE

0

3.80

1

4.25

6

4.75

7

5.15

17

5.85

18

6.55

19

7.25

(Source: www.workworld.org.)

547

548

CHAPTER 6

•

Functions of Several Variables

6.1 OBJECTIVE • Find a function value for a function of several variables.

Functions of Several Variables Suppose that a one-product firm produces x units of its product at a profit of $4 per unit. Then its total profit P is given by P1x2 = 4x. This is a function of one variable. Suppose that a two-product firm produces x units of one product at a profit of $4 per unit and y units of a second product at a profit of $6 per unit. Then its total profit P is a function of the two variables x and y, and is given by P1x, y2 = 4x + 6y.

This function assigns to the input pair 1x, y2 a unique output number, 4x + 6y.

DEFINITION

A function of two variables assigns to each input pair, 1x, y2, exactly one output number, f1x, y2.

We can regard a function of two variables as a machine that has two inputs. Thus, the domain is a set of pairs 1x, y2 in the plane. When such a function is given by a formula, the domain normally consists of all ordered pairs 1x, y2 that are meaningful replacements in the formula.

(x, y) x y

4x  6y P(x, y)  4x  6y

■ EXAMPLE 1 For the above profit function, P1x, y2 = 4x + 6y, find P125, 102. Solution P125, 102 is defined to be the value of the function found by substituting 25 for x and 10 for y: Quick Check 1 A company’s cost function is given by C1x, y2 = 6.5x + 7.25y. Find C110, 152.

P125, 102 = 4 # 25 + 6 # 10 = 100 + 60 = $160.

This result means that by selling 25 units of the first product and 10 of the second, the two-product firm will make a profit of $160.

O Quick Check 1 The following are examples of functions of several variables, that is, functions of two or more variables. If there are n variables, then there are n inputs for such a function.

■ EXAMPLE 2 Business: Monthly Payment on an Amortized Loan. Large purchases are often financed with an amortized loan. Borrowers like to know how much they can expect to pay per month for every thousand dollars borrowed. The monthly

O

6.1

•

Functions of Several Variables

549

payment P depends on the annual percentage rate (APR) r and the term of the loan t (in years). The function P of the two variables r and t is given by 1000ra1 + P1r, t2 =

r 12t b 12

r 12t 12a1 + b - 12 12

.

How much per month can a borrower expect to pay per thousand dollars borrowed at an APR of 6.5% for a 6-yr term? Solution We let r = 0.065 and t = 6 and evaluate P10.065, 62: 100010.0652a1 + Quick Check 2

O

Determine the monthly payment per thousand dollars borrowed at an APR of 7.25% for a term of 8 yr.

P10.065, 62 =

0.065 12162 b 12

0.065 12162 b 12a1 + - 12 12

= $16.81.

The monthly payment is $16.81 per thousand dollars borrowed.

O Quick Check 2 ■ EXAMPLE 3 Business: Payment Tables. The formula in Example 2 is used to generate a table of payments that allows borrowers to easily judge the combined effects of the APR and the term. The table below shows the monthly payments per thousand dollars borrowed at various APRs and terms.

Annual Percentage Rate, r

Term, t (in years) 4

5

6

7

8

0.05

$23.03

$18.87

$16.10

$14.13

$12.66

0.055

$23.26

$19.10

$16.34

$14.37

$12.90

0.06

$23.49

$19.33

$16.57

$14.61

$13.14

0.065

$23.71

$19.57

$16.81

$14.85

$13.39

0.07

$23.95

$19.80

$17.05

$15.09

$13.63

0.075

$24.18

$20.04

$17.29

$15.34

$13.88

a) What can a borrower expect to pay per month at an APR of 5.5% for a 7-yr term? b) What can a borrower expect to pay per month at the same APR as in part (a) but for 6-yr term? Quick Check 3 a) What is the monthly payment per thousand dollars borrowed at an APR of 6.5% for a term of 5 yr? b) How much less per month would the payment be with the same APR but for a term of 6 yr?

Solution The monthly payments are read directly from the table. a) We see that P10.055, 72 = $14.37 per month. b) From the table, P10.055, 62 = $16.34 per month. The borrower pays more per month with a 6-yr term, but less overall than at the same rate for a 7-yr term: $16.34 for 72 months, for a total payment of $1,176.48 for the 6-yr term, or $14.37 for 84 months, for a total of $1,207.08 for the 7-yr term. Tables like the one above allow us to see the behavior of a multivariable function at a glance.

O Quick Check 3

O

550

CHAPTER 6

•

Functions of Several Variables

■ EXAMPLE 4 Business: Total Cost. The total cost to a company, in thousands of dollars, of producing its goods is given by

C1x, y, z, w2 = 4x 2 + 5y + z - ln 1w + 12,

where x dollars are spent for labor, y dollars for raw materials, z dollars for advertising, and w dollars for machinery. This is a function of four variables (all in thousands of dollars). Find C13, 2, 0, 102.

Solution We substitute 3 for x, 2 for y, 0 for z, and 10 for w: C13, 2, 0, 102 = 4 # 32 + 5 # 2 + 0 - ln 110 + 12 = 4 # 9 + 10 + 0 - 2.397895 L $43.6 thousand, or $43,600.

■ EXAMPLE 5 Business: Cost of Storage Equipment. A business purchases a piece of storage equipment that costs C1 dollars and has capacity V1. Later it wishes to replace the original with a new piece of equipment that costs C2 dollars and has capacity V2. Industrial economists have found that in such cases, the cost of the new piece of equipment can be estimated by the function of three variables C2 = a

V2 0.6 b C1. V1

For $45,000, a beverage company buys a manufacturing tank that has a capacity of 10,000 gallons. Later it decides to buy a tank with double the capacity of the original. Estimate the cost of the new tank. Quick Check 4 a) Repeat Example 5 assuming that the company buys a tank with a capacity of 2.75 times that of the original. b) What is the percentage increase in cost for this tank compared to the cost of the original tank?

Solution We substitute 20,000 for V2, 10,000 for V1, and 45,000 for C1: 20,000 0.6 b 145,0002 10,000 = 2 0.6145,0002 L $68,207.25.

C2 = a

Note that a 100% increase in capacity was achieved by about a 52% increase in cost. This is independent of any increase in the costs of labor, management, or other equipment resulting from the purchase of the tank.

O Quick Check 4 ■ EXAMPLE 6 Social Science: The Gravity Model. As the populations of two cities grow, the number of telephone calls between the cities increases, much like the gravitational pull will increase between two growing objects in space. The average number of telephone calls per day between two cities is given by N1d, P1, P22 =

2.8P1P2 d 2.4

,

where d is the distance, in miles, between the cities and P1 and P2 are their populations. The cities of Dallas and Fort Worth are 30 mi apart and have populations of 1,279,910 and 720,250, respectively. (Sources: Population Division, U.S. Census Bureau, 2009 estimates, and Rand McNally.) Find the average number of calls per day between the two cities.

O

•

6.1

Functions of Several Variables

551

Pilot Point Decatur McKinney

Denton

Plano DALLAS FORT WORTH 30 mi

Kaufman

Amarillo

Waxahachie Ennis

Alvarado Cleburne

FORT WORTH

Hillsboro

Midland

DALLAS

Waco Jasper AUSTIN

Alpine

Del Rio

Houston San Antonio

Larado

Quick Check 5

O

Find the average number of calls per day between Phoenix, Arizona (population 1,552,300) and Tucson, Arizona (population 541,800), given that the distance between the two cities is 120 mi. (Source: www.census.gov.)

Solution We evaluate the function with the aid of a calculator: N130, 1,279,910, 720,2502 =

2.811,279,9102 1720,2502 302.4

L 735,749,066.

O Quick Check 5 Geometric Interpretations

z

Visually, a function of two variables, z = f1x, y2,

z1

0

(x1, y1, z1) y1

x1

x

(x1, y1)

y

can be thought of as matching a point 1x 1, y12 in the xy-plane with the number z 1 on a number line. Thus, to graph a function of two variables, we need a three-dimensional coordinate system. The axes are generally placed as shown to the left. The line z, called the z-axis, is placed perpendicular to the xy-plane at the origin. To help visualize this, think of looking into the corner of a room, where the floor is the xy-plane and the z-axis is the intersection of the two walls. To plot a point 1x 1, y1, z 12, we locate the point 1x 1, y12 in the xy-plane and move up or down in space according to the value of z 1. z

■ EXAMPLE 7 Plot these points:

and

5

P112, 3, 52, P212, - 2, - 42, P310, 5, 22, P412, 3, 02.

4 P1(2, 3, 5)

3 2 1 –3

–2

Solution The solution is shown at the right. 5

4

3

P3(0, 5, 2)

–1 2

1

x P2(2, – 2, –4)

1

2

3 4 P4(2, 3, 0)

5

y

552

CHAPTER 6

•

Functions of Several Variables z

The graph of a function of two variables,

z 1  f (x 1, y1)

z = f1x, y2,

consists of ordered triples 1x 1, y1, z 12, where z 1 = f1x 1, y12. This graph takes the form of a surface. The domain of a two-variable function is the set of points in the xy-plane for which f is defined.

S

y

D

x

(x 1, y1) Elliptic paraboloid: z = x 2 + y2 z

■ EXAMPLE 8 Determine the domain of each two-variable function. a) f1x, y2 = x 2 + y 2 b) g1x, y2 = 21 - x 2 - y 2 1 c) h1x, y2 = x 2 + y 2 + 2 x + y2

Solution y

a) Since we can square any real number and sum any two squares, the function f is defined for all x and all y. Therefore, the domain for f is D = 51x, y2 ƒ - q 6 x 6 q ,

x

Hemisphere: z = √ 1 − x 2 − y2

- q 6 y 6 q 6.

The graph of f is a surface called an elliptic paraboloid. Satellite dishes are elliptic paraboloids: the weak incoming signals bounce off the interior surface of the paraboloid and collect at a single point, called the focus, thus amplifying the signal.

z

b) The expression within the radical must be nonnegative. Therefore, 1 - x 2 - y 2 Ú 0, which simplifies to x 2 + y 2 … 1. The domain for g is D y x

Quick Check 6 Determine the domain of each multivariable function. x + y a) f1x, y2 = x - y 1 2 + x - 2 3 + y c) h1x, y2 = ln 1y - x 32 b) g1x, y2 =

z

D = 51x, y2 ƒ x 2 + y 2 … 16.

The graph of g is a surface called a hemisphere, of radius 1. Its domain is a filled-in circle of radius 1. We can think of the domain of g as the “shadow” it casts on the xy-plane. c) Since zero cannot be in the denominator, we must have x 2 + y 2 Z 0. Therefore, x and y cannot be 0 simultaneously. The domain of h is D = 51x, y2 ƒ 1x, y2 Z 10, 026.

y

The graph of h is shown at right.

O Quick Check 6

x

z  x2  y2 

1 x2  y2

O

6.1

•

Functions of Several Variables

TECHNOLOGY CONNECTION

Exploratory

Graph: 1xy22 + 1yz22 + 1zx22 = xyz. This is entered as follows: EXAMPLE 4

Another useful and inexpensive app for the iPhone and iPod Touch is Grafly, a graphing calculator that creates visually appealing 3D graphs of functions of two variables. It has full graphing interactivity, with touch-based zoom and scroll features. Live gravity mode allows graph exploration through simply moving the device “around the figure” to view it from any angle. This app is well explained at the iPhone store. Some functions and their graphs are presented here as examples.

(xy)^2+(yz)^2+(xz)^2=xyz

The graph is shown at the right.

Graph: A 1 - 2x 2 + y 2 B 2 + z 2 = 0.2. This is entered as follows:

EXAMPLE 1

(1-sqrt(x^2+y^2))^2+x^2=0.2

The graph is shown at the right.

Graph: 4x 2 + 2y 2 + z 2 = 1. This is entered as follows:

EXAMPLE 5

4x^2+2y^2+z^2=1

The graph is shown at the right.

Graph: ƒ 12x 2 + 2y 220.25 ƒ + ƒ 2z ƒ = 1. This is entered as follows: EXAMPLE 2

abs((2x^2+2y^2)^0.25)+abs((z)^0.5)=1

The graph is shown at the right.

Graph: z = - 8xe -41x This is entered as follows:

EXAMPLE 6

2

+ y22

.

z=–8xe^(-4(x^2+y^2))

The graph is shown at the right.

Graph: z = e -41x This is entered as follows:

EXAMPLE 3

2

+ y22

.

z=e^(-4(x^2+y^2))

EXERCISE

The graph is shown at the right.

Use Grafly to graph the functions in Exercises 1–12 on p. 554.

553

554

CHAPTER 6

•

Functions of Several Variables

Section Summary • A function of two variables assigns to each input pair, 1x, y2, exactly one output number, f1x, y2. • A function of two variables generates points 1x, y, z2, where z = f1x, y2.

• The graph of a function of two variables is a surface and requires a three-dimensional coordinate system. • The domain of a function of two variables is the set of points in the xy-plane for which the function is defined.

EXERCISE SET

6.1 1. For f 1x, y2 = x 2 - 3xy, find f 10, - 22, f 12, 32, and f 110, - 52.

2. For f 1x, y2 = 1 y 2 + 2xy23, find f 1- 2, 02, f 13, 22, and f 1- 5, 102. 3. For f 1x, y2 = 3x + 7xy, find f 10, - 22, f 1- 2, 12, and f 12, 12.

4. For f 1x, y2 = log 101x + y2 + 3x 2, find f 13, 72, f 11, 992, and f 12, - 12.

5. For f 1x, y2 = ln x + y 3, find f 1e, 22, f 1e 2, 42, and f 1e 3, 52. 6. For f 1x, y2 = 2 x - 3 y, find f 10, 22, f 13, 12, and f 12, 32. 7. For f 1x, y, z2 = x 2 - y 2 + z 2, find f 1- 1, 2, 32 and f 12, - 1, 32.

8. For f 1x, y, z2 = 2 x + 5zy - x, find f 10, 1, - 32 and f 11, 0, - 32. In Exercises 9–12, determine the domain of each function of two variables. 9. f 1x, y2 = 2y - 3x 11. h1x, y2 = xe

2y

10. g1x, y2 =

1 y + x2

y 1 12. k1x, y2 = + x x - 1

APPLICATIONS Business and Economics 13. Price–earnings ratio. The price–earnings ratio of a stock is given by R1P, E2 =

P , E

where P is the price of the stock and E is the earnings per share. The price per share of Hewlett-Packard stock was $32.03, and the earnings per share were $1.25. (Source: yahoo.finance.com.) Find the price–earnings ratio. Use decimal notation rounded to the nearest hundredth. 14. Yield. The yield of a stock is given by Y1D, P2 =

D , P

where D is the dividend per share of stock and P is the price per share. The price per share of Texas Instruments stock was $30, and the dividend per share was $0.12. (Source: yahoo.finance.com.) Find the yield. Use percent notation rounded to the nearest hundredth of a percent. 15. Cost of storage equipment. Consider the cost model in Example 5. For $100,000, a company buys a storage tank that has a capacity of 80,000 gal. Later it replaces the tank with a new tank that has double the capacity of the original. Estimate the cost of the new tank. 16. Savings and interest. A sum of $1000 is deposited in a savings account for which interest is compounded monthly. The future value A is a function of the annual percentage rate r and the term t, in months, and is given by A1r, t2 = 1000a 1 +

r 12t b . 12

a) Determine A10.05, 102. b) What is the interest earned for the rate and term in part (a)? c) How much more interest can be earned over the same term as in part (a) if the APR is increased to 5.75%? 17. Monthly car payments. Kim is shopping for a car. She will finance $10,000 through a lender. Use the table in Example 3 to answer the following questions. a) One lender offers Kim an APR of 6% for a 6-yr term. What would Kim’s monthly payment be? b) A competing lender offers an APR of 5.5% but for a 7-yr term. What would Kim’s monthly payment be? c) Assume that Kim makes the minimum payment each month for the entire term of the loan. Calculate her total payments for both options described in parts (a) and (b). Which option costs Kim less overall?

Life and Physical Sciences 18. Poiseuille’s Law. The speed of blood in a vessel is given by V1L, p, R, r, v2 =

p 1R 2 - r 22, 4Lv

where R is the radius of the vessel, r is the distance of the blood from the center of the vessel, L is the length of the blood vessel, p is the pressure, and v is the viscosity. Find V11, 100, 0.0075, 0.0025, 0.052.

555

Exercise Set 6.1

19. Wind speed of a tornado. Under certain conditions, the wind speed S, in miles per hour, of a tornado at a distance d feet from its center can be approximated by the function S1a, d, V2 =

aV 0.51d 2

,

23. Dewpoint. The dewpoint is the temperature at which moisture in the air condenses into liquid (dew). It is a function of air temperature t and relative humidity h. The table below shows the dewpoints for select values of t and h. Relative Humidity (%)

Air Temperature (degrees Fahrenheit)

where a is a constant that depends on certain atmospheric conditions and V is the approximate volume of the tornado, in cubic feet. Approximate the wind speed 100 ft from the center of a tornado when its volume is 1,600,000 ft3 and a = 0.78.

b) A goaltender gave up 124 goals during the season and had a goals against average of 3.75. How many minutes did he play? (Round to the nearest integer.) c) State the domain for A.

20. Body surface area. The Mosteller formula for approximating the surface area S, in square meters 1m22, of a human is given by S1h, w2 =

2hw , 60

where h is the person’s height in centimeters and w is the person’s weight in kilograms. (Source: www.halls.md.) Use the Mosteller approximation to estimate the surface area of a person whose height is 165 cm and whose weight is 80 kg. 21. Body surface area. The Haycock formula for approximating the surface area S, in square meters 1m22, of a human is given by S1h, w2 = 0.024265h0.3964w 0.5378, where h is the person’s height in centimeters and w is the person’s weight in kilograms. (Source: www.halls.md.) Use the Haycock approximation to estimate the surface area of a person whose height is 165 cm and whose weight is 80 kg.

General Interest 22. Goals against average. A hockey goaltender’s goals against average A is a function of the number of goals g allowed and the number m of minutes played and is given by the formula A1g, m2 =

60g . m

a) Determine the goals against average of a goaltender who allows 35 goals while playing 820 min. Round A to the nearest hundredth.

20

40

60

80

100

70

29

44

55

63

70

80

35

53

65

73

80

90

43

62

74

83

90

100

52

71

84

93

100

a) What is the dewpoint when the air temperature is 80°F with a relative humidity of 60%? b) What is the dewpoint when the air temperature is 90°F with a relative humidity of 40%? c) The air feels humid when the dewpoint reaches about 60. If the air temperature is 100°F, at what approximate relative humidity will the air feel humid? d) Explain why the dewpoint is equal to the air temperature when the relative humidity is 100%.

SYNTHESIS 24. For the tornado described in Exercise 19, if the wind speed measures 200 mph, how far from the center was the measurement taken? 25. According to the Mosteller formula in Exercise 20, if a person’s weight drops 19%, by what percentage does his or her surface area change? 26. Explain the difference between a function of two variables and a function of one variable. 27. Find some examples of functions of several variables not considered in the text, even some that may not have formulas.

TECHNOLOGY CONNECTION General Interest Wind chill temperature. Because wind speed enhances the loss of heat from the skin, we feel colder when there is wind than when there is not. The wind chill temperature is what the temperature would have to be with no wind in order to give the same chilling effect. The wind chill temperature, W, is given by W1v, T2  91.4 

110.45  6.682v  0.447v21457  5T2 110

,

556

CHAPTER 6

•

Functions of Several Variables

where T is the actual temperature measured by a thermometer, in degrees Fahrenheit, and v is the speed of the wind, in miles per hour. Find the wind chill temperature in each case. Round to the nearest degree. 28. T = 30°F, v = 25 mph

32. Use a computer graphics program such as Maple or Mathematica, an Internet site such as www.wolframalpha.com, a graphing calculator, or an iPhone app such us Grafly to view the graph of each function given in Exercises 1–8. Use a 3D graphics program to generate the graph of each function.

29. T = 20°F, v = 20 mph

33. f 1x, y2 = y 2

30. T = 20°F, v = 40 mph

35. f 1x, y2 = 1x 4 - 16x 22e -y

31. T = - 10°F, v = 30 mph

34. f 1x, y2 = x 2 + y 2

2

36. f 1x, y2 = 41x 2 + y 22 - 1x 2 + y 222 37. f 1x, y2 = x 3 - 3xy 2 38. f 1x, y2 =

1 x 2 + 4y 2

Answers to Quick Checks 1. $173.75 2. $13.76 3. (a) $19.57 (b) $2.76 4. (a) $82,568.07 (b) 83.5% increase 5. 24,095,597 calls/day 6. (a) D = 51x, y2 ƒ y Z x6 (b) D = 51x, y2 ƒ x Z 2, y Z - 36 (c) D = 51x, y2 ƒ y 7 x 36

6.2 OBJECTIVES • Find the partial derivatives of a given function. • Evaluate partial derivatives. • Find the four second-order partial derivatives of a function in two variables.

Partial Derivatives Finding Partial Derivatives Consider the function f given by z = f1x, y2 = x 2y 3 + xy + 4y 2. Suppose for the moment that we fix y at 3. Then

f1x, 32 = x 21332 + x132 + 41322 = 27x 2 + 3x + 36.

Note that we now have a function of only one variable. Taking the first derivative with respect to x, we have 54x + 3. In general, without replacing y with a specific number, we can consider y fixed. Then f becomes a function of x alone, and we can calculate its derivative with respect to x. This derivative is called the partial derivative of f with respect to x. Notation for this partial derivative is 0f 0x

or

0z . 0x

Now, let’s again consider the function z = f1x, y2 = x 2y 3 + xy + 4y 2.

6.2

•

Partial Derivatives

557

The color blue indicates the variable x when we fix y and treat it as a constant. The expressions y 3, y, and y 2 are then also treated as constants. We have 0f 0z = = 2xy 3 + y. 0x 0x Similarly, we find 0f>0y or 0z>0y by fixing x (treating it as a constant) and calculating the derivative with respect to y. From z = f1x, y2 = x 2y 3 + xy + 4y 2,

The color blue indicates the variable.

we get 0f 0z = = 3x 2y 2 + x + 8y. 0y 0y A definition of partial derivatives is as follows.

DEFINITION For z = f1x, y2, the partial derivatives with respect to x and y are f1x + h, y2 - f1x, y2 0z = lim h :0 0x h

and

f1x, y + h2 - f1x, y2 0z = lim . h :0 0y h

We can find partial derivatives of functions of any number of variables. Since we can apply the theorems for finding derivatives presented earlier, we will rarely need to use the definition to find a partial derivative.

■ EXAMPLE 1 For w = x 2 - xy + y 2 + 2yz + 2z 2 + z, find 0w , 0x

0w , and 0y

0w . 0z

Solution In order to find 0w>0x, we regard x as the variable and treat y and z as constants. From w = x 2 - xy + y 2 + 2yz + 2z 2 + z, we get 0w = 2x - y. 0x To find 0w>0y, we regard y as the variable and treat x and z as constants. We get 0w = - x + 2y + 2z; 0y Quick Check 1 For u = x 2y 3z 4, find 0u 0u 0u , , and . 0x 0y 0z

To find 0w>0z, we regard z as the variable and treat x and y as constants. We get 0w = 2y + 4z + 1. 0z

O Quick Check 1

O

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•

Functions of Several Variables

We will often make use of a simpler notation: fx for the partial derivative of f with respect to x and fy for the partial derivative of f with respect to y. Similarly, if z = f1x, y2, then z x represents the partial derivative of z with respect to x, and z y represents the partial derivative of z with respect to y.

■ EXAMPLE 2 For f1x, y2 = 3x 2y + xy, find fx and fy. Solution We have Quick Check 2 x For f1x, y2 = 7x 3y 2 - , y find fx and fy.

fx = 6xy + y, fy = 3x 2 + x.

Treating y as a constant Treating x 2 and x as constants

O Quick Check 2

O

For the function in Example 2, let’s evaluate fx at 12, - 32: fx12, - 32 = 6 # 2 # 1- 32 + 1- 32 = - 39. If we use the notation 0f>0x = 6xy + y, where f = 3x 2y + xy, the value of the partial derivative at 12, - 32 is given by 0f 2 = 6 # 2 # 1- 32 + 1- 32 0x 12, -32 = - 39.

However, this notation is not quite as convenient as fx12, - 32.

■ EXAMPLE 3 For f1x, y2 = e xy + y ln x, find fx and fy. TECHNOLOGY CONNECTION

Exploratory Consider finding values of a partial derivative of f 1x, y2 = 3x 3y + 2xy using a calculator that finds derivatives of functions of one variable. How can you find fx1- 4, 12? Then how can you find fy12, 62?

Solution fx = y # e xy + y # = ye xy +

1 x

y , x

fy = x # e xy + 1 # ln x = xe xy + ln x

The Geometric Interpretation of Partial Derivatives The graph of a function of two variables z = f1x, y2 is a surface S, which might have a graph similar to the one shown to the right, where each input pair 1x, y2 in the domain D has only one output, z = f1x, y2.

z

(x, y, z)

S

D

x

(x, y)

y

6.2

Now suppose that we hold x fixed at the value a. The set of all points for which x = a is a plane parallel to the yz -plane; thus, when x is fixed at a, y and z vary along that plane, as shown to the right. The plane in the figure cuts the surface along the curve C1. The partial derivative fy gives the slope of tangent lines to this curve, in the positive y-direction.

•

559

Partial Derivatives

z

C1

Slope is

∂z , or f y ∂y

a y

x

z

Similarly, if we hold y fixed at the value b, we obtain a curve C2, as shown to the right. The partial derivative fx gives the slope of tangent lines to this curve, in the positive x-direction.

Slope is

∂z , or f x ∂x

C2

b x

y

An Economics Application: The Cobb–Douglas Production Function One model of production that is frequently considered in business and economics is the Cobb–Douglas production function: p1x, y2 = Ax ay 1 - a, for A 7 0 and 0 6 a 6 1, where p is the number of units produced with x units of labor and y units of capital. (Capital is the cost of machinery, buildings, tools, and other supplies.) The partial derivatives 0p 0x

and

0p 0y

are called, respectively, the marginal productivity of labor and the marginal productivity of capital.

■ EXAMPLE 4 A cellular phone company has the following production function for a smart phone: p1x, y2 = 50x 2/3y 1/3, where p is the number of units produced with x units of labor and y units of capital. a) Find the number of units produced with 125 units of labor and 64 units of capital. b) Find the marginal productivities. c) Evaluate the marginal productivities at x = 125 and y = 64.

560

CHAPTER 6

•

Functions of Several Variables

Solution

a) p1125, 642 = 50112522/316421/3 = 501252142 = 5000 units 100y 1/3 0p 2 = px = 50a bx -1/3y 1/3 = 0x 3 3x 1/3 0p 1 50x 2/3 Marginal productivity of capital = = py = 50a bx 2/3y -2/3 = 0y 3 3y 2/3

b) Marginal productivity of labor =

Quick Check 3 A publisher’s production function for textbooks is given by p1x, y2 = 72x 0.8y 0.2, where p is the number of books produced, x is units of labor, and y is units of capital. Determine the marginal productivities at x = 90 and y = 50.

c) For 125 units of labor and 64 units of capital, we have Marginal productivity of labor = px1125, 642 10016421/3

=

1/3

311252

=

100142 3152

= 26 23 ,

Marginal productivity of capital = py1125, 642 =

50112522/3 2/3

31642

=

501252 31162

= 26 241 .

O Quick Check 3 Let’s interpret the marginal productivities of Example 4. To visualize the marginal productivity of labor, suppose that capital is fixed at 64 units. Then a one-unit change in labor, from 125 to 126, will cause production to increase by about 26 23 units. To visualize the marginal productivity of capital, suppose that the amount of labor is fixed at 125 units. Then a one-unit change in capital from 64 to 65 will cause production to 1 increase by about 26 24 units. A Cobb–Douglas production function is consistent with the law of diminishing returns. That is, if one input (either labor or capital) is held fixed while the other increases infinitely, then production will eventually increase at a decreasing rate. With such functions, it also turns out that if a certain maximum production is possible, then the expense of more labor, for example, may be required for that maximum output to be attainable.

Higher-Order Partial Derivatives Consider z = f1x, y2 = 3xy 2 + 2xy + x 2. Then

0f 0z = = 3y 2 + 2y + 2x. 0x 0x

Suppose that we continue and find the first partial derivative of 0z/0x with respect to y. This will be a second-order partial derivative of the original function z. Its notation is as follows: 0 0z 0 0f 0 a b = a b = 13y 2 + 2y + 2x2 = 6y + 2. 0y 0x 0y 0x 0y The notation

0 0z a b is often expressed as 0y 0x

0 2z 0y 0x

or

0 2f . 0y 0x

We could also denote the preceding partial derivative using the notation fxy: fxy = 6y + 2.

O

6.2

•

Partial Derivatives

561

Note that in the notation fxy, x and y are in the order (left to right) in which the differentiation is done, but in 0 2f , 0y 0x the order of x and y is reversed. In each case, the differentiation with respect to x is done first, followed by differentiation with respect to y. Notation for the four second-order partial derivatives is as follows.

DEFINITION

Second-Order Partial Derivatives 0 2f 0 f 0 z Take the partial with respect to x, 0 2z 1. = = 2 = 2 = fxx and then with respect to x again. 0x 0x 0x 0x 0x 0x 2

2

2.

0 2f 0 2z = = fxy 0y 0x 0y 0x

Take the partial with respect to x, and then with respect to y.

0 2f 0 2z = = fyx 0x 0y 0x 0y 0 2f 0 2f 0 2z 0 2z 4. = = 2 = 2 = fyy 0y 0y 0y 0y 0y 0y

3.

Take the partial with respect to y, and then with respect to x. Take the partial with respect to y, and then with respect to y again.

■ EXAMPLE 5 For z = f1x, y2 = x 2y 3 + x 4y + xe y, find the four second-order partial derivatives.

Solution a)

0 2f 0x 2

= fxx =

0 12xy 3 + 4x 3y + e y2 0x

Differentiate twice with respect to x.

= 2y 3 + 12x 2y b)

c) Quick Check 4 For z = g1x, y2 = 6x 2 + 3xy 4 - y 2, find the four second-order partial derivatives.

d)

0 2f 0 = fxy = 12xy 3 + 4x 3y + e y2 0y 0x 0y = 6xy 2 + 4x 3 + e y 0 2f 0 = fyx = 13x 2y 2 + x 4 + xe y2 0x 0y 0x = 6xy 2 + 4x 3 + e y 0 2f 0y 2

= fyy =

0 13x 2y 2 + x 4 + xe y2 0y

Differentiate with respect to x and then with respect to y.

Differentiate with respect to y and then with respect to x.

Differentiate twice with respect to y.

= 6x 2y + xe y

O Quick Check 4 We see by comparing parts (b) and (c) of Example 5 that 0 2f 0 2f = 0y 0x 0x 0y

and

fxy = fyx.

Although this will be true for virtually all functions that we consider in this text, it is not true for all functions. One function for which it is not true is given in Exercise 69. In Section 6.3, we will see how higher-order partial derivatives are used in applications to find extrema for functions of two variables.

O

562

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Functions of Several Variables

Section Summary • For z = f1x, y2, the partial derivatives with respect to x and y are, respectively: f1x + h, y2 - f1x,y2 0z = lim and h:0 0x h f1x, y + h2 - f1x, y2 0z = lim . h:0 0y h • Simpler notations for partial derivatives are fx and z x for 0z and fy and z y for . 0y

0z 0x

• For a surface z = f1x, y2 and a point 1x 0, y0, z 02 on this surface, the partial derivative of f with respect to x gives the slope of the tangent line at 1x 0, y0, z 02 in the positive x-direction. Similarly, the partial derivative of f with respect to y gives the slope of the tangent line at 1x 0, y0, z 02 in the positive y-direction. • For z = f1x, y2, the second-order partial derivatives are fxx =

0 2f 0x

,f = 2 xy

0 2f 0 2f 0 2f , fyx = , and fyy = 2 . 0y 0x 0x 0y 0y

Often (but not always), fxy = fyx.

EXERCISE SET

6.2 Find

Find fx, fy, and fL. (The symbol L is the Greek letter lambda.)

z z z z 2 2 , and . , ,  x  y  x 12, 32  y 10, 52

23. f 1x, y, l2 = 5xy - l12x + y - 82

1. z = 2x - 3y

2. z = 7x - 5y

3. z = 3x 2 - 2xy + y

4. z = 2x 3 + 3xy - x

Find fx, fy, fx12, 42, and fy14, 32. 5. f 1x, y2 = 2x - 5xy

6. f 1x, y2 = 5x + 7y

Find fx, fy, fx12, 12, and fy13, 22. 7. f 1x, y2 = 2x + y 2

8. f 1x, y2 = 2x - y 2

9. f 1x, y2 = e

11. f 1x, y2 = e

xy

13. f 1x, y2 = y ln 1x + 2y2 14. f 1x, y2 = x ln 1x - y2 15. f 1x, y2 = x ln 1xy2 17. f 1x, y2 =

y x y 3x

10. f 1x, y2 = e

12. f 1x, y2 = e

3x - 2y 2xy

16. f 1x, y2 = y ln 1xy2 18. f 1x, y2 =

y x + y 5x

19. f 1x, y2 = 312x + y - 522 20. f 1x, y2 = 413x + y - 822 Find

26. f 1x, y, l2 = x 2 - y 2 - l14x - 7y - 102 Find the four second-order partial derivatives. 27. f 1x, y2 = 5xy

29. f 1x, y2 = 7xy 2 + 5xy - 2y

2

2x - y

25. f 1x, y, l2 = x 2 + y 2 - l110x + 2y - 42

28. f 1x, y2 = 2xy

2

Find fx and fy.

24. f 1x, y, l2 = 9xy - l13x - y + 72

f f and . b m

21. f 1b, m2 = m 3 + 4m 2b - b 2 + 12m + b - 522 + 13m + b - 622 22. f 1b, m2 = 5m 2 - mb 2 - 3b + 12m + b - 822 + 13m + b - 922

30. f 1x, y2 = 3x 2y - 2xy + 4y 31. f 1x, y2 = x 5y 4 + x 3y 2

32. f 1x, y2 = x 4y 3 - x 2y 3 Find fxx, fxy, fyx, and fyy. (Remember, fyx means to differentiate with respect to y and then with respect to x.) 33. f 1x, y2 = 2x - 3y

34. f 1x, y2 = 3x + 5y

37. f 1x, y2 = x + e y

38. f 1x, y2 = y - e x

35. f 1x, y2 = e 2xy

39. f 1x, y2 = y ln x

36. f 1x, y2 = e xy

40. f 1x, y2 = x ln y

APPLICATIONS Business and Economics 41. The Cobb–Douglas model. Lincolnville Sporting Goods has the following production function for a certain product: p1x, y2 = 2400x 2/5y 3/5,

Exercise Set 6.2

where p is the number of units produced with x units of labor and y units of capital. a) Find the number of units produced with 32 units of labor and 1024 units of capital. b) Find the marginal productivities. c) Evaluate the marginal productivities at x = 32 and y = 1024. d) Interpret the meanings of the marginal productivities found in part (c).

563

higher humidity. The temperature–humidity index, Th, is what the temperature would have to be with no humidity in order to give the same heat effect. One index often used is given by Th  1.98T  1.0911  H21T  582  56.9, where T is the air temperature, in degrees Fahrenheit, and H is the relative humidity, expressed as a decimal. Find the temperature–humidity index in each case. Round to the nearest tenth of a degree.

42. The Cobb–Douglas model. Riverside Appliances has the following production function for a certain product: p1x, y2 = 1800x 0.621y 0.379, where p is the number of units produced with x units of labor and y units of capital. a) Find the number of units produced with 2500 units of labor and 1700 units of capital. b) Find the marginal productivities. c) Evaluate the marginal productivities at x = 2500 and y = 1700. d) Interpret the meanings of the marginal productivities found in part (c). Nursing facilities. A study of Texas nursing homes found that the annual profit P (in dollars) of profit-seeking, independent nursing homes in urban locations is modeled by the function

45. T = 85°F and H = 60%

P1w, r, s, t2  0.007955w 0.638 r 1.038 s 0.873 t 2.468.

46. T = 90°F and H = 90%

In this function, w is the average hourly wage of nurses and aides (in dollars), r is the occupancy rate (as a percentage), s is the total square footage of the facility, and t is the Texas Index of Level of Effort (TILE), a number between 1 and 11 that measures state Medicaid reimbursement. (Source: K. J. Knox, E. C. Blankmeyer, and J. R. Stutzman, “Relative Economic Efficiency in Texas Nursing Facilities,” Journal of Economics and Finance, Vol. 23, 199–213 (1999).) Use the preceding information for Exercises 43 and 44.

47. T = 90°F and H = 100%

43. A profit-seeking, independent Texas nursing home in an urban setting has nurses and aides with an average hourly wage of $20 an hour, a TILE of 8, an occupancy rate of 70%, and 400,000 ft2 of space. a) Estimate the nursing home’s annual profit. b) Find the four partial derivatives of P. c) Interpret the meaning of the partial derivatives found in part (b).

51. Body surface area. The Mosteller formula for approximating the surface area, S, in m2, of a human is given by

44. The change in P due to a change in w when the other variables are held constant is approximately ¢P L

0P ¢w. 0w

Use the values of w, r, s, and t in Exercise 43 and assume that the nursing home gives its nurses and aides a small raise so that the average hourly wage is now $20.25 an hour. By approximately how much does the profit change?

Life and Physical Sciences Temperature–humidity heat index. In the summer, humidity interacts with the outdoor temperature, making a person feel hotter because of reduced heat loss from the skin caused by

48. T = 78°F and H = 100% 49. Find

0Th , and interpret its meaning. 0H

50. Find

0Th , and interpret its meaning. 0T

S =

2hw , 60

where h is the person’s height in centimeters and w is the person’s weight in kilograms. (Source: www.halls.md.) 0S a) Compute . 0h 0S b) Compute . 0w c) The change in S due to a change in w when h is constant is approximately ¢S L

0S ¢w. 0w

Use this formula to approximate the change in someone’s surface area given that the person is 170 cm tall, weighs 80 kg, and loses 2 kg. 52. Body surface area. The Haycock formula for approximating the surface area, S, in m2, of a human is given by S = 0.024265h0.3964w 0.5378,

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•

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where h is the person’s height in centimeters and w is the person’s weight in kilograms. (Source: www.halls.md.) 0S a) Compute . 0h 0S . b) Compute 0w c) The change in S due to a change in w when h is constant is approximately

Use this formula to approximate the change in someone’s surface area given that the person is 170 cm tall, weighs 80 kg, and loses 2 kg.

Social Sciences Reading ease. The following formula is used by psychologists and educators to predict the reading ease, E, of a passage of words:

xy x - y

65. Do some research on the Cobb–Douglas production function, and explain how it was developed. 66. Explain the meaning of the first partial derivatives of a function of two variables in terms of slopes of tangent lines.

0 2f 0x 2

+

0 2f 0y 2

xfxy - fy = 0. 69. Consider the function f defined as follows: f 1x, y2 =

where w is the number of syllables in a 100-word section and s is the average number of words per sentence. Find the reading ease in each case.

lim

54. w = 180 and s = 6

h :0

56. Find

xy1x 2 - y 22

L

x 2 + y2

, for 1x, y2 Z 10, 02, for 1x, y2 = 10, 02.

0,

a) Find fx10, y2 by evaluating the limit

53. w = 146 and s = 5 0E . 0w

= 0.

68. Consider f 1x, y2 = x 3 - 5xy 2. Show that f is a solution to the partial differential equation

E  206.835  0.846w  1.015s,

55. Find

64. f 1x, y2 =

67. Consider f 1x, y2 = ln 1x 2 + y 22. Show that f is a solution to the partial differential equation

0S ¢w. 0w

¢S L

Find fxx, fxy, fyx, and fyy. y x 63. f 1x, y2 = 2 - 2 y x

f 1h, y2 - f 10, y2 h

.

b) Find fy1x, 02 by evaluating the limit

0E . 0s

lim

h :0

SYNTHESIS

f 1x, h2 - f 1x, 02 h

.

c) Now find and compare fyx10, 02 and fxy10, 02.

Find fx and ft. 57. f 1x, t2 = 59. f 1x, t2 =

x 2 + t2 x - t 2

58. f 1x, t2 =

2

22x - 22t 1 + 22t

x2 + t2 x - t 2

2

b

5

x3 + t

60. f 1x, t2 = 2x t

61. f 1x, t2 = 6x 2/3 - 8x 1/4t 1/2 - 12x - 1/2t 3/2 62. f 1x, t2 = a

x2 - t

4

3 5

Answers to Quick Checks 0u 0u 0u = 2xy 3z 4, = 3x 2y 2z 4, = 4x 2y 3z 3 0x 0y 0z 1 x 2. fx = 21x 2y 2 - , fy = 14x 3y + 2 y y 3. px190, 502 = 51.21 textbooks>unit of labor, py190, 502 = 23.05 textbooks>unit of capital 4. gxx = 12, gyy = 36xy 2 - 2, gxy = 12y 3, gyx = 12y 3 1.

6.3

6.3 OBJECTIVE

•

565

Maximum–Minimum Problems

Maximum–Minimum Problems We will now find maximum and minimum values of functions of two variables.

DEFINITION

• Find relative extrema of a function of two variables.

A function f of two variables: 1. has a relative maximum at 1a, b2 if f1x, y2 … f1a, b2

for all points 1x, y2 in a region containing 1a, b2; 2. has a relative minimum at 1a, b2 if f1x, y2 Ú f1a, b2

for all points 1x, y2 in a region containing 1a, b2.

This definition is illustrated in Figs. 1 and 2. A relative maximum (or minimum) may not be an “absolute” maximum (or minimum), as illustrated in Fig. 3. z

z

fx(a, b) = 0 Relative maximum = f(a, b)

Graph of f

fy(a, b) = 0 Graph of f fx(a, b) = 0 Relative minimum = f(a, b) fy(a, b) = 0 y

y x

x

(a, b)

FIGURE 1

FIGURE 2

Determining Maximum and Minimum Values

z

Absolute maximum

Relative maximum

Graph of f

y

x

Relative minimum

FIGURE 3

(a, b)

Suppose that a function f has a relative maximum or minimum value at some point 1a, b2 inside its domain. (We assume that f and its partial derivatives exist and are “continuous” inside its domain, though we will not formally define continuity.) If we fix y at the value b, then f1x, b2 can be regarded as a function of x. Because a relative maximum or minimum occurs at 1a, b2, we know that f1x, b2 achieves a maximum or minimum at 1a, b2 and f1x2 = 0. Similarly, if we fix x at a, then f1a, y2 can be

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•

Functions of Several Variables

regarded as a function of y that achieves a relative extremum at 1a, b2, and thus fy = 0. In short, since an extremum exists at 1a, b2, we must have fx1a, b2 = 0 and fy1a, b2 = 0.

z

fy(a, b) = 0

fx(a, b) = 0

x

(a, b)

FIGURE 4

(1)

We call a point 1a, b2 at which both partial derivatives are 0 a critical point. This concept of a critical value is comparable to that for functions of one variable. Thus, one strategy for finding relative maximum or minimum values is to solve a system of equations like (1) to find critical points. Just as for functions of one variable, this strategy does not guarantee that we will have a relative maximum or minimum value. We have argued only that if f has a maximum or minimum value at 1a, b2, then both its partial derivatives must be 0 at that point. Look back at Figs. 1 and 2. Then note Fig. 4, which illustrates a case in which the partial derivatives are 0 but the function does not have a relative maximum or minimum value at 1a, b2. Considering Fig. 4, suppose that we fix y at a value b. Then f1x, b2, considered as the output of a function of one variable x, has a minimum at a, but f does not. Similarly, if we fix x at a, then f1a, y2, considered as the output of a function of one variable y, has a maximum at b, but f does not. The point f1a, b2 is called a saddle point. In other words, fx1a, b2 = 0 and fy1a, b2 = 0 [the point 1a, b2 is a critical point], but f does not attain a relative maximum or minimum value at 1a, b2. y A test for finding relative maximum and minimum values that involves the use of first- and second-order partial derivatives is stated below. We will not prove this theorem.

f(x, b) has a minimum

THEOREM 1

f(a, y) has a maximum

The shape of a perfect tent. To give a tent roof the maximum strength possible, designers draw the fabric into a series of three-dimensional shapes that, viewed in profile, resemble a horse’s saddle and that mathematicians call an anticlastic curve. Two people with a stretchy piece of fabric such as Spandex can duplicate the shape, as shown above. One person pulls up and out on two diagonal corners; the other person pulls down and out on the other two corners. The opposing tensions draw each point of the fabric’s surface into rigid equilibrium. The more pronounced the curve, the stiffer the surface.

The D-Test To find the relative maximum and minimum values of f: 1. Find fx, fy, fxx, fyy, and fxy, 2. Solve the system of equations fx = 0, fy = 0. Let 1a, b2 represent a solution. 3. Evaluate D, where D = fxx1a, b2 # fyy1a, b2 - 3fxy1a, b242. 4. Then a) f has a maximum at 1a, b2 if D 7 0 and fxx1a, b2 6 0. b) f has a minimum at 1a, b2 if D 7 0 and fxx1a, b2 7 0. c) f has neither a maximum nor a minimum at 1a, b2 if D 6 0. The function has a saddle point at 1a, b2. See Fig. 4. d) This test is not applicable if D = 0.

The D-test is somewhat analogous to the Second Derivative Test (Section 2.2) for functions of one variable. Saddle points are analogous to critical values at which concavity changes and there are no relative maximum or minimum values. A relative maximum or minimum may or may not be an absolute maximum or minimum value. Tests for absolute maximum or minimum values are rather complicated. We will restrict our attention to finding relative maximum or minimum values. Fortunately, in most of our applications, relative maximum or minimum values turn out to be absolute as well.

6.3

•

Maximum–Minimum Problems

567

■ EXAMPLE 1 Find the relative maximum and minimum values of f1x, y2 = x 2 + xy + y 2 - 3x.

Solution 1. Find fx, fy, fxx, fyy, and fxy: fx = 2x + y - 3, fy = x + 2y, fxx = 2; fyy = 2; fxy = 1. 2. Solve the system of equations fx = 0, fy = 0:

(1) (2)

2x + y - 3 = 0, x + 2y = 0. TECHNOLOGY CONNECTION

Exploratory Examine the graph of the equation in Example 1 using Grafly to visualize the relative minimum.

Solving equation (2) for x, we get x = - 2y. Substituting - 2y for x in equation (1) and solving, we get 21- 2y2 + y - 3 - 4y + y - 3 - 3y y

= = = =

0 0 3 - 1.

To find x when y = - 1, we substitute - 1 for y in equation (1) or equation (2). We choose equation (2): x + 21- 12 = 0 x = 2.

Thus, 12, - 12 is the only critical point, and f12, - 12 is our candidate for a maximum or minimum value. 3. We must check to see whether f12, - 12 is a maximum or minimum value: D = fxx12, - 12 # fyy12, - 12 - 3fxy12, - 1242 = 2 # 2 - 3142 Using step 1 = 3.

z

4. Thus, D = 3 and fxx12, - 12 = 2. Since D 7 0 and fxx12, - 12 7 0, it follows from the D-test that f has a relative minimum at 12, - 12. That minimum value is found as follows: Quick Check 1 Find the relative maximum and minimum values of f1x, y2 = x 2 + xy + 2y 2 - 7x.

f12, - 12 = 2 2 + 21- 12 + 1- 122 - 3 # 2 = 4 - 2 + 1 - 6 = - 3. This is the relative

y

minimum.

O Quick Check 1

x

(2, −1, −3), relative minimum

z = f (x, y) = x2 + xy + y2 − 3x

■ EXAMPLE 2 Find the relative maximum and minimum values of f1x, y2 = xy - x 3 - y 2.

Solution 1. Find fx, fy, fxx, fyy, and fxy: fx = y - 3x 2, fxx = - 6x; fxy

fy = x - 2y, fyy = - 2; = 1.

O

568

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•

Functions of Several Variables

2. Solve the system of equations fx = 0, fy = 0:

(1) (2)

y - 3x 2 = 0, x - 2y = 0.

Solving equation (1) for y, we get y = 3x 2. Substituting 3x 2 for y in equation (2) and solving, we get

TECHNOLOGY CONNECTION

Exploratory

x - 213x 22 = 0 x - 6x 2 = 0 x11 - 6x2 = 0.

Examine the graph of the equation in Example 2 on Grafly to visualize the relative maximum and the saddle point.

Factoring

Setting each factor equal to 0 and solving, we have x = 0

1 - 6x = 0 x = 16.

or or

x = 0

To find y when x = 0, we substitute 0 for x in equation (1) or equation (2). We choose equation (2): 0 - 2y = 0 - 2y = 0 y = 0.

Thus, 10, 02 is a critical point, and f10, 02 is one candidate for a maximum or minimum value. To find the other, we substitute 16 for x in either equation (1) or equation (2). We choose equation (2): 1 6

z

- 2y = 0 - 2y = - 16 1 y = 12 .

1 Thus, A 16, 12 B is another critical point, and f A 16, 121 B is another candidate for a maximum or minimum value. 1 3–4. We must check both 10, 02 and A 16, 12 B to see whether they yield maximum or minimum values.

(0, 0, 0), saddle point 1 1 , (,16 , 12 432 )

For 10, 02: D = fxx10, 02 # fyy10, 02 - 3fxy10, 0242

relative maximum

= 1- 6 # 02 # 1- 22 - 3142 = - 1.

y

Using step 1

Since D 6 0, it follows that f10, 02 is neither a maximum nor a minimum value, but a saddle point. x

For z = f(x, y) = xy − x 3 − y2

A 16, 121 B : D = fxx A 16, 121 B # fyy A 16, 121 B - 3fxy A 16, 121 B 42 = A - 6, 16 B # 1- 22 - 3142 = - 11- 22 - 1 = 1.

Quick Check 2 Find the critical points of g1x, y2 = x 3 + y 2 - 3x - 4y + 3. Then use the D-test to classify each point as a relative maximum, a relative minimum, or a saddle point.

Using step 1

1 Thus, D = 1 and fxx A 16, 12 B = - 1. Since D 7 0 and fxx A 16, 121 B 6 0, it follows that 1 f has a relative maximum at A 16, 12 B ; that maximum value is 1 f A 16, 12 B =

=

O Quick Check 2

#

1 1 6 12

-

1 72

1 216

-

A 16 B 3 - A 121 B 2 -

1 144

=

1 432 .

This is the relative maximum.

O

6.3

•

Maximum–Minimum Problems

569

■ EXAMPLE 3

Business: Maximizing Profit. A firm produces two kinds of golf ball, one that sells for $3 and one priced at $2. The total revenue, in thousands of dollars, from the sale of x thousand balls at $3 each and y thousand at $2 each is given by R1x, y2 = 3x + 2y. The company determines that the total cost, in thousands of dollars, of producing x thousand of the $3 ball and y thousand of the $2 ball is given by C1x, y2 = 2x 2 - 2xy + y 2 - 9x + 6y + 7.

How many balls of each type must be produced and sold in order to maximize profit?

Solution The total profit P1x, y2 is given by P1x, y2 = R1x, y2 - C1x, y2 = 3x + 2y - 12x 2 - 2xy + y 2 - 9x + 6y + 72 P1x, y2 = - 2x 2 + 2xy - y 2 + 12x - 4y - 7. 1. Find Px, Py, Pxx, Pyy, and Pxy: Px = - 4x + 2y + 12, Pxx = - 4; Pxy

Py = 2x - 2y - 4, Pyy = - 2; = 2.

2. Solve the system of equations Px = 0, Py = 0: - 4x + 2y + 12 = 0, 2x - 2y - 4 = 0.

(1) (2)

Adding these equations, we get - 2x + 8 = 0. z

Then

(4, 2, 13), relative maximum

- 2x = - 8 x = 4. To find y when x = 4, we substitute 4 for x in equation (1) or equation (2). We choose equation (2): 2 # 4 - 2y - 4 - 2y + 4 - 2y y

y

x

z = P(x, y) =

−2x 2

+ 2xy −

y2

+ 12x − 4y − 7

= = = =

0 0 -4 2.

Thus, 14, 22 is the only critical point, and P14, 22 is a candidate for a maximum or minimum value. 3. We must check to see whether P14, 22 is a maximum or minimum value: D = Pxx14, 22 # Pyy14, 22 - 3Pxy14, 2242 = 1- 421- 22 - 2 2 Using step 1 = 4.

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•

Functions of Several Variables

4. Thus, D = 4 and Pxx14, 22 = - 4. Since D 7 0 and Pxx14, 22 6 0, it follows that P has a relative maximum at 14, 22. So, in order to maximize profit, the company must produce and sell 4 thousand of the $3 golf balls and 2 thousand of the $2 golf balls. The maximum profit will be

Quick Check 3

O

Repeat Example 3 using the same cost function and assuming that the company’s total revenue, in thousands of dollars, comes from the sale of x thousand balls at $3.50 each and y thousand at $2.75 each.

P14, 22 = - 2 # 4 2 + 2 # 4 # 2 - 2 2 + 12 # 4 - 4 # 2 - 7 = 13,

or $13 thousand.

O Quick Check 3

Section Summary

• A two-variable function f has a relative maximum at 1a, b2 if f1x, y2 … f1a, b2 for all points in a region containing 1a, b2 and has a relative minimum at 1a, b2 if f1x, y2 Ú f1a, b2 for all points in a region containing 1a, b2.

• The D-test is used to classify a critical point as a relative minimum, a relative maximum, or a saddle point.

EXERCISE SET

6.3 Find the relative maximum and minimum values. 1. f 1x, y2 = x 2 + xy + y 2 - y

2. f 1x, y2 = x + xy + y - 5y 2

2

3. f 1x, y2 = 2xy - x 3 - y 2 4. f 1x, y2 = 4xy - x 3 - y 2

5. f 1x, y2 = x 3 + y 3 - 3xy 6. f 1x, y2 = x 3 + y 3 - 6xy

7. f 1x, y2 = x 2 + y 2 - 2x + 4y - 2

8. f 1x, y2 = x 2 + 2xy + 2y 2 - 6y + 2 9. f 1x, y2 = x 2 + y 2 + 2x - 4y

10. f 1x, y2 = 4y + 6x - x 2 - y 2 11. f 1x, y2 = 4x 2 - y 2 12. f 1x, y2 = x 2 - y 2 13. f 1x, y2 = e x

2

14. f 1x, y2 = e x

2

+y +1 2

- 2x + y2 - 4y + 2

APPLICATIONS Business and Economics In Exercises 15–22, assume that relative maximum and minimum values are absolute maximum and minimum values. 15. Maximizing profit. Safe Shades produces two kinds of sunglasses; one kind sells for $17, and the other for $21. The total revenue in thousands of dollars from the sale of x thousand sunglasses at $17 each and y thousand at $21 each is given by R1x, y2 = 17x + 21y. The company determines that the total cost, in thousands of dollars, of producing x thousand of the $17 sunglasses and y thousand of the $21 sunglasses is given by C1x, y2 = 4x 2 - 4xy + 2y 2 - 11x + 25y - 3. Find the number of each type of sunglasses that must be produced and sold in order to maximize profit. 16. Maximizing profit. A concert promoter produces two kinds of souvenir shirt; one kind sells for $18, and the other for $25. The total revenue from the sale of x

Exercise Set 6.3

thousand shirts at $18 each and y thousand at $25 each is given by R1x, y2 = 18x + 25y. The company determines that the total cost, in thousands of dollars, of producing x thousand of the $18 shirt and y thousand of the $25 shirt is given by C1x, y2 = 4x 2 - 6xy + 3y 2 + 20x + 19y - 12. How many of each type of shirt must be produced and sold in order to maximize profit?

1 2 10 a p

where a is the amount spent on advertising, in millions of dollars, and n is the number of items sold, in thousands. Find the maximum value of P and the values of a and n at which it is attained. 19. Minimizing the cost of a container. A trash company is designing an open-top, rectangular container that will have a volume of 320 ft3. The cost of making the bottom of the container is $5 per square foot, and the cost of the sides is $4 per square foot. Find the dimensions of the container that will minimize total cost. (Hint: Make a substitution using the formula for volume.) 20. Two-variable revenue maximization. Boxowitz, Inc., a computer firm, markets two kinds of calculator that compete with one another. Their demand functions are expressed by the following relationships: (1) (2)

where p1 and p2 are the prices of the calculators, in multiples of $10, and q1 and q2 are the quantities of the calculators demanded, in hundreds of units. a) Find a formula for the total-revenue function, R, in terms of the variables p1 and p2. [Hint: R = p1q1 + p2q2; then substitute expressions from equations (1) and (2) to find R1p1, p22.] b) What prices p1 and p2 should be charged for each product in order to maximize total revenue? c) How many units will be demanded? d) What is the maximum total revenue? 21. Two-variable revenue maximization. Repeat Exercise 20, using

q2 = 56 - 2p1 - 4p2.

Find the minimum temperature and where it occurs. Is there a maximum temperature?

24. f 1x, y2 = xy +

4 2 + x y

25. f 1x, y2 = 2y 2 + x 2 - x 2y

26. S1b, m2 = 1m + b - 7222 + 12m + b - 7322 + 13m + b - 7522

P1a, n2 = - 5a 2 - 3n 2 + 48a - 4n + 2an + 290,

q1 = 64 - 4p1 - 2p2

T1x, y2 = x 2 + 2y 2 - 8x + 4y.

23. f 1x, y2 = e x + e y - e x + y

- 80,

18. Maximizing profit. A one-product company finds that its profit, P, in millions of dollars, is given by

and

22. Temperature. A flat metal plate is located on a coordinate plane. The temperature of the plate, in degrees Fahrenheit, at point 1x, y2 is given by

Find the relative maximum and minimum values and the saddle points.

where a is the amount spent on advertising, in millions of dollars, and p is the price charged per item of the product, in dollars. Find the maximum value of P and the values of a and p at which it is attained.

q1 = 78 - 6p1 - 3p2, q2 = 66 - 3p1 - 6p2,

Life and Physical Sciences

SYNTHESIS

17. Maximizing profit. A one-product company finds that its profit, P, in millions of dollars, is given by P1a, p2 = 2ap + 80p - 15p 2 -

571

27. Is a cross-section of an anticlastic curve always a parabola? Why or why not? 28. Explain the difference between a relative minimum and an absolute minimum of a function of two variables.

TECHNOLOGY CONNECTION Use a 3D graphics program to graph each of the following functions. Then estimate any relative extrema. 29. f 1x, y2 =

-5 x 2 + 2y 2 + 1

30. f 1x, y2 = x 3 + y 3 + 3xy 31. f 1x, y2 = 32. f 1x, y2 =

3xy1x 2 - y 22 x 2 + y2

y + x 2y 2 - 8x xy

Answers to Quick Checks

1. 14, - 1, - 142, relative minimum 2. 11, 2, - 32, relative minimum; 1- 1, 2, 12, saddle point 3. Maximum profit is $17.031 thousand when x = $4.625 thousand and y = $3 thousand

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•

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Functions of Several Variables

6.4 OBJECTIVES • Find a regression line. • Solve applied problems involving regression lines.

An Application: The Least-Squares Technique We have made frequent use in this book of a graphing calculator to perform regression. The purpose of this section is to develop an understanding of the process of regression by using the method for finding the minimum value for a function of two variables developed in the preceding section. We first considered regression in Section R.6. An equation found by regression provides a model of the phenomenon that the data measure, from which predictions can be made. For example, in business, one might want to predict future sales on the basis of past data. In ecology, one might want to predict future demand for natural gas on the basis of past usage. Suppose that we wish to find a linear equation, y = mx + b, to fit some data. To determine this equation is to determine the values of m and b. But how? Let’s consider some factual data. Suppose that a car rental company that offers hybrid (gas–electric) vehicles charts its revenue as shown in Fig. 1 and the accompanying table. How best could we predict the company’s revenue for the year 2016?

Revenue (in millions of dollars)

YEARLY REVENUE OF SKY BLUE CAR RENTALS 20.0 19.0 18.0 17.0 16.0 15.0 14.0 13.0 12.0 11.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0

Year, x

1996

2001

2006

2011

2016

5.2

8.9

11.7

16.8

?

Yearly Revenue, y (in millions of dollars)

$16.8

$11.7 $8.9

Suppose that we plot these points and try to draw a line through them that fits. Note that there are several ways in which this might be done (see Figs. 2 and 3). Each would give a different estimate of the company’s total revenue for 2016. Note that the years for which revenue is given follow 5-yr increments. Thus, computations can be simplified if we use the data points 11, 5.22, 12, 8.92, 13, 11.72, and 14, 16.82, as plotted in Fig. 3, where each horizontal unit represents 5 years and x = 1 is 1996.

$5.2

1996

2001 2006 Year

2011

20.0

15.0

10.0

Another line of “fit” 20.0

15.0

10.0

5.0 1

2016

2011

2006

2001

5.0

Yearly revenue (in millions of dollars)

Yearly revenue (in millions of dollars)

A line of “fit”

1996

Yearly revenue (in millions of dollars)

FIGURE 1

2

3

4

20.0 (4, 16.8) 15.0

10.0

(3, y3 ) (4, y4 ) (2, y2 ) (3, 11.7) (1, y1) (2, 8.9)

1

5

Year

FIGURE 2

FIGURE 3

(1, 5.2)

5.0

FIGURE 4

2

3

4

5

6.4

•

An Application: The Least-Squares Technique

573

To determine the equation of the line that “best” fits the data, we note that for each data point there will be a deviation, or error, between the y-value at that point and the y-value of the point on the line that is directly above or below the point. Those deviations, in this case, y1 - 5.2, y2 - 8.9, y3 - 11.7, and y4 - 16.8, will be positive or negative, depending on the location of the line (see Fig. 4). We wish to fit these data points with a line, y = mx + b, that uses values of m and b that, somehow, minimize the y-deviations in order to have a good fit. One way of minimizing the deviations is based on the least-squares assumption.

The Least-Squares Assumption The line of best fit is the line for which the sum of the squares of the y-deviations is a minimum. This is called the regression line.

Note that squaring each y-deviation gives us a series of nonnegative terms that we can sum. Were we to simply add the y-deviations, positive and negative deviations would cancel each other out. Using the least-squares assumption with the yearly revenue data, we want to minimize 1 y1 - 5.222 + 1 y2 - 8.922 + 1 y3 - 11.722 + 1 y4 - 16.822.

(1)

Also, since the points 11, y12, 12, y22, 13, y32, and 14, y42 must be solutions of y = mx + b, it follows that y1 y2 y3 y4

= = = =

m112 m122 m132 m142

+ + + +

b b b b

= = = =

m + b, 2m + b, 3m + b, 4m + b.

Substituting m + b for y1, 2m + b for y2, 3m + b for y3, and 4m + b for y4 in equation (1), we now have a function of two variables:

S1m, b2 = 1m + b - 5.222 + 12m + b - 8.922 + 13m + b - 11.722 + 14m + b - 16.822 . Thus, to find the regression line for the given set of data, we must find the values of m and b that minimize the function S given by the sum in this last equation. To apply the D-test, we first find the partial derivatives 0S>0b and 0S>0m: 0S = 21m + b - 5.22 + 212m + b - 8.92 + 213m + b - 11.72 + 214m + b - 16.82 0b = 20m + 8b - 85.2, and 0S = 21m + b - 5.22 + 212m + b - 8.922 + 213m + b - 11.723 + 214m + b - 16.824 0m = 60m + 20b - 250.6.

Yearly revenue (in millions of dollars)

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We set these derivatives equal to 0 and solve the resulting system: 20m + 8b - 85.2 = 0, 5m + 2b = 21.3, or 60m + 20b - 250.6 = 0; 15m + 5b = 62.65.

Predicted value = 20.05

20.0

It can be shown that the solution of this system is

15.0

b = 1.25,

10.0

5.0 1

2

3

4

5

m = 3.76.

(See the Technology Connection below.)

We leave it to the student to complete the D-test to verify that 11.25, 3.762 does, in fact, yield the minimum of S. There is no need to compute S11.25, 3.762. The values of m and b are all we need to determine y = mx + b. The regression line is y = 3.76x + 1.25.

FIGURE 5

Substituting for m and b

The graph of this “best-fit” regression line together with the data points is shown in Fig. 5. Compare it to Figs. 2, 3, and 4. TECHNOLOGY CONNECTION

Solving Linear Systems Using Matrices In this Technology Connection, we explore a method of solution called reduced row echelon form (rref). We can use this method to solve the system of equations discussed above: 5m + 2b = 21.3, 15m + 5b = 62.65. From this system, we can write the matrix c

5 15

2 5

21.3 d. 62.65

The first column is called the m-column (because the entries are the coefficients of the variable m), the second column is called the b-column, and the final column is the constants column. Before entering the numbers of the system of equations into a matrix, it is crucial that the m and b terms are in the correct positions to the left of the equal signs and the constants are to the right of the equal signs. When a matrix is in reduced row echelon form, it has the following appearance: 1 c 0

0 1

a d. b

With this form, the system is considered solved, as we can rewrite this matrix as the system 1x + 0y = a. 0x + 1y = b. Therefore, x = a and y = b.

On your calculator, select MATRIX (it may be a 2nd function on some models). Under EDIT, select [A]. With this setting, matrix [A] has 2 rows and 3 columns, so it is of size 2 * 3. Enter these values, pressing ENTER after each one. Then enter the values of the matrix into the matrix field, pressing ENTER after each one. After you have entered the matrix values, press 2ND and QUIT to exit. Matrix [A] is now stored in the calculator’s memory. To convert matrix [A] into reduced row echelon form, press MATRIX, and under MATH, scroll down to rref. Press ENTER. Now press MATRIX once again, and under NAMES, select [A], and press ENTER. The result will be the reduced row echelon form equivalent to the original matrix [A]: rref([A]) [ [ 1 0 3.76 ] [ 0 1 1.25 ] ]

Therefore, m = 3.76 and b = 1.25.

EXERCISES Use the reduced row echelon form of a matrix to solve the following systems of equations with your calculator. 1. 2x + 6y = 14

x - 5y = - 17

2.

3x + y = 7 10x + 3y = 11

3. 2x + y + 7 = 0

x = 6 - y

•

6.4

An Application: The Least-Squares Technique

575

We can now extrapolate from the data to predict the car rental company’s yearly revenue in 2016: y = 3.76152 + 1.25 = 20.05.

Quick Check 1

O

Use the method of least squares to determine the regression line for the data points 11, 252, 12, 482, 13, 76.72, and 14, 104.82.

The yearly revenue in 2016 is predicted to be about $20.05 million. How might you check this prediction? The method of least squares is a statistical process illustrated here with only four data points in order to simplify the explanation. Most statistical researchers would warn that many more than four data points should be used to get a “good” regression line. Furthermore, making predictions too far in the future from any mathematical model may not be valid. The further into the future a prediction is made, the more dubious one should be about the prediction.

O Quick Check 1

TECHNOLOGY CONNECTION

Exploratory As we have seen in Section R.6 and in other parts of the book, graphing calculators can perform linear regression, as well as quadratic, exponential, and logarithmic regression. Use such a calculator now to fit a linear equation to the yearly revenue data for the car rental company. With some calculators, you will also obtain a number r, called the coefficient of correlation. Although we cannot develop that concept in detail in this text, keep in mind that r is used to describe the strength of the linear relationship between x and y. The closer ƒ r ƒ is to 1, the better the correlation. For the yearly revenue data, r L 0.993, which indicates a fairly good linear relationship. Keep in mind that a high linear correlation does not necessarily indicate a “cause-and-effect” connection between the variables.

*The Regression Line for an Arbitrary Collection of Data Points (c1, d1), (c2, d2), . . . , (cn, dn) Look again at the regression line y = 3.76x + 1.25

for the data points 11, 5.22, 12, 8.92, 13, 11.72, and 14, 16.82. Let’s consider the arithmetic averages, or means, of the x-coordinates, denoted x, and of the y-coordinates, denoted y: x =

1 + 2 + 3 + 4 = 2.5, 4

y =

5.2 + 8.9 + 11.7 + 16.8 = 10.65. 4

It turns out that the point 1x, y2, or 12.5, 10.652, is on the regression line since 10.65 = 3.7612.52 + 1.25 . Thus, the equation for the regression line can be written y - y = m1x - x2, or, in this case, y - 10.65 = m1x - 2.52 . All that remains, in general, is to determine m. Suppose that we want to find the regression line for an arbitrary number of points 1c1, d12, 1c2, d22, Á , 1cn, dn2. To do so, we find the values m and b that minimize the function S given by n

S1b, m2 = 1 y1 - d122 + 1 y2 - d222 + Á + 1 yn - dn22 = a 1 yi - di22 , i=1

where yi = mci + b.

*This subsection is considered optional and can be omitted without loss of continuity.

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Using a procedure like the one we used earlier to minimize S, we can show that y = mx + b takes the form y - y = m1x - x2, n

where x =

n

a ci

i=1

n

, y =

n

a di

i=1

n

, and

m =

a 1ci - x21di - y2

i=1

n

a 1ci - x2

.

2

i=1

Let’s see how this works out for the yearly revenue data from our earlier example. ci

di

ci - x

1ci - x22

1di - y2

1ci - x2 1di - y2

1 2 3 4

5.2 8.9 11.7 16.8

- 1.5 - 0.5 0.5 1.5

2.25 0.25 0.25 2.25

- 5.45 - 1.75 1.05 6.15

8.175 0.875 0.525 9.225

4

a ci = 10

i=1

4

4

i=1

x = 2.5

4

2 a 1ci - x2 = 5

a di = 42.6

a 1ci - x21di - y2 = 18.8

i=1

i=1

y = 10.65

m =

18.8 = 3.76 5

Thus, the regression line is y - 10.65 = 3.761x - 2.52, which simplifies to y = 3.76x + 1.25.

Section Summary • Regression is a technique for determining a continuous function that “best fits” a set of data points.

• For linear regression, the method of least squares uses calculus on a function of two variables to find the values m and b that determine the regression line y = mx + b, the line of best fit.

EXERCISE SET

6.4 For each data set, find the regression line without using a calculator. 1.

2.

x

1

2

4

5

y

1

3

3

4

x

1

3

5

y

2

4

7

3.

4.

x

1

2

3

5

y

0

1

3

4

x

1

2

4

y

3

5

8

Exercise Set 6.4

All of the following exercises can be done with a graphing calculator if your instructor so directs. The calculator can also be used to check your work.

577

a) Find the regression line, y = mx + b. b) Use the regression line to predict the average ticket price for an NFL game in 2012 and in 2015.

Life and Physical Sciences

APPLICATIONS Business and Economics 5. Labor force. The minimum hourly wage in the United States has grown over the years, as shown in the table below. Number of Years, x, since 1990

Minimum Hourly Wage

0 1 6 7 17 18 19

$3.80 4.25 4.75 5.15 5.85 6.55 7.25

(Source: www.workworld.org.)

7. Life expectancy of women. Consider the data in the following table showing the average life expectancy of women in various years. Note that x represents the actual year.

Year, x

Life Expectancy of Women, y (years)

1950 1960 1970 1980 1990 2000 2003

71.1 73.1 74.7 77.4 78.8 79.5 80.1

(Source: Centers for Disease Control, June 2006.)

a) Find the regression line, y = mx + b. b) Use the regression line to predict the life expectancy of women in 2010 and 2015. 8. Life expectancy of men. Consider the following data showing the average life expectancy of men in various years. Note that x represents the actual year.

a) For the data in the table, find the regression line, y = mx + b. b) Use the regression line to predict the minimum hourly wage in 2015 and 2020. 6. Football ticket prices. Ticket prices for NFL football games have experienced steady growth, as shown in the following table. Number of Years, x, since 1999 Season

Average Ticket Price (dollars)

0 1 2 3 4 5 6

$45.03 49.35 47.49 50.02 52.95 54.75 58.95

(Source: Team Marketing Report.)

Year, x

Life Expectancy of Men, y (years)

1950 1960 1970 1980 1990 2000 2003

65.6 66.6 67.1 70.0 71.8 74.1 74.8

(Source: Centers for Disease Control, June 2006.)

a) Find the regression line, y = mx + b. b) Use the regression line to predict the life expectancy of men in 2010 and 2015.

General Interest 9. Grade predictions. A professor wants to predict students’ final examination scores on the basis of their midterm test scores. An equation was determined on the basis of data on the scores of three students who took the same course with the same instructor the previous semester (see the following table). Midterm Score, x

Final Exam Score, y

70% 60 85

75% 62 89

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a) Find the regression line, y = mx + b. (Hint: The y-deviations are 70m + b - 75, 60m + b - 62, and so on.) b) The midterm score of a student was 81%. Use the regression line to predict the student’s final exam score. 10. Predicting the world record in the high jump. It has been established that most world records in track and field can be modeled by a linear function. The table below shows world high-jump records for various years. Note that x represents the actual year.

Year, x

World Record in High Jump, y (in inches)

1912 (George Horme) 1956 (Charles Dumas) 1973 (Dwight Stones) 1989 (Javier Sotomayer) 1993 (Javier Sotomayer)

78.0 84.5 90.5 96.0 96.5

(Source: www.wikipedia.org.)

12. Discuss the idea of linear regression with a professor from another discipline in which regression is used. Explain how it is used in that field.

TECHNOLOGY CONNECTION 13. General interest: predicting the world record for running the mile. Note that x represents the actual year in following table.

Year, x

World Record, y (in minutes:seconds)

1875 (Walter Slade) 1894 (Fred Bacon) 1923 (Paavo Nurmi) 1937 (Sidney Wooderson) 1942 (Gunder Hägg) 1945 (Gunder Hägg) 1954 (Roger Bannister) 1964 (Peter Snell) 1967 (Jim Ryun) 1975 (John Walker) 1979 (Sebastian Coe) 1980 (Steve Ovett) 1985 (Steve Cram) 1993 (Noureddine Morceli)

4:24.5 4:18.2 4:10.4 4:06.4 4:06.2 4:01.4 3:59.6 3:54.1 3:51.1 3:49.4 3:49.0 3:48.40 3:46.31 3:44.39

(Source: USA Track & Field and infoplease.com.)

a) Find the regression line, y = mx + b. b) Use the regression line to predict the world record in the high jump in 2010 and in 2050. c) Does your answer in part (b) for 2050 seem realistic? Explain why extrapolating so far into the future could be a problem. 11. How would you explain the concept of linear regression to a friend?

a) Find the regression line, y = mx + b, that fits the data in the table. (Hint: Convert each time to decimal notation; for instance, 4:24.5 = 4 24.5 60 = 4.4083.) b) Use the regression line to predict the world record in the mile in 2010 and in 2015. c) In July 1999, Hicham El Guerrouj set the current (as of December 2006) world record of 3:43.13 for the mile. (Source: USA Track & Field and infoplease.com.) How does this compare with what is predicted by the regression line?

Answer to Quick Check 1. y = 26.81x - 3.4

6.5

6.5 OBJECTIVES • Find maximum and minimum values using Lagrange multipliers.

•

Constrained Optimization

579

Constrained Optimization In Section 6.3, we discussed a method for determining maximum and minimum values on a surface represented by a two-variable function z = f1x, y2. If restrictions are placed on the input variables x and y, we can determine the maximum and minimum values on the surface subject to the restrictions. This process is called constrained optimization.

Path Constraints: Lagrange Multipliers

• Solve constrained optimization problems involving Lagrange multipliers.

Imagine that you are hiking up a mountain. If there are no constraints on your movement, you may seek out the mountain’s summit—its “maximum point.” 6903 ft The figure at the right shows a relief M map of a mountaintop; its unconstrained maximum point occurs at the , labeled with a spot elevation of 6903 ft. A hiking trail, marked as a black dashed line, bypasses the summit. If you were constrained to this hiking path, you could not (Source: USGS maps at www.mytopo.com.) reach the summit. You could, however, achieve a maximum elevation along the path. This constrained maximum point is approximated at M. In many applications modeled by two-variable functions, constraints on the input variables are necessary. If the input variables are related to one another by an equation, it is called a constraint. Let’s return to a problem we considered in Chapter 2: A hobby store has 20 ft of fencing to fence off a rectangular electric-train area in one corner of its display room. The two sides up against the wall require no fence. What dimensions of the rectangle will maximize the area? We maximize the function

#

A = xy subject to the condition, or constraint, x + y = 20. Note that A is a function of two variables. When we solved this earlier, we first solved the constraint for y: y = 20 - x. 20  x

x

We then substituted 20 - x for y to obtain A1x, y2 = x120 - x2 = 20x - x 2, which is a function of one variable. Next, we found a maximum value using Maximum– Minimum Principle 1 (see Section 2.4). By itself, the function of two variables A1x, y2 = xy

580

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•

Functions of Several Variables

has no maximum value. This can be checked using the D-test. With the constraint x + y = 20, however, the function does have a maximum. We see this in the following graph. Constraint, x + y = 20

z

Surface, A = xy Constrained maximum = 100

Path constraint

y

10 10

x

It may be quite difficult to solve a constraint for one variable. The method outlined below allows us to proceed without doing so.

The Method of Lagrange Multipliers To find a maximum or minimum value of a function f1x, y2 subject to the constraint g1x, y2 = 0: 1. Form a new function, called the Lagrange function: F1x, y, l2 = f1x, y2 - lg1x, y2. The variable l (lambda) is called a Lagrange multiplier. 2. Find the first partial derivatives Fx, Fy, and Fl. 3. Solve the system Fx = 0,

Fy = 0, and Fl = 0.

Let 1a, b, l2 represent a solution of this system. We normally must determine whether 1a, b2 yields a maximum or minimum of the function f. For the problems in this text, we will specify that a maximum or minimum exists. The method of Lagrange multipliers can be extended to functions of three (or more) variables.

We can illustrate the method of Lagrange multipliers by resolving the electric-train area problem.

■ EXAMPLE 1 Find the maximum value of A1x, y2 = xy subject to the constraint x + y = 20.

Solution Note first that x + y = 20 is equivalent to x + y - 20 = 0. 1. We form the Lagrange function F, given by F1x, y, l2 = xy - l # 1x + y - 202.

6.5

•

Constrained Optimization

581

2. We find the first partial derivatives: Fx = y - l, Fy = x - l, Fl = - 1x + y - 202. 3. We set each derivative equal to 0 and solve the resulting system: y - l = 0, x - l = 0, - 1x + y - 202 = 0, or x + y - 20 = 0.

(1) (2) (3)

From equations (1) and (2), it follows that x = y = l. Substituting x for y in equation (3), we get x + x - 20 = 0 2x = 20 x = 10. Quick Check 1 Find the maximum value of A1x, y2 = xy subject to the constraint x + 2y = 30.

Thus, y = x = 10. The maximum value of A subject to the constraint occurs at 110, 102 and is A110, 102 = 10 # 10 = 100.

O Quick Check 1 ■ EXAMPLE 2 Find the maximum value of f1x, y2 = 3xy subject to the constraint 2x + y = 8. Note: f might be interpreted, for example, as a production function with a budget constraint 2x + y = 8.

Solution Note that first we express 2x + y = 8 as 2x + y - 8 = 0. 1. We form the Lagrange function F, given by F1x, y, l2 = 3xy - l12x + y - 82. 2. We find the first partial derivatives: Fx = 3y - 2l, Fy = 3x - l, Fl = - 12x + y - 82. 3. We set each derivative equal to 0 and solve the resulting system: 3y - 2l = 0, 3x - l = 0, - 12x + y - 82 = 0, or 2x + y - 8 = 0. Solving equation (2) for l, we get l = 3x .

(1) (2) (3)

O

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•

Functions of Several Variables

Substituting in equation (1) for l, we get

TECHNOLOGY CONNECTION

3y - 2 # 3x = 0, or 3y = 6x,

Exploratory

(4)

or y = 2x .

Substituting 2x for y in equation (3), we get

Use a 3D graphics program to graph both f 1x, y2 = 3xy and 2x + y = 8. Then check the results of Example 2 visually.

2x + 2x - 8 = 0 4x = 8 x = 2. Then, using equation (4), we have y = 2 # 2 = 4.

The maximum value of f subject to the constraint occurs at 12, 42 and is

Quick Check 2 Find the minimum value of g1x, y2 = x 2 + y 2 subject to the constraint 3x - y = 1.

f12, 42 = 3 # 2 # 4 = 24.

O Quick Check 2 ■ EXAMPLE 3 Business: The Beverage-Can Problem. The standard beverage can holds 12 fl. oz, or has a volume of 21.66 in3. What dimensions yield the minimum surface area? Find the minimum surface area. (Assume that the shape of the can is a right circular cylinder.)

Solution We want to minimize the function s, given by s1h, r2 = 2prh + 2pr2

h

subject to the volume constraint pr 2h = 21.66, or

pr 2h - 21.66 = 0.

r

Note that s does not have a minimum without the constraint. 1. We form the Lagrange function S, given by S1h, r, l2 = 2prh + 2pr 2 - l1pr 2h - 21.662. 2. We find the first partial derivatives: 0S = 2pr - lpr 2, 0h 0S = 2ph + 4pr - 2lprh, 0r 0S = - 1pr 2h - 21.662. 0l 3. We set each derivative equal to 0 and solve the resulting system: 2pr - lpr 2 = 0, 2ph + 4pr - 2lprh = 0, - 1pr 2h - 21.662 = 0, or pr 2h - 21.66 = 0. Note that, since p is a constant, we can solve equation (1) for r: pr12 - lr2 = 0 pr = 0 or 2 - lr = 0 2 r = 0 or r = . l

We assume l Z 0.

(1) (2) (3)

O

6.5

•

Constrained Optimization

583

Since r = 0 cannot be a solution to the original problem, we continue by substituting 2>l for r in equation (2): 2ph + 4p #

2 2 - 2lp # # h = 0 l l

2ph +

8p - 4ph = 0 l 8p - 2ph = 0 l - 2ph = h =

so

8p , l

4 . l

Since h = 4>l and r = 2>l, it follows that h = 2r. Substituting 2r for h in equation (3) yields pr 212r2 - 21.66 2pr 3 - 21.66 2pr 3 pr 3

= = = =

0 0 21.66 10.83 10.83 r3 = p r =

Quick Check 3 Repeat Example 3 for a right circular cylinder with a volume of 500 mL. (Hint: 1 mL = 1 cm3.) (This was Example 3 in Section 2.5.)

3 10.83 L 1.51 in. A p

Thus, when r = 1.51 in., we have h = 3.02 in. The surface area is then a minimum and is approximately 2p11.51213.022 + 2p11.5122, or about 42.98 in2.

O Quick Check 3 The actual dimensions of a standard-sized 12-oz beverage can are r = 1.25 in. and h = 4.875 in. A natural question arising from the solution of Example 3 is, “Why don’t beverage companies make cans using the dimensions found in that example?” To do this would mean an enormous cost for retooling. New can-making machines and new beverage-filling machines would have to be designed and purchased. Vending machines would no longer be the correct size. A partial response to the desire to save aluminum has been found in recycling and in manufacturing cans with bevelled edges. These cans require less aluminum. As a result of many engineering advances, the amount of aluminum required to make 1000 cans has been reduced over the years from 36.5 lb to 28.1 lb. Consumer preference is another very important factor affecting the shape of the can. Market research has shown that a can with the dimensions found in Example 3 is not as comfortable to hold and might not be accepted by consumers.

Closed and Bounded Regions: The Extreme-Value Theorem In Examples 1, 2, and 3, all the constraints were given as equations. Constraints may also be stated as inequalities. If there are multiple constraints on the input variables x and y, these may be plotted on the xy-plane to form a region of feasibility, which contains the x and y values that satisfy all the constraints simultaneously. If the constraints form a closed and bounded region (closed meaning it includes the boundaries, and bounded meaning it has finite area, with no portions tending to infinity), then the Extreme-Value Theorem can be adapted for the two-variable function.

O

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THEOREM

Extreme-Value Theorem for Two-Variable Functions If f1x, y2 is continuous for all 1x, y2 within a region of feasibility that is closed and bounded, then f is guaranteed to have both an absolute maximum value and an absolute minimum value.

Critical points may occur at a vertex, along a boundary, or in the interior. Therefore, all these parts of a region must be checked for critical points.

■ EXAMPLE 4 Business: Maximizing Revenue. Kim likes to create stylish tee shirts, one style with a script x on the front and another with a script y on the front. She sells them to her math students as a fundraiser for her favorite charity. Kim determines that her weekly revenue is modeled by the two-variable function R1x, y2 = - x2 - xy - y2 + 20x + 22y - 25, where x is the number of x-shirts sold, and y is the number of y-shirts sold. Kim spends 2 hr working on each x-shirt and 4 hr working on each y-shirt, and she works no more than 40 hr per week on this project. How many of each style should she produce in order to maximize her weekly revenue? Assume x Ú 0 and y Ú 0; in other words, she cannot produce negative quantities of the tee shirts.

Solution The number of hours Kim

y

works per week is a constraint: 2x + 4y … 40. We write the inequality with a less-than-or-equal-to sign since she may 2x + 4y ≤ 40 not work the full 40 hr. Along with the x ≥ 0 constraints x Ú 0 and y Ú 0, this conRegion of feasibility straint allows us to sketch the region of x feasibility. This is a closed and bounded y≥0 region. Since the revenue function R is continuous for all x and all y, the Extreme-Value Theorem guarantees an absolute minimum and an absolute maximum point. In this example, we are interested in the absolute maximum (revenue). We determine that the three vertex points of the region are 10, 0, - 252, 120, 0, - 252, and 10, 10, 952. These are all critical points. Next, we check the interior of the region. We find the partial derivatives of R with respect to x and with respect to y: Rx = - 2x - y + 20, Ry = - x - 2y + 22. Setting these expressions equal to 0, we solve the system for x and y:

or

- 2x - y + 20 = 0, - x - 2y + 22 = 0; 2x + y = 20, x + 2y = 22. After simplification

The system is solved when x = 6 and y = 8. However, this point is outside the region of feasibility; Kim would have to work 2162 + 4182 = 44 hr, which is not allowed under the given constraint. Therefore, this solution must be ignored. (We address this issue at the end of this example.) The boundaries of the region must also be checked for possible critical points: • To check along the y-axis, we substitute x = 0 into the revenue function: R10, y2 = - y 2 + 22y - 25.

6.5

•

Constrained Optimization

585

The derivative is Ry = - 2y + 22. Setting this expression equal to 0, we obtain y = 11. However, this is outside the region of feasibility and is ignored. • To check along the x-axis, we substitute y = 0 into the revenue function: R1x, 02 = - x 2 + 20x - 25. The derivative is Rx = - 2x + 20. Setting this expression equal to 0, we get x = 10. This is a feasible solution, and thus is a critical value. The critical point is 110, 0, 752. • To check along the line 2x + 4y = 40, we use the method of Lagrange multipliers to determine possible critical values. The constraint is written as 2x + 4y - 40 = 0, and the Lagrange function is formed: L1x, y, l2 = - x 2 - xy - y 2 + 20x + 22y - 25 - l12x + 4y - 402. Its first partial derivatives are as follows: L x = - 2x - y + 20 - 2l, L y = - x - 2y + 22 - 4l, L l = - 2x - 4y + 40. We set each partial derivative equal to 0:

(1) (2) (3)

- 2x - y + 20 - 2l = 0 - x - 2y + 22 - 4l = 0 - 2x - 4y + 40 = 0 We solve equations (1) and (2) for l: l = - x - 12 y + 10 and l = - 14 x - 12 y +

11 2.

Equating the right-hand sides of these two equations gives us a single equation in terms of x and y. Note that the - 12 y terms cancel (sum to zero): - x - 12 y + 10 = - 14 x - 12 y +

11 2

- 34 x = - 92 x = 6. We now substitute x = 6 into the constraint, 2x + 4y = 40, to determine y: 2162 + 4y 12 + 4y 4y y

Quick Check 4 Repeat Example 4, using the same revenue function but assuming that each x-shirt requires 4 hr to create, each y-shirt requires 2 hr to create, and Kim is willing to work 36 hr per week at most. How many of each style of shirt should Kim produce to maximize her weekly revenue?

= = = =

40 40 28 7.

y This is a feasible solution. Therefore, x = 6 and y = 7 (0, 10, 95) yield a critical point: 16, 7, 1222. (6, 7, 122) In the graph to the right, all critical points (with their revenue values) are plotted on the Region of feasibility region of feasibility. Therefore, x Kim should produce 6 of the (0, 0, −25) (10, 0, 75) (20, 0, −25) x-shirts and 7 of the y-shirts to maximize her weekly revenue at $122. If there were no constraints, the maximum weekly revenue would occur at x = 6 and y = 8, for a total of $123. Kim might think that working an extra 4 hr for one more dollar of revenue is not worth it.

O Quick Check 4

O

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Functions of Several Variables

Section Summary • If input variables x and y for a function f1x, y2 are related by another equation, that equation is a constraint. • Constrained optimization is a method of determining maximum and minimum points on a surface represented by z = f1x, y2, subject to given restrictions (constraints) on the input variables x and y. • The method of Lagrange multipliers allows us to find a maximum or minimum value of a function f1x, y2 subject to the constraint g1x, y2 = 0.

• If the constraints are inequalities, the set of points that satisfy all the constraints simultaneously is called the region of feasibility. • If the region of feasibility is closed and bounded and the surface z = f1x, y2 is continuous over the region, then the Extreme-Value Theorem guarantees that f will have both an absolute maximum and an absolute minimum value. • Critical points may be located at vertices, along a boundary, or in the interior of a region of feasibility.

EXERCISE SET

6.5 Find the maximum value of f subject to the given constraint. 1. f 1x, y2 = xy;

3x + y = 10

2. f 1x, y2 = 2xy; 4x + y = 16

3. f 1x, y2 = 4 - x - y ; x + 2y = 10 2

2

4. f 1x, y2 = 3 - x 2 - y 2; x + 6y = 37

APPLICATIONS Business and Economics 17. Maximizing typing area. A standard piece of printer paper has a perimeter of 39 in. Find the dimensions of the paper that will give the most area. What is that area? Does standard 8 12 * 11 in. paper have maximum area?

6. f 1x, y2 = x 2 + y 2; x + 4y = 17

18. Maximizing room area. A carpenter is building a rectangular room with a fixed perimeter of 80 ft. What are the dimensions of the largest room that can be built? What is its area?

8. f 1x, y2 = 2x 2 + y 2 - xy; x + y = 8

19. Minimizing surface area. An oil drum of standard size has a volume of 200 gal, or 27 ft3. What dimensions yield the minimum surface area? Find the minimum surface area.

Find the minimum value of f subject to the given constraint. 5. f 1x, y2 = x + y ; 2x + y = 10 2

2

7. f 1x, y2 = 2y 2 - 6x 2; 2x + y = 4

9. f 1x, y, z2 = x 2 + y 2 + z 2; y + 2x - z = 3

10. f 1x, y, z2 = x 2 + y 2 + z 2; x + y + z = 2

Use the method of Lagrange multipliers to solve each of the following. 11. Of all numbers whose sum is 50, find the two that have the maximum product. 12. Of all numbers whose sum is 70, find the two that have the maximum product. 13. Of all numbers whose difference is 6, find the two that have the minimum product. 14. Of all numbers whose difference is 4, find the two that have the minimum product. 15. Of all points 1x, y, z2 that satisfy x + 2y + 3z = 13, find the one that minimizes 1x - 122 + 1 y - 122 + 1z - 122.

16. Of all points 1x, y, z2 that satisfy 3x + 4y +2z = 52, find the one that minimizes 1x - 122 + 1 y - 422 + 1z - 222.

Do these drums appear to be made in such a way as to minimize surface area?

20. Juice-can problem. A standard-sized juice can has a volume of 99 in3. What dimensions yield the minimum surface area? Find the minimum surface area. 21. Maximizing total sales. The total sales, S, of a oneproduct firm are given by S1L, M2 = ML - L2,

Exercise Set 6.5

where M is the cost of materials and L is the cost of labor. Find the maximum value of this function subject to the budget constraint M + L = 90. 22. Maximizing total sales. The total sales, S, of a oneproduct firm are given by S1L, M2 = ML - L2, where M is the cost of materials and L is the cost of labor. Find the maximum value of this function subject to the budget constraint M + L = 70. 23. Minimizing construction costs. A company is planning to construct a warehouse whose interior volume is to be 252,000 ft3. Construction costs per square foot are estimated to be as follows: Walls: Floor: Ceiling:

$3.00 $4.00 $3.00

z

x

y

a) The total cost of the building is a function C1x, y, z2, where x is the length, y is the width, and z is the height. Find a formula for C1x, y, z2. b) What dimensions of the building will minimize the total cost? What is the minimum cost? 24. Minimizing the costs of container construction. A container company is going to construct a shipping crate of volume 12 ft3 with a square bottom and top. The cost of the top and the sides is $2 per square foot, and the cost for the bottom is $3 per square foot. What dimensions will minimize the cost of the crate? 25. Minimizing total cost. Each unit of a product can be made on either machine A or machine B. The nature of the machines makes their cost functions differ: Machine A:

C1x2 = 10 +

Machine B:

C1 y2 = 200 +

x2 , 6 y3 9

587

In Exercises 26–29, find the absolute maximum and minimum values of each function, subject to the given constraints. 26. f 1x, y2 = x 2 + y 2 - 2x - 2y; x Ú 0, y Ú 0, x … 4, and y … 3 27. g1x, y2 = x 2 + 2y 2;

- 1 … x … 1 and - 1 … y … 2

28. h1x, y2 = x + y - 4x - 2y + 1; x Ú 0, y Ú 0, and x + 2y … 5 2

2

29. k1x, y2 = - x 2 - y 2 + 4x + 4y; 0 … x … 3, y Ú 0, and x + y … 6 30. Business: maximizing profits with constraints. A manufacturer of decorative end tables produces two models, basic and large. Its weekly profit function is modeled by P1x, y2 = - x 2 - 2y 2 - xy + 140x + 210y - 4300, where x is the number of basic models sold each week and y is the number of large models sold each week. The warehouse can hold at most 90 tables. Assume that x and y must be nonnegative. How many of each model of end table should be produced to maximize the weekly profit, and what will the maximum profit be? 31. Business: maximizing profits with constraints. A farmer has 300 acres on which to plant two crops, celery and lettuce. Each acre of celery costs $250 to plant and tend, and each acre of lettuce costs $300 to plant and tend. The farmer has $81,000 available to cover these costs. a) Suppose the farmer makes a profit of $45 per acre of celery and $50 per acre of lettuce. Write the profit function, determine how many acres of celery and lettuce he should plant to maximize profit, and state the maximum profit. (Hint: Since the graph of the profit function is a plane, you will not need to check the interior for possible critical points.) b) Suppose the farmer’s profit function is instead P1x, y2 = - x 2 - y 2 + 600y - 75,000. Assuming the same constraints, how many acres of celery and lettuce should he plant to maximize profit, and what is that maximum profit?

SYNTHESIS Find the indicated maximum or minimum values of f subject to the given constraint. 32. Minimum: f 1x, y2 = xy;

x 2 + y2 = 9

33. Minimum: f 1x, y2 = 2x 2 + y 2 + 2xy + 3x + 2y; y2 = x + 1

34. Maximum: f 1x, y, z2 = x + y + z; x 2 + y 2 + z 2 = 1 35. Maximum: f 1x, y, z2 = x 2y 2z 2; .

Total cost is given by C1x, y2 = C1x2 + C1 y2. How many units should be made on each machine in order to minimize total costs if x + y = 10,100 units are required?

x 2 + y2 + z2 = 2

36. Maximum: f 1x, y, z2 = x + 2y - 2z; x 2 + y 2 + z 2 = 4

37. Maximum: f 1x, y, z, t2 = x + y + z + t; x 2 + y2 + z2 + t 2 = 1

38. Minimum: f 1x, y, z2 = x 2 + y 2 + z 2; x - 2y + 5z = 1

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Functions of Several Variables

39. Economics: the Law of Equimarginal Productivity. Suppose that p1x, y2 represents the production of a two-product firm. The company produces x units of the first product at a cost of c1 each and y units of the second product at a cost of c2 each. The budget constraint, B, is a constant given by

41. Discuss the difference between solving a maximum– minimum problem using the method of Lagrange multipliers and the method of Section 6.3. 42. Write a brief report on the life and work of the mathematician Joseph Louis Lagrange (1736–1813).

B = c1 x + c2 y.

TECHNOLOGY CONNECTION

Use the method of Lagrange multipliers to find the value of l in terms of px, py, c1, and c2. The resulting equation holds for any production function p and is called the Law of Equimarginal Productivity.

43–50. Use a 3D graphics program to graph both equations in each of Exercises 1–8. Then visually check the results that you found analytically.

40. Business: maximizing production. A computer company has the following Cobb–Douglas production function for a certain product:

Answers to Quick Checks 225 15 at x = 15, y = 2 2 1 3 1 2. g = at x = ,y = 10 10 10 3. r L 4.3 cm, h L 8.6 cm, s L 348.73 cm2 4. x = 5, y = 8, maximum revenue = $122 1. A =

p1x, y2 = 800x 3/4y 1/4, where x is the labor, measured in dollars, and y is the capital, measured in dollars. Suppose that the company can make a total investment in labor and capital of $1,000,000. How should it allocate the investment between labor and capital in order to maximize production?

6.6 OBJECTIVE • Evaluate a double integral.

Double Integrals So far in this chapter, we have discussed functions of two variables and their partial derivatives. In this section, we consider integration of a function of two variables, in a process called iterated integration. The following is an example of a double integral: 6

2

6

10xy 2 dx dy, or

L3 L-1

L3

a

2

L-1

10xy 2 dxb dy.

Evaluating a double integral is somewhat similar to “undoing” a second partial derivative. We first evaluate the inside integral, indicated by the innermost differential (here dx), and treat the other variable(s) (here y) as constant(s): 2

L-1

10xy 2 dx = 10y 2 c

x2 2 d = 5y 23x 242-1 = 5y 232 2 - 1- 1224 = 15y 2. 2 -1

Color indicates the variable. All else is constant.

Then we evaluate the outside integral, associated with the differential dy: 6

L3

15y 2 dy = 15c =

y3 3

d

53y 3463

6 3

= 5163 - 332 = 945.

More precisely, the given double integral is called a double iterated integral. The word “iterate” means “to do again.” If dx and dy, as well as the limits of integration, are interchanged, we have 2

L-1 L3

6

10xy 2 dy dx.

6.6

•

Double Integrals

589

We first evaluate the inside, y-integral, treating x as a constant: 6

10xy dy = 10x c 2

L3

y3 3

d

6 3

=

10x 3 6 cy d 3 3

=

10 x163 - 332 = 630x. 3

Then we evaluate the outside, x-integral: 2

L-1

630x dx = 630 c

x2 2 d 2 -1

= 3153x 242-1

= 31532 2 - 1- 1224 = 945. Note that we get the same result.

DEFINITION

If f 1x, y2 is defined over the rectangular region R bounded by a … x … b and c … y … d, then the double integral of f 1x, y2 over R is given by d

b

Lc La

b

f1x, y2 dx dy or

La Lc

d

f1x, y2 dy dx.

In a more technical definition of the double integral, Riemann sums are used. However, for the functions in this text, the above definition is sufficient. Sometimes double integrals are defined over a nonrectangular region, in which case the bounds of integration may contain variables.

■ EXAMPLE 1 Evaluate 1

x

L0 Lx2

xy 2 dy dx.

Solution We first evaluate the inside integral with respect to y, treating x as a constant: x

Lx2

xy 2 dy = xc

y3 3

d

x x2

=

1 x3x 3 - 1x 2234 3

=

1 4 1x - x 72. 3

Then we evaluate the outside integral: 1

1 1 x5 x8 1 1x 4 - x 72 dx = c d 3 L0 3 5 8 0 =

Quick Check 1 1

Evaluate 4

2x

L0 L12 x

2xy dy dx.

Thus,

x

L0 Lx

2

O Quick Check 1

xy 2 dy dx =

1 15 18 05 08 1 ca - b - a bd = . 3 5 8 5 8 40

1 . 40

O

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Functions of Several Variables

The Geometric Interpretation of Multiple Integrals Suppose that the region D in the xy-plane is bounded by the functions y1 = g1x2 and y2 = h1x2 and the lines x 1 = a and x 2 = b. We want the volume, V, of the solid above D and under the surface z = f1x, y2. We can think of the solid as composed of many vertical columns, one of which is shown in Fig. 1 in red. The volume of this column can be thought of as l # w # h, or z # ¢y # ¢x. Integrating such columns in the y-direction, we obtain y2

Ly1

z dy, so c

y2

Ly1

z dy d ¢x

can be pictured as a “slab,” or slice. Then integrating such slices in the x-direction, we obtain the entire volume: b

V =

La

c

b

or

V =

This can be thought of as a collection of slices that fills the volume.

y2

Ly1

z dy d dx,

h1x2

La Lg1x2

z

z dy dx, z = f(x, y)

where z = f1x, y2.

z x1 = a

y

x2 = b D

x

Δy

Δx

y1 = g(x) y2 = h(x)

FIGURE 1

In Example 1, the region of integration D is the plane region between the graphs of y = x 2 and y = x, as shown in Figs. 2 and 3. z

z = xy2 y

y1 = x 2 y2 = x

y x

D

1

FIGURE 2

x

y2 = x

y1 = x 2

FIGURE 3

When we evaluated the double integral in Example 1, we found the volume of the solid based on D and capped by the surface z = xy 2, as shown in Fig. 3.

6.6 y

•

Double Integrals

591

■ EXAMPLE 2 Business: Demographics and Vehicle Ownership. The density of

(0, 4)

(5, 4)

(5, 0) x

privately owned vehicles in a city is given by the two-variable function p1x, y2 = 18 xy, where x is miles in the east–west direction, y is miles in the north–south direction, and p is the number of privately owned vehicles per square mile, in thousands. If the city limits are as shown in the figure to the left, what is the total number of privately owned vehicles in the city?

Solution We must decide on the order of integration. If we decide to integrate with respect to y first, then x second, the iterated integral is g21x2 1 8 xy La Lg11x2 b

y (0, 4)

y = g2( x)

(5, 4)

y = g1( x)

(5, 0) x

dy dx.

Since we are integrating with respect to y first, a helpful visual method for determining the bounds of integration is to draw an arrow in the positive y-direction, intersecting the shaded region representing the city. The arrow enters the region at the diagonal boundary first; this boundary is g11x2. The arrow exits the region through the horizontal boundary, and this boundary is g21x2. The bounds for the outer integral, with respect to x, are constants: the region extends from x = 0 to x = 5. The diagonal boundary is a line with slope - 45 and a y-intercept of 4. Therefore, g11x2 = 4 - 45 x. The horizontal boundary is g21x2 = 4. The integral is thus 5

4

L0 L

4- 45 x

1 8 xy

dy dx.

We integrate the inside integral first, with respect to y: 4

L4 - 45 x

4

1 1 1 xy dy = x B y 2 R 8 8 2 4- 45 x =

1 1 1 4 2 xc 1422 - a4 - xb d 8 2 2 5

=

1 16 8 2 xa x x b 8 5 25

=

2 2 1 3 x x . 5 25

After simplification

We now integrate with respect to x: 5

Quick Check 2 Redo Example 2, integrating with respect to x first, then y second. (Hint: Draw the arrow in the positive x-direction, and define the boundaries as functions of y.)

2 1 3 2 1 4 5 a x2 x b dx = c x 3 x d 25 15 100 L0 5 0 2 1 = a 1523 1524 b - 0 15 100 =

125 L 10.417. 12

Therefore, the city has about 10,417 privately owned vehicles.

O Quick Check 2 An Application to Probability Suppose that we throw a dart at a region R in a plane. We assume that the dart lands on a point 1x, y2 in R (see Fig. 4). We can think of 1x, y2 as a continuous random variable whose coordinates can be those of any ordered pair in region R. A function f is said to be a joint probability density function if f1x, y2 Ú 0, for all 1x, y2 in R,

O

592

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•

Functions of Several Variables

and

L LR

f1x, y2 dx dy = 1,

where 1 1 R refers to the double integral evaluated over the region R. y

y R

R

d G

(x, y) c

a

x

x

b

FIGURE 5

FIGURE 4

Suppose that we want to know the probability that the dart hits a point 1x, y2 in a rectangular subregion G of R, where G is the set of points for which a … x … b and c … y … d (Fig. 5). This probability is given by d

L LG

f1x, y2 dx dy =

b

Lc La

f1x, y2 dx dy.

Section Summary • The double integral of a two-variable function f1x, y2 over a rectangular region R bounded by a … x … b and c … y … d is written d

b

b

f1x, y2 dx dy

Lc La

or

• If the region of integration is not rectangular, the double integral may have variables in its bounds.

d

f1x, y2 dy dx.

La Lc

EXERCISE SET

6.6 2

Evaluate 3

1.

1

2y dx dy

L0 L0 3

3.

7. 9.

13x + y2 dx dy

x

L0 Lx2

1x + y2 dy dx

8.

x y dy dx

L1 L-2

1x + 5y2 dx dy

2

1x + y2 dy dx

L-1 Lx L0 L0 1

12.

3

L-4 L1

2

10.

ex

1 dy dx L0 L1 y

1 3

1

xy dy dx

3x dx dy

-1

6.

1

L-1 Lx

1

11.

-1

L0 L- 2

1

4.

13.

4

L0 L0 4

x y dy dx

L- 1 L1

1

2.

2 2

5

5.

1

x

e x + y dy dx

x

1x 2 + y 22 dy dx

L0 L- 1

x

L0 L0

1x + y 22 dy dx

3

14.

x 2

L1 L0

2e x dy dx

15. Find the volume of the solid capped by the surface z = 1 - y - x 2 over the region bounded on the xy-plane by y = 1 - x 2, y = 0, x = 0, and x = 1, by evaluating the integral 1

1 - x2

11 - y - x 22 dy dx.

L0 L0

16. Find the volume of the solid capped by the surface z = x + y over the region bounded on the xy-plane by y = 1 - x, y = 0, x = 0, and x = 1, by evaluating the integral 1

L0 L0

1-x

1x + y2 dy dx.

Exercise Set 6.6

For Exercises 17 and 18, suppose that a continuous random variable has a joint probability density function given by f 1x, y2  x2  13 xy, 0 ◊ x ◊ 1, 0 ◊ y ◊ 2. 17. Find 2

L0 L0

L1 L0

A triple iterated integral such as s

d

b

f 1x, y, z2 dx dy dz

Lr Lc La

18. Find the probability that a point 1x, y2 is in the region bounded by 0 … x … 12, 1 … y … 2, by evaluating the integral 2

SYNTHESIS

1

f 1x, y2 dx dy.

is evaluated in much the same way as a double iterated integral. We first evaluate the inside x-integral, treating y and z as constants. Then we evaluate the middle y-integral, treating z as a constant. Finally, we evaluate the outside z-integral. Evaluate these triple integrals.

1/2

f 1x, y2 dx dy.

1

23.

For Exercises 19 and 20, suppose that a continuous random variable has a joint probability density function given by 1 3 xy

 f 1x, y2  x2  3x  1 ◊ x ◊ 2, 3 ◊ y ◊ 5

1 3y

21. Life sciences: population. The population density of 1 2 fireflies in a field is given by p1x, y2 = 100 x y, where 0 … x … 30 and 0 … y … 20, x and y are in feet, and p is the number of fireflies per square foot. Determine the total population of fireflies in this field. 22. Life sciences: population. The population density of a city is given by p1x, y2 = 2x 2 + 5y, where x and y are in miles and p is the number of people per square mile, in hundreds. The city limits are as shown in the graph below. Determine the city’s population. y

4

12x + 3y - z2 dx dy dz

2

L0 L1 L-1

18x - 2y + z2 dx dy dz

1-x

L0 L0 2

26.

2

L0 L1 L-1

1

25.

20. Find the probability that a point is in the region bounded by 1 … x … 2 and 4 … y … 5.

3

2

24.

 2,

19. Find the probability that a point is in the region bounded by 1 … x … 2 and 3 … y … 4.

2-x

L0 6 - 2y

L0 L2 - y

xyz dz dy dx 24 - y2

L0

z dz dx dy

27. Describe the geometric meaning of the double integral of a function of two variables. 28. Explain how Exercise 17 can be answered without performing any calculations or finding any antiderivatives.

TECHNOLOGY CONNECTION 29. Use a calculator that does multiple integration to evaluate some double integrals found in this exercise set.

Answers to Quick Checks 4

(3, 4)

(1, 0)

(3, 0)

1. 16 3

x

593

2.

5

L0 L5 - 5>4y

1 8 xy

dx dy = 10.417, or 10,417 vehicles

CHAPTER 6 SUMMARY KEY TERMS AND CONCEPTS

EXAMPLES

SECTION 6.1 A function of two variables assigns to each input pair, 1x, y2, exactly one output number, f 1x, y2.

Business. A company produces two products. The first product costs $5.25 per unit to produce, and the second costs $7.50 per unit to produce. If x is the number of units of the first product and y is the number of units of the second product, the cost function C is given by C1x, y2 = 5.25x + 7.50y. If the company produces 30 units of the first product and 45 units of the second product, the total cost of producing these products is C130, 452 = 5.251302 + 7.501452 = $495

The graph of a two-variable function is a surface; graphing such a function requires a three-dimensional coordinate system. Points on the surface are expressed as ordered triples 1x, y, z2, where z = f 1x, y2.

z

The function g1x, y2 = 24 - x 2 - y 2 is a hemisphere of radius 2. Examples of points on the surface of this hemisphere are 10, 0, 22, 12, 0, 02, and 10, 2, 02.

2

2 2 x

The domain of a two-variable function is the set of points in the xy-plane for which f is defined.

The function f 1x, y2 = x 3 + y 3 - 3x - 27y - 2 is defined for all x and for all y. Therefore, the domain of f is D = 51x, y2 ƒ - q 6 x 6 q ,- q 6 y 6 q 6.

The function g1x, y2 = 24 - x 2 - y 2 is defined as long as the expression inside the radical is nonnegative. We have 4 - x 2 - y 2 Ú 0, which simplifies to x 2 + y 2 … 4. Therefore, the domain of g is D = 51x, y2 ƒ x 2 + y 2 … 46.

For the cost function C1x, y2 = 5.25x + 7.50x, the variables x and y represent quantities of products. Thus, they cannot be negative. Therefore, the domain for C is D = 51x, y2 ƒ x Ú 0, y Ú 06.

SECTION 6.2 Let f be a function of two variables, x and y. The partial derivative of f with respect to x is defined as

Let f 1x, y2 = x 2 + 2xy 3 + 2y. The partial derivative of f with respect to x is

f 1x + h, y2 - f 1x, y2 0f = lim . h : 0 0x h

The variable y is treated as a constant during the differentiation steps.

594

CHAPTER 6

•

0f = 2x + 2y 3. 0x The partial derivative of f with respect to y is

Functions of Several Variables

0f 1 = 6xy 2 + . 0y 22y

y

595

Chapter 6 Summary

KEY TERMS AND CONCEPTS

EXAMPLES

The partial derivative of f with respect to y is defined as

When x = 2 and y = 1, the slope of the tangent line at 12, 1, 132 on the surface representing the graph of f in the positive x-direction is

f 1x, y + h2 - f 1x, y2 0f = lim . h:0 0y h

The variable x is treated as a constant during the differentiation steps.

0f ` = 2122 + 21123 = 6. 0x 12, 12 The slope of the tangent line at that point on the surface in the positive y-direction is 0f 1 ` = 61221122 + = 12.5. 0y 12, 12 22112

Other common notations for partial derivatives are fx for the partial derivative of f with respect to x and fy for the partial derivative of f with respect to y. The partial derivative 0f>0x is interpreted as the slope of the tangent line at a point 1x, y, z2 on the surface representing the graph of f in the positive x-direction. Similarly, 0f>0y is interpreted as the slope of the tangent line at a point 1x, y, z2 on the surface in the positive y-direction. For functions of many variables, the partial derivative with respect to one of the variables is found by treating all the other variables as constants and differentiating using normal techniques. Let f be a function of two variables, x and y. Its second-order partial derivatives are fxx = fyx

0 2f

0 2f = , 0y 0x

, fxy 0x2 0 2f 0 2f = , and fyy = 2 . 0x 0y 0y

Let w1x, y, z2 = 3x 2y 3z 7. The partial derivatives of w are wx = 6xy 3z 7, wy = 9x 2y 2z 7, wz = 21x 2y 3z 6. Let f 1x, y2 = x 2 + 2xy 3 + 2y. Its first partial derivatives are fx =

0f 0f 1 = 2x + 2y 3 and fy = = 6xy 2 + . 0x 0y 21y

Its second-order partial derivatives are fxx = 2,

fxy = 6y 2,

fyx = 6y 2, and fyy = 12xy -

1 42y 3

.

Often, fxy = fyx.

SECTION 6.3 If f is a function of two variables, x and y, it has a relative maximum at 1a, b2 if f 1x, y2 … f 1a, b2

for all points 1x, y2 in a region containing 1a, b2. Similarly, f has a relative minimum at 1a, b2 if

Let f 1x, y2 = x 3 + y 3 - 3x - 27y - 2.

The first partial derivatives are fx1x, y2 = 3x 2 - 3 and fy1x, y2 = 3y 2 - 27. When fx = 0, we have x = ;1. Similarly, when fy = 0, we have y = ;3. There are four critical points:

f 1x, y2 Ú f 1a, b2

for all points 1x, y2 in a region containing 1a, b2.

A critical point occurs at 1a, b2 if both partial derivatives of f at 1a, b2 are 0. That is, fx1a, b2 = 0 and fy1a, b2 = 0.

and

11, 3, - 582, where f 11, 32 = - 58, 11, - 3, 502, where f 11, - 32 = 50, 1- 1, 3, - 542, where f 1- 1, 32 = - 54, 1- 1,- 3, 542, where f 1- 1, - 32 = 54.

The four second-order partial derivatives are fxx = 6x, fyy = 6y, and fxy = fyx = 0. Therefore, D = 16x216y2 - 02 = 36xy.

(continued)

596

CHAPTER 6

•

Functions of Several Variables

KEY TERMS AND CONCEPTS SECTION 6.3 (continued ) The D-test is used to determine whether critical points are relative maxima or minima. If 1a, b2 is a critical point, then

D = fxx1a, b2 # fyy1a, b2 - 3 fxy1a, b242.

And:

1. If D 7 0 and fxx1a, b2 6 0, then f has a maximum at 1a, b2. 2. If D 7 0 and fxx1a, b2 7 0, then f has a minimum at 1a, b2. 3. If D 6 0, then f has a saddle point at 1a, b2. 4. The D-test is not applicable if D = 0.

SECTION 6.4 A line of best fit for a set of data points is called a regression line. The method of least squares uses partial derivatives to determine this line.

EXAMPLES At 11, 32, we have D = 108 7 0, and fxx11, 32 = 6 7 0. Therefore, 11, 3, - 582 is a relative minimum.

At 11, - 32 and at 1- 1, 32, we have D = - 108 6 0. Therefore, 11, - 3, 502 and 1- 1, 3, - 542 are saddle points. At 1- 1, - 32, we have D = 108 7 0, and fxx1- 1, - 32 = - 6 6 0. Therefore, 1- 1, - 3, 542 is a relative maximum.

Relative maximum, z (−1, −3, 54)

Saddle point, (−1, 3, −54)

Saddle point, (1, −3, 50)

y x

Relative minimum, (1, 3, −58)

A researcher obtains the data points 11, 32, 14, 52, and 16, 72. Find the regression line. Using the method of least squares, we must minimize 1y1 - 322 + 1y2 - 522 + 1y3 - 722,

where

y1 = m112 + b = m + b, y2 = m142 + b = 4m + b, y3 = m162 + b = 6m + b.

After substitution, we have a two-variable function:

S1m, b2 = 1m + b - 322 + 14m + b - 522 + 16m + b - 722.

The partial derivatives are (after simplification): Sm = 106m + 22b - 130, Sb = 22m + 6b - 30. Setting the partial derivatives equal to 0 and solving the system, we get m =

40 15 , b = . 19 19

Therefore, the regression line is 40 y = 15 19 x + 19 .

y 7 6 5 4 3 2 1 0

(6, 7) 15 40 y  x   19 19 (4, 5)

(1, 3)

1

2

3

4

5

6

x

Chapter 6 Summary

KEY TERMS AND CONCEPTS SECTION 6.5 If the input variables x and y of a twovariable function f 1x, y2 are themselves related by an equation g1x, y2 = 0, then g1x, y2 = 0 is a constraint. The process of determining maximum and minimum values of f subject to the constraint g is called constrained optimization.

EXAMPLES Maximize f 1x, y2 = xy, subject to the constraint x + 2y = 1. We can substitute x = 1 - 2y into f:

f 11 - 2y, y2 = 11 - 2y2y = y - 2y 2.

Differentiating f with respect to y, we get fy = 1 - 4y. Setting this equal to 0, we get y = 14. Therefore, x = 1 - 2 A 14 B = 12. The function f has a maximum value of 81 at x = the constraint.

The method of Lagrange multipliers is one way to determine a maximum or minimum value of a function f subject to a constraint g. 1. Form the Lagrange function

F1x, y, l2 = f 1x, y2 - lg1x, y2.

The variable l is a Lagrange multiplier. 2. Find the first partial derivatives Fx, Fy, and Fl. 3. Solve the system Fx = 0, Fy = 0, and Fl = 0. A suggested method of solution is to isolate l in the equations Fx = 0 and Fy = 0, substitute to cancel out l, and simplify the resulting equation in terms of x and y. Make another substitution into Fl = 0, and determine the values of x and y.

597

1 2

and y = 14, which lies on the line given by

Maximize f 1x, y2 = xy, subject to the constraint x + 2y = 1. We write the constraint as x + 2y - 1 = 0 and form the Lagrange function: F1x, y, l2 = xy - l1x + 2y - 12. Differentiating F with respect to its three input variables, we have Fx = y - l, Fy = x - 2l, and Fl = - x - 2y + 1. We set all three expressions equal to 0. From Fx = 0 and Fy = 0, we isolate l: l = y and l = 12 x. By substitution, we have y = 12 x. Substituting for y in Fl and setting the expression equal to 0 gives - x - 2 A 12 x B + 1 = 0

Solving for x, we have x = 12. Therefore, y = constrained maximum value of 18 at A 12, 14 B .

A B = 14, and f has a

1 1 2 2

SECTION 6.6

If f 1x, y2 is defined over a rectangular region R bounded by a … x … b and c … x … d, then the double integral of f 1x, y2 over R is d

b

Lc La b

or

La Lc

d

f 1x, y2 dx dy f 1x, y2 dy dx.

If the region is not rectangular, then the bounds of integration may contain variables. Iterated integrals are evaluated by first integrating the inside integral, indicated by the innermost differential, and then integrating the outer integral. If f 1x, y2 is continuous over a region of feasibility that is closed and bounded, then the Extreme-Value Theorem guarantees the existence of an absolute maximum value and an absolute minimum value of f.

Evaluate 2

L0 L1

3

x2y dy dx.

The inside integral is integrated first. We integrate with respect to y, treating x as a constant: x2y dy = x2 C 12 y2 D 1 = x2 A 92 -

3

L1

3

1 2

B = 4x2.

We integrate the result with respect to x: 2

L0

4x2 dx =

C 43 x3 D 0 = 4318 - 02 = 2

32 3.

Find the absolute maximum and minimum values of f 1x, y2 = x 2 + y 2 - 2x - 2y, subject to the constraints 0 … x … 2 and 0 … y … 3. Since f is continuous on 0 … x … 2 and on 0 … y … 3, we check for critical points at all vertices and boundaries and in the interior. We find that f has an absolute maximum at both 10, 3, 32 and 12, 3, 32 and an absolute minimum at 11, 1, - 22.

CHAPTER 6 REVIEW EXERCISES These review exercises are for test preparation. They can also be used as a practice test. Answers are at the back of the book. The blue bracketed section references tell you what part(s) of the chapter to restudy if your answer is incorrect.

CONCEPT REINFORCEMENT Match each expression in column A with an equivalent expression in column B. Assume that z  f 1x, y2. [6.2, 6.6] Column A 0z 1. 0x

a)

L2

3 1 2x

0z 0y

b) fyx

3.

0 15x3y72 0x

c) fxy

4.

0 15x3y72 0y

d)

5.

0 z 0x 0y

e) fx

6.

0 2z 0y 0x

f) 15x2y7

3

8.

27. Consider the data in the following table regarding enrollment in colleges and universities during a recent 3-year period. [6.4]

3

2xy3 dx dy

L2

y3 dy

2xy3 dy dx

9. f 12, 02

12. fxy

h) fy

10. fx

11. fy

13. fyx

14. fxx

15. fyy

0 (1999) 2 (2001) 4 (2003) 5 (2004) 6 (2005)

$129 149 201 222 226

(Source: Kaiser Family Foundation and The New York Times, 10/23/05.)

2

18.

0z 0y 0 z

2

0x 2

0 z 0x 0y 2

19.

2

21.

f 1x, y2 = x 2 - 2xy + 2y 2 + 20

2 + 2y - 2. [6.1] x - 1

Given z  2x ln y  xy , find each of the following. [6.2] 3

0 z 2

22.

0y 2

subject to the constraint 2x - 6y = 15. [6.5]

30. Find the maximum value of f 1x, y2 = 6xy subject to the constraint 2x + y = 20. [6.5] 31. Find the absolute maximum and minimum values of f 1x, y2 = x 2 - y 2 subject to the constraints - 1 … x … 3 and - 1 … y … 2. [6.5] Evaluate. [6.6]

Find the relative maximum and minimum values. [6.3] 23. f 1x, y2 = x - 6xy + y + 6x + 3y 3

598

Year, x

Workers’ Average Monthly Out-of-Pocket Premium for Health Insurance for a Family

29. Find the minimum value of

16. State the domain of f 1x, y2 =

0 z 0y 0x

7.2 8.0 8.4

g) 35x3y6

Given f1x, y2  e y  3xy3  2y, find each of the following. [6.1, 6.2]

20.

1 2 3

28. Consider the data in the table below regarding workers’ average monthly out-of-pocket premium for health insurance for a family. [6.4] a) Find the regression line, y = mx + b. b) Use the regression line to predict workers’ average monthly out-of-pocket premium for health insurance for a family in 2012.

REVIEW EXERCISES

0z 0x

Enrollment, y (in millions)

1

L2 L0

17.

Year, x

a) Find the regression line, y = mx + b. b) Use the regression line to predict enrollment in the fourth year.

1

L2 L0

26. f 1x, y2 = x 4 + y 4 + 4x - 32y + 80

dx

2

3

25. f 1x, y2 = 3x - 6y - x 2 - y 2

Column B

2.

7.

24. f 1x, y2 = x 2 - xy + y 2 - 2x + 4y

CHAPTER 6

2

•

1 5

Functions of Several Variables

1

32.

L0 L1

2

x2 y3 dy dx

1

33.

x

L0 Lx 2

1x - y2 dy dx

Chapter 6 Test

34. Business: demographics. The density of students living near a university is modeled by p1x, y2 = 9 - x 2 - y 2, where x and y are in miles and p is the number of students per square mile, in hundreds. Assume the university is located at 10, 02. Find the number of students who live in the shaded region shown below. [6.6] y (0, 2)

(0, 0)

599

SYNTHESIS 35. Evaluate 2

1-x

22 - x2

z dz dy dx. [6.6]

L0 L1 - 2x L0

36. Business: minimizing surface area. Suppose that beverages could be packaged in either a cylindrical container or a rectangular container with a square top and bottom. Each container is designed to have the minimum surface area for its shape. If we assume a volume of 26 in3, which container would have the smaller surface area? [6.3, 6.5]

TECHNOLOGY CONNECTION

(2, 0) x

37. Use a 3D graphics program to graph f 1x, y2 = x 2 + 4y 2. [6.1]

CHAPTER 6 TEST Given f1x, y2  e x  2x 3y  y, find each of the following. 0f 2. 0x

1. f 1- 1, 22 4.

7.

0 2f 0x

5.

2

0f 3. 0y

0 2f 0x 0y

6.

0 2f 0y 0x

0 2f

11. Find the maximum value of

f 1x, y2 = 6xy - 4x 2 - 3y 2

subject to the constraint x + 3y = 19. 12. Evaluate 3

3

L0 L1

0y 2

Find the relative maximum and minimum values. 8. f 1x, y2 = x 2 - xy + y 3 - x

4x3y2 dx dy.

SYNTHESIS 13. Business: maximizing production. Southwest Appliances has the following Cobb–Douglas production function for a certain product:

9. f 1x, y2 = 4y 2 - x 2

10. Business: predicting total sales. Consider the data in the following table regarding the total sales of a company during the first three years of operation.

Year, x

Sales, y (in millions)

1 2 3

$10 15 19

p1x, y2 = 50x 2>3y 1>3, where x is labor, measured in dollars, and y is capital, measured in dollars. Suppose that Southwest can make a total investment in labor and capital of $600,000. How should it allocate the investment between labor and capital in order to maximize production? 14. Find fx and ft :

a) Find the regression line, y = mx + b. b) Use the regression line to predict sales in the fourth year.

f 1x, t2 =

x 2 - 2t x 3 + 2t

.

TECHNOLOGY CONNECTION 15. Use a 3D graphics program to graph f 1x, y2 = x -

1 2

y 2 - 13 x 3.

Extended Technology Application Minimizing Employees’Travel Time in a Building If employees spend considerable time moving between offices, designing a building to minimize travel time can reap enormous savings for a company. For a multilevel building with a square base, one design concern is minimizing travel time between the most remote points. We will make use of Lagrange multipliers to help design such a building. Let’s assume that each floor has a square grid of hallways, as shown in the figure at the lower right. Suppose that you are standing at point P in the top northeast corner of the twelfth floor of this building. How long will it take to reach the most remote point at the southwest corner on the first floor—that is, point Q?

Let’s call the time t. We find a formula for t in two steps: 1. You are to go from the twelfth floor to the first floor. This is a move in a vertical direction. 2. You need to cross horizontally from one corner of the building to the other. The vertical time is h, the height of point P from the ground, divided by a, the speed at which you can travel in a vertical direction (elevator speed). Thus, vertical time is given by h>a. The horizontal time is the time it takes to go across one level, by way of the square grid of hallways (from R to Q in the figure). If each floor is a square with side of length k, then the distance from R to Q is 2k. If the walking speed is b, then the horizontal time is given by 2k>b. Thus, the time it will take to go from P to Q is a function of two variables, h and k, given by t1h, k2 = vertical time + horizontal time h 2k = + , a b where a and b, the elevator speed and walking speed, are constants. What happens if we must choose between two (or more) building plans with the same floor area, but with different dimensions?

P Q

h

Q

R k

600

CHAPTER 6

•

Functions of Several Variables

R

Extended Technology Application

Will the travel time be the same? Or will it be different for the two buildings? First, what is the total floor area of a given building? Suppose that the building has n floors, each a square of side k. Then the total floor area is given by

2. Do different dimensions, with a fixed floor area, yield different travel times? In Exercises 3–5, assume that you are finding the dimensions of a multilevel building with a square base that will minimize travel time t between the most remote points in the building. Each floor has a square grid of hallways. The height of point P is h, and the length of a side of each floor is k. The elevator speed is 10 ft>sec and the average speed of a person walking is 4 ft>sec. The total floor area of the building is 40,000 ft2. The height of each floor is 12 ft.

A = nk 2. Note that the area of the roof is not included. If h is the height of point P and c is the height of each floor—that is, the distance from the carpeting on one floor to the carpeting on the floor above—then n = 1 + h>c, with A = 11 + h/c2 k 2.

3. Use the information given to find a formula for the function t1h, k2.

Let’s return to the problem of two buildings with the same total floor area, but with different dimensions, and see what happens to t1h, k2.

4. Find a formula for the constraint. 5. Use the method of Lagrange multipliers to find the dimensions of the building that will minimize travel time t between the most remote points in the building.

EXERCISES 1. Use the TABLE feature on your calculator or spreadsheet software to complete the table below. For each case in the table, let the elevator speed a = 10 ft>sec, the walking speed b = 4 ft>sec, and the height of each floor c = 15 ft. Each case in the table covers two situations, though the floor area stays essentially the same for a particular case.

CASE BUILDING

1 2 3 4 5 6 7 8 9 10

B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2

6. Use a 3D graphics program to graph both equations in Exercises 3 and 4. Then visually check the results you found analytically.

n

k

2 3 2 3 4 5 5 10 5 10 10 17 10 17 17 26 17 26 26 50

40 32.66 60 48.99 40 35.777 60 42.426 150 106.066 40 30.679 80 61.357 40 32.344 50 40.43 77 55.525

A

h

t(h, k)

3200 3200 7200

15 21.5 30 19.4

601

Cumulative Review 1. Write an equation of the line with slope - 4 and containing the point 1- 7, 12. 2. For f 1x2 = x 2 - 5, find f 1x + h2. f 1x2 = e

5 - x, for x Z 2, - 3, for x = 2.

b) Find lim f 1x2. x :2

c) Find f 122. d) Is f continuous at 2?

lim

x : -4

f(x)

–2 –1 –1 –2 –3 –4

1 2 3 4 5 6 7 8 x

x 2 - 16 x + 4

7.

4 x - 3

lim

x: q

8. lim

x:q

18. f 1x2 =

19. y = ln 1x 2 + 52

20. y = e ln x

21. y = e 3x + x 2

22. y = e 2x - 3

24. For y = x 2 -

2 , find d 2y>dx 2. x

26. Differentiate implicitly to find dy>dx if x 3 + x>y = 7.

12x - 7 3x + 2

27. Find an equation of the tangent line to the graph of y = e x - x 2 - 3 at the point 10, - 22.

2x 3 - x

28. Find the x-value(s) at which the tangent lines to f 1x2 = x 3 - 2x 2 have a slope of - 1.

8x - x 2 + 1 5

9. If f 1x2 = x 2 + 3, find f ¿1x2 by determining f 1x + h2 - f 1x2 lim . h :0 h For exercises 10–12, refer to the following graph of y  h1x2.

Sketch the graph of each function. List and label the coordinates of any extrema and points of inflection. State where the function is increasing or decreasing, where it is concave up or concave down, and where any asymptotes occur. 29. f 1x2 = x 3 - 3x + 1

30. f 1x2 = 2x 2 - x 4 - 3

y y = h(x)

31. f 1x2 = 32. f 1x2 =

a

b

c

d

e

f

x

10. Identify the input values for which h has no limit. 11. Identify the input values for which h is discontinuous. 12. Identify the input values for which the derivative of h does not exist.

602

5x 3 + 4 2x - 1

17. f 1x2 = 22x 5 - 8

25. Business: average cost. Doubletake Clothing finds that the cost, in dollars, of producing x pairs of jeans is given by C1x2 = 320 + 92x. Find the rate at which the average cost is changing when 100 pairs of jeans have been produced.

x :1

x :3

16. f 1x2 = x -6

23. f 1x2 = ln 1e x - 42

5. lim 2x 3 + 8 6. lim

14. y = x 2 - 7x + 3

3

Find each limit, if it exists. If a limit does not exist, state that fact. 4.

13. y = - 9x + 3 15. y = x 1/4

y 6 5 4 3 2 1

3. a) Graph:

Differentiate.

8x x + 1 2

8 x - 4 2

Find the absolute maximum and minimum values, if they exist, over the indicated interval. If no interval is indicated, consider the entire real number line. 33. f 1x2 = 3x 2 - 6x - 4 34. f 1x2 = - 5x + 1

35. f 1x2 = 13 x 3 - x 2 - 3x + 5;

3- 2, 04

Cumulative Review

36. Business: maximizing profit. For custom sweatshirts, Detailed Clothing’s total revenue and total cost, in dollars, are given by

Find the number of sweatshirts, x, that must be produced and sold in order to maximize profit. 37. Business: minimizing inventory costs. An appliance store sells 450 MP3 players each year. It costs $4 to store a player for a year. When placing an order, there is a fixed cost of $1 plus $0.75 for each player. How many times per year should the store reorder MP3 players, and in what lot size, in order to minimize inventory costs? 38. Let y = 3x - 2x + 1. Use differentials to find the approximate change in y when x = 2 and ¢x = 0.05. 2

39. Business: exponential growth. A national frozen yogurt firm is experiencing growth of 10% per year in the number, N, of franchises that it owns; that is, dN = 0.1N, dt where N is the number of franchises and t is the time, in years, from 2001. a) Given that there were 8000 franchises in 2001, find the solution of the equation, assuming that N0 = 8000 and k = 0.1. b) How many franchises were there in 2009? c) What is the doubling time of the number of franchises? 40. Economics: elasticity of demand. Consider the demand function q = D1x2 = 240 - 20x, where q is the quantity of coffee mugs demanded at a price of x dollars. a) Find the elasticity. b) Find the elasticity at x = $2, and state whether the demand is elastic or inelastic. c) Find the elasticity at x = $9, and state whether the demand is elastic or inelastic. d) At a price of $2, will a small increase in price cause total revenue to increase or decrease? e) Find the value of x for which the total revenue is a maximum. 41. Business: approximating cost overage. A large square plot of ground measures 75 ft by 75 ft, with a tolerance of ;4 in. Landscapers are going to cover the plot with grass sod. Each square of sod costs $8 and measures 3 ft by 3 ft. a) Use differentials to estimate the change in area when the measurement tolerance is taken into acount. b) How many extra squares of sod should the landscapers bring to the job, and how much extra will this cost? Evaluate.

3x dx 5

17 -x 3x2 dx (Use Table 1 on pp. 454–455.) 45. x e dx (Do not use Table 1.) 75 46. 1x + 32 ln x dx 47. dx x 48. 32x dx 44.

2

3 x4

R1x2 = 4x 2 + 11x + 110, C1x2 = 4.2x 2 + 5x + 10.

42.

603



0

43.

-1

12e x + 12 dx

1

0

49. Find the area under the graph of y = x 2 + 3x over the interval 31, 54. 50. Business: present value. Find the present value of $250,000 due in 30 yr at 6%, compounded continuously. 51. Business: value of a fund. Leigh Ann wants to have $50,000 saved in 10 yr. a) She could make a one-time deposit at an APR of 5.45%, compounded continuously. Find the amount she should deposit. b) She could instead make an investment that would yield a constant revenue stream of R1t2 dollars per year, at 5.45% compounded continuously. Find R1t2. c) Calculate the interest earned in part (a) and in part (b). 52. Business: contract buyout. An executive works under an 8-yr contract that pays him $200,000 per year. He invests the money at an APR of 4.85%, compounded continuously. After 5 yr, the company offers him a buyout of the contract. What is the lowest amount he should accept, if the continuously compounded APR is the same? 53. Determine whether the following improper integral is convergent or divergent, and calculate its value if it is convergent:



q

3

1 x7

dx.

54. Given the probability density function f 1x2 =

3 2x 2

over 31, 34,

find E1x2. 55. Let x be a continuous random variable that is normally distributed with mean m = 3 and standard deviation s = 5. Using Table A (p. 621), find P1- 2 … x … 82. 56. Business: distribution of salaries. The salaries paid by a large corporation are normally distributed with a mean m = $45,000 and a standard deviation s = $6,000. a) Find the probability that a randomly chosen employee earns between $42,000 and $55,000 per year. b) An executive of the corporation earns $60,000 per year. In what percentile of the salaries does this salary place him? c) A new employee insists on a salary that is in the top 2% of salaries. What is the minimum salary that this employee would accept?

604

Cumulative Review

57. Economics: supply and demand. Demand and supply functions are given by p = D1x2 = 1x - 2022

Given f 1x, y2  e y  4x 2y 3  3x, find each of the following. 62. fx

63. fyy

64. Find the relative maximum and minimum values of f 1x, y2 = 8x 2 - y 2.

and p = S1x2 = x 2 + 10x + 50, where p is the price per unit, in dollars, when x units are sold. Find the equilibrium point and the consumer’s surplus.

65. Maximize f 1x, y2 = 4x + 2y - x 2 - y 2 + 4, subject to the constraint x + 2y = 9. 66. Evaluate

3

58. Find the volume of the solid of revolution generated by rotating the region under the graph of -x

0

2

-1

e x dy dx.

y = e , from x = 0 to x = 5, about the x-axis. 59. Solve the differential equation dy>dx = xy. 60. Let f 1x, y2 = 216 - x 2 - y 2. a) Evaluate f 13, 12. b) State the domain of f.

61. Consider the data in the following table. Age of business (in years)

1

3

5

Profit (in tens of thousands of dollars)

4

7

9

a) Find the regression line, y = mx + b. b) Use the regression line to predict the profit when the business is 10 years old.

Distance to north of mall (in miles)

67. Business: demographics. The number of shoppers, in hundreds per square mile, who frequent a mall is modeled by the two-variable function f 1x, y2 = 10 - x - y 2, where x is miles from the mall toward the east and y is miles from the mall toward the north. The graph below shows a shaded region to the northeast of the mall, which is at 10, 02. Find the total number of frequent mall shoppers in the region. y

(0, 2)

(0, 0)

(2, 2)

Distance to east x of mall (in miles)

APPEN DIX

A

Review of Basic Algebra

This appendix covers most of the algebraic topics essential to a study of calculus. It might be used in conjunction with Chapter R or as the need for certain skills arises throughout the book.

Exponential Notation

OBJECTIVES

Let’s review the meaning of an expression

• Manipulate exponential expressions.

a n,

• Multiply and factor algebraic expressions.

where a is any real number and n is an integer; that is, n is a number in the set 5 Á , - 3, - 2, - 1, 0, 1, 2, 3, Á 6. The number a is called the base and n is called the exponent. If n is greater than 1, then

• Solve equations, inequalities, and applied problems.

⎫⎢ ⎢ ⎢ ⎬ ⎢ ⎢⎢ ⎭⎢

a n = a # a # a Á a. n factors

In other words, a n is the product of n factors, each of which is a.

■ EXAMPLE 1 Express each of the following without exponents: a) 4 3

b) 1- 225

c) 1- 224

d) - 2 4

e) 11.0822

1 3 f) a b 2

Solution

4 3 = 4 # 4 # 4 = 64 1- 225 = 1- 221- 221- 221- 221- 22 = - 32 1- 224 = 1- 221- 221- 221- 22 = 16 - 2 4 = - 12 42 = - 122122122122 = - 16 The base is 2, not - 2. 2 11.082 = 1.08 * 1.08 = 1.1664 1 3 1 1 1 1 f) a b = # # = 2 2 2 2 8

a) b) c) d) e)

We define an exponent of 1 as follows: a 1 = a, for any real number a. In other words, any real number to the first power is that number itself. We define an exponent of 0 as follows: a 0 = 1, for any nonzero real number a. That is, any nonzero real number a to the zero power is 1.

605

606

APPENDIX A

■ EXAMPLE 2 Express without exponents: a) 1- 2x20

1 0 c) a b 2

b) 1- 2x21

d) e 0

e) e 1

1 1 f) a b 2

Solution a) 1- 2x20 = 1

b) 1- 2x21 = - 2x

d) e 0 = 1

e) e 1 = e

1 0 c) a b = 1 2 1 1 1 f) a b = 2 2

The meaning of a negative integer as an exponent is as follows:

a -n =

1 1 n n = a b , for any nonzero real number a. a a

That is, any nonzero real number a to the - n power is the reciprocal of a n, or equivalently, (the reciprocal of a)n.

■ EXAMPLE 3 Express without negative exponents: a) 2 -5

1 -2 c) a b 4

b) 10-3

d) x -5

e) e -k

f) t -1

Solution a) 2 -5 =

1 1 1 = # # # # = 5 2 2 2 2 2 32 2

b) 10-3 =

1 -2 4 2 c) a b = a b = 4 2 = 16 4 1 e) e -k =

1 ek

1 1 1 = = , or 3 # # 10 10 10 1000 10

d) x -5 =

1 x5

f) t -1 =

1 1 = 1 t t

Properties of Exponents Note the following: b 5 # b -3 = 1b # b # b # b # b2 # b#b#b# # b b b#b#b = 1 # b # b = b 2.

1

b#b#b

=

We can obtain the same result by adding the exponents. This is true in general.

THEOREM 1 For any nonzero real number a and any integers n and m, a n # a m = a n + m.

(To multiply when the bases are the same, add the exponents.)

0.001

Review of Basic Algebra

■ EXAMPLE 4 Multiply: a) x 5 # x 6

b) x -5 # x 6

c) 2x -3 # 5x -4

Solution

a) x 5 # x 6 = x 5 + 6 = x 11

607

d) r 2 # r b) x -5 # x 6 = x -5 + 6 = x

c) 2x -3 # 5x -4 = 10x -3 + 1-42 = 10x -7, or

d) r 2 # r = r 2 + 1 = r 3

10 x7

Note the following: b5 , b2 =

b5 b#b#b#b#b = 2 b#b b

b#b# # # b b b b#b = 1 # b # b # b = b 3. =

We can obtain the same result by subtracting the exponents. This is true in general.

THEOREM 2 For any nonzero real number a and any integers n and m, an = a n - m. am (To divide when the bases are the same, subtract the exponent in the denominator from the exponent in the numerator.)

■ EXAMPLE 5 Divide: a)

a3 a2

b)

x7 x7

c)

e3 e -4

d)

e -4 e -1

Solution a)

a3 = a3 - 2 = a1 = a a2

b)

x7 = x7-7 = x0 = 1 x7

c)

e3 = e 3 - 1-42 = e 3 + 4 = e 7 e -4

d)

e -4 = e -4 - 1-12 = e -4 + 1 = e -3, or e -1

1 e3

Note the following:

1b 223 = b 2 # b 2 # b 2 = b 2 + 2 + 2 = b 6.

We can obtain the same result by multiplying the exponents. The other results in Theorem 3 can be similarly motivated.

THEOREM 3 For any nonzero real numbers a and b, and any integers n and m, 1a n2m = a nm,

1ab2n = a nb n, and

a n an a b = n. b b

608

APPENDIX A

■ EXAMPLE 6 Simplify: a) 1x -223

b) 1e x22

c) 12x 4y -5z 32-3

#3

1 x6

d) a

x2 3 b p 4q 5

Solution a) 1x -223 = x -2

= x -6, or

b) 1e x22 = e 2x

c) 12x 4y -5z 32-3 = 2 -31x 42-31y -52-31z 32-3 =

1 -12 15 -9 x y z , or 23

d) a y 15

1x 223 x2 3 x6 b = = p 4q 5 1p 4q 523 1p 4231q 523 =

8x 12z 9

x6 p 12q 15

Multiplication The distributive law is important when multiplying. This law is as follows.

The Distributive Law For any numbers A, B, and C, A1B + C2 = AB + AC. Because subtraction can be regarded as addition of an additive inverse, it follows that A1B - C2 = AB - AC.

■ EXAMPLE 7 Multiply: a) 31x - 52

b) P11 + i2

c) 1x - 521x + 32

Solution

a) 31x - 52 = 3 # x - 3 # 5 = 3x - 15

c) 1x - 521x + 32 = 1x - 52x + 1x - 523 = x # x - 5x + 3x - 5 # 3 = x 2 - 2x - 15

d) 1a + b21a + b2

b) P11 + i2 = P # 1 + P # i = P + Pi

d) 1a + b21a + b2 = 1a + b2a + 1a + b2b = a # a + ba + ab + b # b = a 2 + 2ab + b 2

The following formulas, which are obtained using the distributive law, are also useful when multiplying. All three are used in Example 8, which follows. 1A + B22 = A2 + 2AB + B 2 1A - B22 = A2 - 2AB + B 2 1A - B21A + B2 = A2 - B 2

■ EXAMPLE 8 Multiply: a) 1x + h22

Solution

b) 12x - t22

a) 1x + h22 = x 2 + 2xh + h2 b) 12x - t22 = 12x22 - 212x2t + t 2 = 4x 2 - 4xt + t 2 c) 13c + d213c - d2 = 13c22 - d 2 = 9c 2 - d 2

c) 13c + d213c - d2

Review of Basic Algebra

609

Factoring Factoring is the reverse of multiplication. That is, to factor an expression, we find an equivalent expression that is a product. Always remember to look first for a common factor.

■ EXAMPLE 9 Factor: a) P + Pi d) x 2 - 5x - 14

b) 2xh + h2 e) 6x 2 + 7x - 5

c) x 2 - 6xy + 9y 2 f ) x 2 - 9t 2

Solution

a) P + Pi = P # 1 + P # i = P11 + i2 b) 2xh + h2 = h12x + h2 c) x 2 - 6xy + 9y 2 = 1x - 3y22 d) x 2 - 5x - 14 = 1x - 721x + 22 e) 6x 2 + 7x - 5 = 12x - 1213x + 52 f) x 2 - 9t 2 = 1x - 3t21x + 3t2

We used the distributive law.

We looked for factors of - 14 whose sum is - 5. We first considered ways of factoring the first coefficient—for example, 12x 213x 2. Then we looked for factors of - 5 such that when we multiply, we obtain the given expression. We used the formula 1A - B21A + B2 = A2 - B 2.

Some expressions with four terms can be factored by first looking for a common binomial factor. This is called factoring by grouping.

■ EXAMPLE 10 Factor: a) t 3 + 6t 2 - 2t - 12

b) x 3 - 7x 2 - 4x + 28

Solution

a) t 3 + 6t 2 - 2t - 12 = t 21t + 62 - 21t + 62

Factoring the first two terms and then the second two terms Factoring out the common binomial factor, t + 6

= 1t 2 - 221t + 62 b) x 3 - 7x 2 - 4x + 28 = x 21x - 72 - 41x - 72 = 1x - 721x 2 - 42 = 1x - 721x - 221x + 22

Factoring the first two terms and then the second two terms Factoring out the common binomial factor, x - 7 Using 1A - B21A + B2 = A2 - B 2

Solving Equations Basic to the solution of many equations are the Addition Principle and the Multiplication Principle. We can add (or subtract) the same number on both sides of an equation and obtain an equivalent equation, that is, a new equation that has the same solutions as the original equation. We can also multiply (or divide) by a nonzero number on both sides of an equation and obtain an equivalent equation.

The Addition Principle

The Multiplication Principle

For any real numbers a, b, and c,

For any real numbers a, b, and c, with c Z 0,

a = b is equivalent to a + c = b + c.

a = b is equivalent to a # c = b # c.

When solving a linear equation, we use these principles and other properties of real numbers to get the variable alone on one side. Then it is easy to determine the solution.

610

APPENDIX A

■ EXAMPLE 11 Solve: - 65 x + 10 = 12 x + 2. Solution We first multiply by 6 on both sides to clear the fractions: 6 A - 65 x + 10 B = 6 A 12 x + 2 B

6A-

5 6x

B + 6 # 10 = 6 A B + 6 # 2 1 2x

- 5x + 60 = 3x + 12 60 = 8x + 12

Using the distributive law Simplifying Using the Addition Principle: We add 5x on both sides.

48 = 8x

1 8

Using the Multiplication Principle

Adding - 12 on both sides

# 48 = 18 # 8x

Multiplying by 18 on both sides

6 = x. The variable is now alone on one side, and we see that 6 is the solution. We can check by substituting 6 into the original equation. The third principle for solving equations is the Principle of Zero Products.

The Principle of Zero Products For any numbers a and b, if ab = 0, then a = 0 or b = 0; and if a = 0 or b = 0, then ab = 0. To solve an equation using this principle, we must have a 0 on one side and a product on the other. The solutions are then obtained by setting each factor equal to 0 and solving the resulting equations.

■ EXAMPLE 12 Solve: 3x1x - 2215x + 42 = 0. Solution We have 3x1x - 2215x 3x = 0 or x - 2 1# 1# x 3 3x = 3 0 or x = 0 or x

+ = = =

42 0 2 2

= 0 or 5x + 4 = 0 or 5x = - 4 or x = - 45.

Using the Principle of Zero Products Solving each separately

The solutions are 0, 2, and - 45. Note that the Principle of Zero Products applies only when a product is 0. For example, although we may know that ab = 8, we do not know that a = 8 or b = 8.

■ EXAMPLE 13 Solve: 4x 3 = x. Solution We have

4x 3 4x 3 - x x14x 2 - 12 x12x - 1212x + 12 x = 0 or 2x - 1 = 0 or x = 0 or 2x = 1 or x = 0 or x = 12 or

= x = 0 = 0 = 0 2x + 1 2x x

Adding - x to both sides Factoring

= 0 = -1 = - 12.

Using the Principle of Zero Products

The solutions are 0, 12, and - 12.

Rational Equations Expressions like the following are polynomials in one variable: x 2 - 4,

x 3 + 7x 2 - 8x + 9,

t - 19.

Review of Basic Algebra

611

The least common multiple, LCM, of two polynomials is found by factoring and using each factor the greatest number of times that it occurs in any one factorization.

■ EXAMPLE 14 Find the LCM: x 2 + 2x + 1, 5x 2 - 5x, and x 2 - 1. Solution x 2 + 2x + 1 5x 2 - 5x x2 - 1 LCM = 5x1x

= = = +

1x + 5x1x 1x + 121x

121x + 12; ⎫⎢ ⎢ ⎬ - 12; ⎢ 121x - 12 ⎭⎢⎢⎢ + 121x - 12

Factoring

A rational expression is a ratio of polynomials. Each of the following is a rational expression: x 2 - 6x + 9 , x2 - 4

x - 2 , x - 3

a + 7 , a 2 - 16

5 . 5t - 15

A rational equation is an equation containing one or more rational expressions. Here are some examples: 2 5 1 = , x 3 6

x +

6 = 5, x

2x 6 18 = 2 . x - 3 x x - 3x

To solve a rational equation, we first clear the equation of fractions by multiplying on both sides by the LCM of all the denominators. The resulting equation might have solutions that are not solutions of the original equation. Thus, we must check all possible solutions in the original equation.

■ EXAMPLE 15 Solve:

2x 6 18 = 2 . x x - 3x x - 3

Solution Note that x 2 - 3x = x1x - 32. The LCM of the denominators is x1x - 32. We multi-

ply by x1x - 32.

2x 6 18 - b = x1x - 32a 2 b x - 3 x x - 3x

Multiplying by the LCM on both sides

2x 6 18 - x1x - 32 # = x1x - 32a 2 b x - 3 x x - 3x

Using the distributive law

x1x - 32a x1x - 32 #

2x 2 - 61x - 32 2x 2 - 6x + 18 2x 2 - 6x 2x1x - 32 2x = 0 x = 0

= 18 = 18 = 0 = 0 or x - 3 = 0 or x = 3

Simplifying

The numbers 0 and 3 are possible solutions. We look at the original equation and see that each makes a denominator 0. We can also carry out a check, as follows. Check For 0:

2x 6 18 = 2 x x - 3x x - 3 2102 6 18 ? 2 0 - 3 0 0 - 3102 0 -

6 0

18 0

The equation has no solution.

UNDEFINED; FALSE

For 3:

2x 6 18 = 2 x x - 3x x - 3 2132 18 6 ? 2 3 - 3132 3 - 3 3 6 - 2 0

18 0

UNDEFINED; FALSE

612

APPENDIX A

■ EXAMPLE 16 Solve:

x2 4 = . x - 2 x - 2

Solution The LCM of the denominators is x - 2. We multiply by x - 2. 1x - 22 #

x2 x - 2 x2 x2 - 4 1x + 221x - 22 x = -2

= 1x - 22 # = 4 = 0 = 0 or x = 2

4 x - 2 Simplifying

Using the Principle of Zero Products

Check For 2:

For - 2:

x2 4 = x - 2 x - 2

x2 4 = x - 2 x - 2 1- 222 4 ? -2 - 2 -2 - 2

22 4 ? 2 - 2 2 - 2 4 4 0 0

UNDEFINED; FALSE

4 -4 -1

4 -4 -1

TRUE

The number - 2 is a solution, but 2 is not (it results in division by 0).

Solving Inequalities Two inequalities are equivalent if they have the same solutions. For example, the inequalities x 7 4 and 4 6 x are equivalent. Principles for solving inequalities are similar to those for solving equations. We can add the same number to both sides of an inequality. We can also multiply on both sides by the same nonzero number, but if that number is negative, we must reverse the inequality sign. The following are the inequality-solving principles.

The Inequality-Solving Principles For any real numbers a, b, and c, a 6 b is equivalent to a + c 6 b + c. For any real numbers a, b, and any positive number c, a 6 b is equivalent to ac 6 bc. For any real numbers a, b, and any negative number c, a 6 b is equivalent to ac 7 bc. Similar statements hold for … and Ú.

Review of Basic Algebra

613

■ EXAMPLE 17 Solve: 17 - 8x Ú 5x - 4. Solution We have 17 - 8x - 8x - 13x 1 - 131- 13x2

Ú Ú Ú …

x …

5x - 4 5x - 21 - 21 1 - 13 1- 212

Adding - 17 to both sides Adding - 5x to both sides Multiplying both sides by - 131 and reversing the inequality sign

21 13 .

Any number less than or equal to 21 13 is a solution.

Applications To solve applied problems, we first translate to mathematical language, usually an equation. Then we solve the equation and check to see whether the solution to the equation is a solution to the problem.

■ EXAMPLE 18 Life Science: Weight Gain. After a 5% gain in weight, a grizzly bear weighs 693 lb. What was its original weight?

Solution We first translate to an equation:

⎫ ⎢ ⎢ ⎢ ⎬ ⎢ ⎢⎢ ⎢ ⎭

⎫ ⎢ ⎢ ⎢ ⎬ ⎢ ⎢⎢ ⎢ ⎭

1Original weight2 + 5%1Original weight2 = 693 w + 5% w = 693.

Now we solve the equation: w + 5%w 1 # w + 0.05w 11 + 0.052w 1.05w

= = = =

693 693 693 693 693 w = = 660. 1.05

Check: 600 + 5% # 660 = 660 + 0.05 # 660 = 660 + 33 = 693. The original weight of the bear was 660 lb.

■ EXAMPLE 19 Business: Total Sales. Raggs, Ltd., a clothing firm, determines that its total revenue, in dollars, from the sale of x suits is given by 200x + 50. Determine the number of suits that the firm must sell to ensure that its total revenue will be more than $70,050.

Solution We translate to an inequality and solve: 200x + 50 7 70,050 200x 7 70,000 x 7 350.

Adding - 50 to both sides 1 Multiplying both sides by 200

Thus the company’s total revenue will exceed $70,050 when it sells more than 350 suits.

614

APPENDIX A

EXERCISE SET

A Express as an equivalent expression without exponents. 1. 53

2. 72

4. 1 - 523

5. 11.0122

1 7. a b 2 10. 16x2

4

1 8. a b 4

1

1 13. a b 3

11. t

0

3. 1- 722

6. 11.0123

3

9. 16x20

1

12. t

1 14. a b 3

0

Multiply. 61. 51x - 72

63. 1x - 521x - 22

65. 1a - b21a + ab + b 2 2

66. 1x 2 - xy + y 221x + y2 67. 12x + 521x - 12

68. 13x + 421x - 12

71. 15x + 2215x - 22

72. 1t - 121t + 12

73. 1a - h22

15. 3 -2

16. 4 -2

75. 15x + t22

1 -3 17. a b 2

1 -2 18. a b 2

77. 5x1x 2 + 322

19. 10-1

20. 10-4

21. e -b

22. t -k

23. b -1

24. h-1

25. x 2 # x 3

26. t 3 # t 4

27. x -7 # x

31. x -4 # x 7 # x

32. x -3 # x # x 3

33. e -t # e t

28. x 5 # x

34. e k # e -k

29. 5x 2 # 7x 3

30. 4t 3 # 2t 4

38. 41. 44.

47.

x5

36.

2

x

x3

39.

x7 e e

70. 13x - 1213x + 12 74. 1a + h22

76. 17a - c22

78. - 3x 21x 2 - 421x 2 + 42 Use the following equation for Exercises 79–82. 1x  h23  1x  h21x  h22  1x  h21x 2  2xh  h22  1x  h2x 2  1x + h22xh  1x  h2h2  x 3  x 2h  2x 2h  2xh2  xh2  h3  x 3  3x 2h  3xh2  h3

79. 1a + b23

81. 1x - 523

80. 1a - b23

82. 12x + 323

Factor.

Divide. 35.

64. 1x - 421x - 32

2

69. 1a - 221a + 22

1

Express as an equivalent expression without negative exponents.

Multiply.

62. x11 + t2

t

42.

4

t

5

45.

t -7 ab1a 2b23 ab -1

Simplify.

49. 1t -223 55. 13x -2y -5z 42-4 57. 1 - 3x -8y 7z 222 59. a

2q 2

b

37.

3

x

ek

40.

ek e

k

e

3

t

43.

-9

46.

t -11

4

83. x - xt

84. x + xh

85. x 2 + 6xy + 9y 2

86. x 2 - 10xy + 25y 2

tk

87. x 2 - 2x - 15

88. x 2 + 8x + 15

tk

89. x 2 - x - 20

90. x 2 - 9x - 10

91. 49x 2 - t 2

92. 9x 2 - b 2

93. 36t 2 - 16m 2

94. 25y 2 - 9z 2

95. a 3b - 16ab 3

96. 2x 4 - 32

97. a 8 - b 8

98. 36y 2 + 12y - 35

x2 x

5

t t

6

-8

t -11 t -7

x 2y 31xy 322

99. 10a 2x - 40b 2x

x -3y 2

50. 1t -324

52. 1e x25

cd 3

48.

x7

51. 1e x24

53. 12x 2y 423

54. 12x 2y 425

56. 15x 3y -7z -52-3

58. 1- 5x 4y -5z -324 60. a

4x 2y a 3b 3

b

3

100. x 3y - 25xy 3

101. 2 - 32x 4

102. 2xy 2 - 50x

103. 9x 2 + 17x - 2

104. 6x 2 - 23x + 20

105. x 3 + 8 (Hint: See Exercise 66.) 106. a 3 - 27 (Hint: See Exercise 65.) 107. y 3 - 64t 3 109. 3x 3 - 6x 2 - x + 2 110. 5y 3 + 2y 2 - 10y - 4

108. m 3 + 1000p 3

Exercise Set A

111. x 3 - 5x 2 - 9x + 45

112. t 3 + 3t 2 - 25t - 75

615

APPLICATIONS Business and Economics

Solve. 113. - 7x + 10 = 5x - 11

114. - 8x + 9 = 4x - 70

115. 5x - 17 - 2x = 6x - 1 - x 116. 5x - 2 + 3x = 2x + 6 - 4x 117. x + 0.8x = 216

118. x + 0.5x = 210

119. x + 0.08x = 216

120. x + 0.05x = 210

122. 7x1x - 2212x + 32 = 0

3x + 1000.

123. x + 1 = 2x + 1

124. 2t = 9 + t

125. t 2 - 2t = t

126. 6x - x 2 = x

127. 6x - x 2 = - x

128. 2x - x 2 = - x

129. 9x 3 = x

130. 16x 3 = x

131. 1x - 322 = x 2 + 2x + 1

2

4x 20 100 + = 2 x x + 5 x + 5x

134.

x 3x + 5 2 + 2 = x + 1 x + 3 x + 4x + 3

137. 0 = 2x -

250 x2

139. 3 - x … 4x + 7

2

132. 1x - 522 = x 2 + x + 3

133.

4 50 50 135. = x x x - 2

158. Investment increase. An investment is made at 7%, compounded annually. It grows to $856 at the end of 1 yr. How much was invested originally? 159. Total revenue. Sunshine Products determines that the total revenue, in dollars, from the sale of x flowerpots is

121. 2x1x + 3215x - 42 = 0

2

157. Investment increase. An investment is made at 8 12 %, compounded annually. It grows to $705.25 at the end of 1 yr. How much was invested originally?

Determine the number of flowerpots that must be sold so that the total revenue will be more than $22,000. 160. Total revenue. Beeswax Inc. determines that the total revenue, in dollars, from the sale of x candles is 5x + 1000. Determine the number of candles that must be sold so that the total revenue will be more than $22,000.

Life and Physical Sciences 161. Weight gain. After a 6% gain in weight, an elk weighs 508.8 lb. What was its original weight?

60 2 60 136. + = x x x - 5 138. 5 -

35 x2

= 0

140. x + 6 … 5x - 6

162. Weight gain. After a 7% gain in weight, a deer weighs 363.8 lb. What was its original weight?

Social Sciences 163. Population increase. After a 2% increase, the population of a city is 826,200. What was the former population?

142. 3x - 3 + 3x 7 1 - 7x - 9

164. Population increase. After a 3% increase, the population of a city is 741,600. What was the former population?

143. - 7x 6 4

144. - 5x Ú 6

General Interest

145. 5x + 2x … - 21

146. 9x + 3x Ú - 24

147. 2x - 7 6 5x - 9

148. 10x - 3 Ú 13x - 8

149. 8x - 9 6 3x - 11

150. 11x - 2 Ú 15x - 7

165. Grade average. To get a B in a course, a student’s average must be greater than or equal to 80% (at least 80%) and less than 90%. On the first three tests, Claudia scores 78%, 90%, and 92%. Determine the scores on the fourth test that will guarantee her a B.

151. 8 6 3x + 2 6 14

152. 2 6 5x - 8 … 12

153. 3 … 4x - 3 … 19

154. 9 … 5x + 3 6 19

155. - 7 … 5x - 2 … 12

156. - 11 … 2x - 1 6 - 5

141. 5x - 5 + x 7 2 - 6x - 8

166. Grade average. To get a C in a course, a student’s average must be greater than or equal to 70% and less than 80%. On the first three tests, Horace scores 65%, 83%, and 82%. Determine the scores on the fourth test that will guarantee him a C.

APPEN DIX

B

Regression and Microsoft Excel Using Excel 2007

We can use Microsoft Excel to enter and plot data and to find lines of best fit using regression. Suppose we are given the following data:

OBJECTIVE • Use Microsoft Excel to perform regression

x

0

2

4

5

6

y

3

4.7

6

6.8

8

Step 1: We enter the data into two columns, as shown in Fig. 1.

FIGURE 2

FIGURE 1

Step 2: We highlight the columns of data (see Fig. 2). Then we go to the Insert tab and select Scatter. Next, we choose the first option, with the markers shown as distinct points. Step 3: A graph of the data points appears, as shown in Fig. 3. Under the Layout tab, we select Trendline and then choose More Trendline Options.

FIGURE 3

616

Regression and Microsoft Excel

617

Step 4: A box opens, with Linear preselected as the type of regression to be used; see Fig. 4. At the bottom, we check the box next to Display Equation on chart. As an option, we can also check the box next to Display R-squared value on chart.

FIGURE 4

FIGURE 5

Step 5: The line of best fit is now displayed on the graph, along with the R 2 value, as shown in Fig. 5. We can visually inspect how closely the line models the data. The R 2 value is called the squared correlation coefficient: an R 2 value close to 1 indicates that the line fits the data well, or, equivalently, that the data have a strong linear trend. In this book, regression is also used to fit exponential and polynomial functions to data. The student is invited to experiment with the various regression options in Excel.

Using Excel for Mac 2008 The steps for finding the line of best fit using Excel for Mac 2008 are given below. (Note: there are three toolbar levels in Excel for Mac, the primary toolbar along the top of the screen, the secondary set of options directly below the primary toolbar, and a tertiary toolbar that is connected to the spreadsheet cells directly, along the top.) Step 1: We enter the data into two columns. Step 2: Under the Charts option (tertiary toolbar), we select X Y (Scatter), then click on the first choice. A scatterplot appears on the screen. Step 3: To add a trendline, we click on one of the points on the scatterplot to “activate” them (they will appear as X shapes). Step 4: We go to the Chart option on the primary toolbar, and select Add Trendline from the drop-down menu. A Format Trendline window appears. We click on Options. Step 5: We check the boxes next to Display equation on chart and Display R-squared value on chart. These will appear on the scatterplot, along with the trendline. Other types of regression can be viewed by clicking on Type in the Format Trendline window.

EXERCISE SET

B 1. Use Excel to find the line of best fit for the following data.

2. Use Excel to find the line of best fit for the following data.

x

4

6

8

10

12

x

-5

-1

3

8

y

15

22

27

33

44

y

2

10

19

35

3. Use Excel to find a quadratic function (polynomial of power 2) that best fits the data in Exercise 2.

APPEN DIX

MathPrint Operating System for TI-84 and TI-84 Plus Silver Edition

OBJECTIVE • Upgrade a TI calculator with MathPrint

C

The graphing calculator screens in this text display math in the format of the TI MathPrint operating system. With MathPrint, the math looks more like that seen in a printed book. You can obtain MathPrint and install it by following the instructions given below. Only the TI-84 family of graphing calculators can be updated with the MathPrint operating system. If you own a TI-83 graphing calculator, you can use this brief appendix to help you “translate” what you see in the Classic mode shown on your calculator.

How to Get MathPrint Mode on a TI Graphing Calculator Before you upgrade your operating system, you must archive all items (programs, lists etc.) stored in random access memory (RAM); otherwise, they will be lost. Follow these steps to accomplish this task: 1. Press F and Mem. Select 2:Mem Mgmt/Del and then 1:All on the next screen. 2. Press [ with the cursor next to any item you wish to archive (it will be marked with an asterisk). 3. Upgrade the operating system (see method 1 or method 2 below). 4. After you upgrade the operating system, repeat steps 1 and 2 to move items out of the archive back into RAM. Unarchived items no longer have an asterisk next to them. There are several ways to upgrade a TI-84 calculator to the latest operating system, which includes MathPrint mode. Two of these methods are presented here.

Method 1: Using TI-Connect You can install the latest operating system by following these steps: 1. Launch TI-Connect software. 2. Connect your calculator to your computer using the Silver USB cable. 3. On a PC, click the Update button on the TI-Connect main menu page. On a Macintosh, doubleclick the TI Software Update button and wait for the software to recognize the device. Click C to reveal the contents of your calculator, scroll down and check the box next to TI-84 Plus family Operating System, and click the Update button at the top of the window. 4. Follow the prompts to upgrade the operating system. Make sure that your calculator’s batteries are fresh.

Method 2: Transfering the MathPrint Operating System from Another Graphing Calculator Press F and Mem and then [ on the sending calculator and check that the latest version of the operating system (2.53 MP or higher) is installed. Then follow these steps: 1. Connect the two graphing calculators with a unit-to-unit link cable. 2. On the receiving calculator, press F and LINK, followed by D and [. The screen will display Waiting....

618

619

MathPrint Operating System for TI-84 and TI-84 Plus Silver Edition

3. On the sending calculator, press F and LINK, scroll down and highlight G:SendOS, and then press [. 4. Follow the prompts to upgrade the operating system. You must also install version 1.1 of the application CatalogHelp on your graphing calculator. This application can be downloaded at no charge from education.ti.com and transferred from a computer to your calculator using TI-Connect software or from another calculator using the preceding steps.

Switching between MathPrint Mode and Classic Mode BACK MATHPRINT CLASSIC n/d On/d ANSWERS: AUTO DEC FRAC GO TO FORMAT GRAPH: NO YES STAT DIAGNOSTICS: OFF ON SET CLOCK 07/20/10 3:07 PM

A TI-84 graphing calculator with MathPrint can be switched from MathPrint to Classic mode by changing the mode settings. To do this, press the . key and scroll to the second page, shown at the left. Use the arrow keys to highlight MATHPRINT or CLASSIC and press [. A TI-84 graphing calculator loaded with the MathPrint operating system and running in Classic mode will show many of the MathPrint features. Below are two examples Feature

MathPrint

Improper fractions

2 5

–

MathPrint in Classic Mode

1 3

2/5 – 1/3 1/15

1 15

Logarithms

log 2 (32)

logBASE(32,2) 5

5

Translating between MathPrint Mode and Classic Mode The following table compares displays of several types in MathPrint mode and Classic mode (on a calculator without MathPrint installed). Feature Improper fractions

MathPrint 2 5

–

Classic (MathPrint not installed)

1 3

2/5–1/3 Frac 1/15

1 15

Enter an expression and press L and select 1:Frac. Mixed fractions

1

Not Supported

2

2 5 * (3 3 ) 121 15

Absolute values

10 – 15

abs(10–15) 5

Press I and F2 and select 1:abs(.

5

Press L and N and select 1:abs(. (continued)

620

APPENDIX C

Feature Summation

MathPrint

Classic (MathPrint not installed)

10

sum(seq(I 2 ,I,1,10)

Σ ( I2)

385

I=1

385

Press F and LIST and then N and select 5:seq( for seq and press F and LIST and then N twice and select 5:sum( for sum.

Press I and F2 and select 2: © (.

Numerical derivatives

d dX

(X2)

nDeriv(X 2 ,X,3) 6

X=3

6

Press L and select 8:nDeriv(.

Press I and F2 and select 3:nDeriv(. Numerical values of integrals

5 1

fnInt(X 2 ,X,1,5)

( X 2 ) dX

41.33333333 41.33333333

Press L and select 9:fnInt(.

Press I and F2 and select 4:fnInt(. Logarithms

log 2 (32)

5

Press I and F2 and select 5:logBASE(.

Evaluating logs with bases other than 10 or e cannot be done on a graphing calculator if the MathPrint operating system is not installed. To evaluate log 2 32, use the change-of-base formula: log(32)/log(2) 5

The &Editor MathPrint features can be accessed from the & editor as well as from the home screen. The following table shows examples that illustrate differences between MathPrint in the & editor and Classic mode. Feature Graphing the derivative of y = x 2

MathPrint Plot 1 Plot 2 Plot 3 Y1 =

d dX

( X2)

X=X

Y2 =

Graphing an antiderivative of y = x 2

Plot 1 Plot 2 Plot 3 Y1 = Y2 =

x 0

( X 2 ) dX

Classic Mode (MathPrint not installed) Plot 1 Plot 2 Plot 3 Y 1 = nDeriv(X 2 ,X,X) Y2 =

Plot 1 Plot 2 Plot 3 Y 1 = fnInt(X 2 ,X,0,X) Y2 =

TAB LE

A

Areas for a Standard Normal Distribution

Entries in the table represent area under the curve between z = 0 and a positive value of z. Because of the symmetry of the curve, area under the curve between z = 0 and a negative value of z are found in a similar manner. Area = Probability

z

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.0 0.1 2 e -x >2 dx 0.2 22p 0.3 0.4

.0000 .0398 .0793 .1179 .1554

.0040 .0438 .0832 .1217 .1591

.0080 .0478 .0871 .1255 .1628

.0120 .0517 .0910 .1293 .1664

.0160 .0557 .0948 .1331 .1700

.0199 .0596 .0987 .1368 .1736

.0239 .0636 .1026 .1406 .1772

.0279 .0675 .1064 .1443 .1808

.0319 .0714 .1103 .1480 .1844

.0359 .0753 .1141 .1517 .1879

0.5 0.6 0.7 0.8 0.9

.1915 .2257 .2580 .2881 .3159

.1950 .2291 .2611 .2910 .3186

.1985 .2324 .2642 .2939 .3212

.2019 .2357 .2673 .2967 .3238

.2054 .2389 .2704 .2995 .3264

.2088 .2422 .2734 .3023 .3289

.2123 .2454 .2764 .3051 .3315

.2157 .2486 .2794 .3078 .3340

.2190 .2517 .2823 .3106 .3365

.2224 .2549 .2852 .3133 .3389

1.0 1.1 1.2 1.3 1.4

.3413 .3643 .3849 .4032 .4192

.3438 .3665 .3869 .4049 .4207

.3461 .3686 .3888 .4066 .4222

.3485 .3708 .3907 .4082 .4236

.3508 .3729 .3925 .4099 .4251

.3531 .3749 .3944 .4115 .4265

.3554 .3770 .3962 .4131 .4279

.3577 .3790 .3980 .4147 .4292

.3599 .3810 .3997 .4162 .4306

.3621 .3830 .4015 .4177 .4319

1.5 1.6 1.7 1.8 1.9

.4332 .4452 .4554 .4641 .4713

.4345 .4463 .4564 .4649 .4719

.4357 .4474 .4573 .4656 .4726

.4370 .4484 .4582 .4664 .4732

.4382 .4495 .4591 .4671 .4738

.4394 .4505 .4599 .4678 .4744

.4406 .4515 .4608 .4686 .4750

.4418 .4525 .4616 .4693 .4756

.4429 .4535 .4625 .4699 .4761

.4441 .4545 .4633 .4706 .4767

2.0 2.1 2.2 2.3 2.4

.4772 .4821 .4861 .4893 .4918

.4778 .4826 .4864 .4896 .4920

.4783 .4830 .4868 .4898 .4922

.4788 .4834 .4871 .4901 .4925

.4793 .4838 .4875 .4904 .4927

.4798 .4842 .4878 .4906 .4929

.4803 .4846 .4881 .4909 .4931

.4808 .4850 .4884 .4911 .4932

.4812 .4854 .4887 .4913 .4934

.4817 .4857 .4890 .4916 .4936

2.5 2.6 2.7 2.8 2.9 3.0

.4938 .4953 .4965 .4974 .4981 .4987

.4940 .4955 .4966 .4975 .4982 .4987

.4941 .4956 .4967 .4976 .4982 .4987

.4943 .4957 .4968 .4977 .4983 .4988

.4945 .4959 .4969 .4977 .4984 .4988

.4946 .4960 .4970 .4978 .4984 .4989

.4948 .4961 .4971 .4979 .4985 .4989

.4949 .4962 .4972 .4979 .4985 .4989

.4951 .4963 .4973 .4980 .4986 .4990

.4952 .4964 .4974 .4981 .4986 .4990

= P10 … x … z2 =



z

0

1

0 z

621

Answers Chapter R

21.

Technology Connection, p. 6 1–20. Left to the student

−5 −4 −3 −2 −1−1

Exercise Set R.1, p. 10 y 5 4 3 2 1

1.

3. y=x+4

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

7.

– 5 – 4 –3 –2 –1–1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

–2 –3 –4 –5

11. 8y – 2x = 4

– 5 – 4 –3 –2 –1–1

1 2 3 4 5 x

y 5 4 3 2 1 −2 −3 −4 −5

15.

1 2 3 4 5 x

– 5 – 4 –3 –2 –1–1 –2 –3 –4 –5

– 5 – 4 –3 –2 –1–1

1 2 3 4 5 x

19.

1 2 3 4 5 x y= x

y 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2 −3

−200

200

–10

10

–15

−200

x = 4 + y2

47.

y1 = x − 4 and y2 = − x − 4 3

30

5x − 6y = 12

1 2 3 4 5 x

y = –9.6 x – 100 4.2

–25

25 –4

8

–30 –3 x = 2 – y2

Technology Connection, p. 16 1. 951; 42,701

1 2 3 4 5 x

2. 21.813

Technology Connection, p. 17 1. 6; 3.99; 150; - 1.5, or - 14 2. - 21.3; - 18.39; - 117.3; 9 3. - 75; - 65.466; - 420.6; 1.68, or 76 3.25, or 293 90 45

Technology Connection, p. 20 y = 7 − x2

1–3. Left to the student

Exercise Set R.2, p. 21 1 2 3 4 5 x

1. Yes 3. Yes 5. Yes 7. Yes 13. Yes 15. No 17. Yes 19. (a) x 5.1 5.01 f 1x2

A

15

200

45.

–2 –3 –4 –5

y = x2 − 5

y 5 4 3 2 1

y 5 4 3 2 1

1 2 3 4 5 x

23. 3.98 min (1954), 3.66 min (2008), 3.64 min (2012) 25. About 27.25 mi/hr, or mph 27. (a) 1.8 million, 3.7 million, 4.4 million, 4.5 million; (b) 44 and 70; (c) about 58; (d)

29. (a) $102,800.00; (b) $102,819.60; (c) $102,829.54; (d) $102,839.46; (e) $102,839.56 31. (a) $31,200.00; (b) $31,212.00; (c) $31,218.12; (d) $31,224.25; (e) $31,224.32 33. $550.86 35. $97,881.97 37. (a) 1996–2000, 2002; (b) 1987, 1990; (c) 1999; (d) 1987, 1990 39. (a) $206,780.16; (b) $42,000; $164,780.16 y = x − 150 41. 43. y = x 3 + 2x 2 − 4x − 13

x+y=5

−2 −3 −4 −5

−5 −4 −3 −2 −1−1

17.

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

–2 –3 –4 –5

13.

y = –3x

−2 −3 −4 −5

y=2 –x – 4 3

y 5 4 3 2 1

y 5 4 3 2 1

y + 1 = x3

−2 −3 −4 −5

–2 –3 –4 –5

y 5 4 3 2 1

9.

y 5 4 3 2 1 – 5 – 4 –3 –2 –1–1

−2 −3 −4 −5

5.

y 5 4 3 2 1

17.4

17.04

9. Yes

11. Yes

5.001

5

17.004

17

indicates that the exercise asks for a written interpretation or explanation; answers will vary.

A-1

A-2

ANSWERS

(b) f 142 = 13, f 132 = 9, f 1- 22 = - 11, f 1k2 = 4k - 3, f 11 + t2 = 4t + 1, f 1x + h2 = 4x + 4h - 3 21. g1- 12 = - 2, g102 = - 3, g112 = - 2, g152 = 22, g1u2 = u2 - 3, g1a + h2 = a 2 + 2ah + h2 - 3, and g1a + h2 - g1a2

1 , f 1- 32 h 49 1 1 1 is undefined, f 102 = , f 1a2 = , f 1t + 42 = , 9 1a + 322 1t + 722 f 1x + h2 - f 1x2 1 f 1x + h2 = , and = 2 h 1x + h + 32 - 2x - h - 6 , h Z 0 (b) Take an input, square it, add 1x + h + 3221x + 322 six times the input, add 9, and then take the reciprocal of the result. y y 25. 27. f(x) = 2x − 5 = 2a + h, h Z 0

−5 −4 −3 −2 −1−1

4 3 2 1

–2 –1–1

1 2 3 4 5 x

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

33.

f(x) = x2 − 2

−5 −4 −3 −2 −1−1

f(x) = 6 − x2

g(x) = x 3

35. Yes 41. No 47. (a)

37. Yes 43. Yes y 5 4 3 2 1

4

1 2 3 4 5 x

1

−5 −4 −3 −2 −1−1

1 2 x

39. No 45. Yes (b) No

x=

y2

−2

1. Domain = ; range = 3- 4, q 2 2. Domain = ; range =  3. Domain = 5x | x is a real number and x Z 06; range = 5x | x is a real number and x Z 06 4. Domain = ; range = 3- 8, q 2 5. Domain = 3- 4, q 2; range = 30, q 2 6. Domain = 3- 3, 34; range = 30, 34 7. Domain = 3- 3, 34; range = 3- 3, 04 8. Domain = ; range =  3. 10, 52 5. 3- 9, - 42 11. 3- 2, 24

15. 1- q , - 24 17. 1- 2, 34

f 1x + h2 - f 1x2

5 4 3 2 1

– 5 – 4 –3 –2 –1–1

9 8 7 6 5 4 3 2 1

y = f(x)

1 2 3 4 5 x

–2 –3 –4 –5 y 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2 −3

−5 −4 −3 −2 −1−1

61. y = f(x)

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

2. Answers may vary.

Y1 .2 .4444 ERR: −.5714 −.3333 −.2667 −.25

13. 3- 4, - 12

= 2x + h - 3, h Z 0 h 51. f 1- 12 = 3, f 112 = - 2 53. f 102 = 17, f 1102 = 6 y y 55. 57.

59.

X −3 −2.5 −2 −1.5 −1 −.5 0 X = −3

1. 3- 2, 44 9. 1p, q 2

–8

49.

.6 ERR: −1 −.75 −1 ERR: .6

−3 −2 −1 0 1 2 3 X = −3

Exercise Set R.3, p. 31 1 2 3 4 5 x

−2 −3 −4 −5

–4

x + 5 ; this is not a function. A 2 3 73. y = 2 # 2x; this is a function. 75. 77. 79. Left to the student X Y1

Technology Connection, p. 29

−2 −3

y 8

–2 –1–1

y 7 6 5 4 3 2 1

1 2 3 4 5 x

−2 −3 −4 −5

y = f(x)

71. y = ;

1.

g(x) = –4x

31.

1 2 3 4 5 x

−2 −3 −4 −5 −6 −7

Technology Connection, p. 28

1 2 x

–2 –3 –4

−2 −3 −4 −5

29.

−5 −4 −3 −2 −1−1

23. (a) f 142 =

5 4 3 2 1

65. $563.25 67. (a) 1.818 m2; (b) 2.173 m2; (c) 1.537 m2 69. (a) Yes; a unique “scale of impact” number is assigned to each event. (b) The inputs are the events; the outputs are the scale of impact numbers.

y 3 2 1

63.

−2 −3 −4 −5

y = f(x)

1 2 3 4 5 x

y = g(x) 1 2 3 4 5 x

7. 3x, x + h4

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 0 1 2 3 4 5

−5 −4 −3 −2 −1 0 1 2 3 4 5

19. 1- q , 12.52

−10

−5

0

10 12.5 15

21. (a) 3;

(b) 5- 3, - 1, 1, 3, 56; (c) 3; (d) 5- 2, 0, 2, 3, 46 23. (a) 4; (b) 5- 5, - 3, 1, 2, 3, 4, 56; (c) 5- 5, - 3, 46; (d) 5- 3, 2, 4, 56 25. (a) - 1; (b) 3- 2, 44; (c) 3; (d) 3- 3, 34 27. (a) - 2; (b) 3- 4, 24; (c) - 2; (d) 3- 3, 34 29. (a) 3; (b) 3- 3, 34; (c) about - 1.4 and 1.4; (d) 3- 5, 44 31. (a) 1; (b) 3- 5, 52; (c) 33, 52; (d) 5- 2, - 1, 0, 1, 26 33. 5x ƒ x is a real number and x Z 26 35. 5x ƒ x Ú 06 37.  39. 5x ƒ x is a real number and x Z 26 41.  43. 5x ƒ x is a real number and x Z 3.56 45. E x ƒ x Ú - 45 F 47.  49. 5x ƒ x is a real number and x Z 5, x Z - 56 51.  53. 5x ƒ x is a real number and x Z 5, 0.08 2t 55. 3- 1, 24 57. (a) A1t2 = 5000a1 + x Z 16 b ; 2 (b) 5t ƒ t Ú 06 59. (a) 30, 84.74; (b) 30, 4,600,0004; (c) 61. (a) 30, 704; (b) 38, 754 63. 65. 67. 1- q , 02 ´ 10, q 2; 30, q 2; ; 31, q 2; 

ANSWERS

Technology Connection, p. 34 1. The line will slant up from left to right, will intersect the y-axis at 10, 12, and will be steeper than y = 10x + 1. 2. The line will slant up from left to right, will pass through the origin, and 2 will be less steep than y = x. 3. The line will slant down 31 from left to right, will pass through the origin, and will be steeper than y = - 10x 4. The line will slant down from left to right, will intersect the y-axis at 10, - 12, and will be less steep than 5 y = - 32 x - 1.

Technology Connection, p. 37 1. The graph of y2 is a shift 3 units up of the graph of y1, and y2 has y-intercept 10, 32. The graph of y3 is a shift 4 units down of the graph of y1, and y3 has y-intercept 10, - 42. The graph of y = x - 5 is a shift 5 units down of the graph of y = x, and y = x - 5 has y-intercept 10, - 52. All lines are parallel. 2. For any x-value, the y2-value is 3 more than the y1-value and the y3-value is 4 less than the y1-value.

Exercise Set R.4, p. 45 y 5 4 3 2 1

1.

3. x=3

−5 −4 −3 −2 −1−1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

−2 −3 −4 −5

y 5 4 x = −4.5 3 2 1

5.

7.

−5 −4 −3 −2 −1−1

11. m = −2 y-intercept: (0, 0)

−5 −4 −3 −2 −1−1

y = 3x − 4

−2 −3 −4 −5

y 5 4 3 2 1

−5 −4 −3 −2 −1−1 −2 −3 −4 −5

17.

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

13.

1 2 3 4 5 x y = −2

y 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2

15.

1 2 3 4 5 x m=3 y-intercept: (0, −4) y=7

m=0 y-intercept: (0, 7)

1 2 3 4 5 x

y 5 4 3 g(x) = −x + 3 2 1 −5 −4 −3 −2 −1−1 −2 −3 −4 −5

31. y - 0 = 21x - 32, or y = 2x - 6 33. y + 6 = 121x - 02, or y = 12 x - 6 35. y - 3 = 0 # 1x - 22, or y = 3 37. - 47 1 39. 3 41. Undefined slope 43. - 34 45. 0 47. 3 3 49. 2 51. y = - 47 x - 17 53. y = 13 x - 11 55. x = 3 3 91 2 57. y = - 34 59. 61. 63. 3.5% x + y = 3 3 15 5 65. About $623.75 per year 67. $90.25 per year 69. (a) R = 4.17T; (b) R L 25 71. (a) M = 0.4W; (b) 0.4 = 40%, so we have M = 40%W. The weight of the muscles is 40% of the body weight. (c) M = 48 lb 73. (a) T1w2 = 0.000806w; (b) $2.98 y 75. (a) C1x2 = 20x + 100,000; $200,000 (b) R1x2 = 45x; (c) P1x2 = R1x2 - C1x2 = C $100,000 25x - 100,000; P (d) a profit of $3,650,000; R (e) 4000 calculators 2000 4000 x −$100,000

77. (a) V1t2 = 5200 - 512.5t; (b) V102 = $5200, V112 = $4687.50, V122 = $4175, V132 = $3662.50, V142 = $3150, V172 = $1612.50, V182 = $1100 79. About $743,590 81. (a) The number - 700 indicates that the value of the photocopier decreases by $700 per year, and the number 3500 indicates that the original value of the copier was $3500. (b) 5 years; (c) 83. (a) D152 = 6 ft, (b) D(r) (c) D1102 = 11.5 ft, 30 D1202 = 22.5 ft, 11r + 5 20 D1502 = 55.5 ft, D(r) = 10 D1652 = 72 ft 10

y = 3.75

1 2 3 4 5 x

−2 −3 −4 −5

y 5 4 3 2 1

y = −2x

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

9.

y 5 4 3 2 1

A-3

10

85. (a) Approximately y =

20 9 x

20

30

r

- 4372.44; (b) approximately

87. (a) N = 1.02P; (b) N = 204,000 94.7%; (c) 2012; (d) people; (c) P = 360,000 people 89. 91. (a) Graph III; (b) graph IV; (c) graph I; (d) graph II f(x) = 0.5x

93. Answers may vary.

Technology Connection, p. 51 1–2. Left to the student

1 2 3 4 5 x

Technology Connection, p. 53

m = 0.5 y-intercept: (0, 0)

1. (a) 2 and 4; (b) 2 and - 5; (c) and (d) left to the student

Technology Connection, p. 54 m = −1 y-intercept: (0, 3)

1 2 3 4 5 x

19. m = 3, y-intercept: 10, 62 21. m = - 2, y-intercept: 10, 32 23. m = - 1, y-intercept: 10, - 42 25. m = 13, y-intercept: A 0, - 73 B 27. y + 3 = - 51x + 22, or y = - 5x - 13 29. y - 3 = - 21x - 22, or y = - 2x + 7

1. - 5 and 2 2. - 4 and 6 3. - 2 and 1 4. 0, - 1.414, and 1.414 (approx.) 5. 0 and 700 6. - 2.079, 0.463, and 3.116 (approx.) 7. - 3.096, - 0.646, 0.646, and 3.096 (approx.) 8. - 1 and 1 9. - 0.387 and 1.721 10. 6.133 11. - 2, - 1.414, 1, and 1.414 12. - 3, - 1, 2, and 3

Technology Connection, p. 55 Left to the student

Technology Connection, p. 58 1. Left to the student

2. For x = 3, y1 is ERR and y2 is 6.

Technology Connection, p. 59 1. 1–8. Left to the student

A-4

ANSWERS

33.

Technology Connection, p. 60 1–2. Left to the student

Technology Connection, p. 64

y 5 4 x2 + 5x + 6 3 f(x) = 2 x+3 1 −7 −6 −5 −4 −3 −2 −1−1

1. Approximately (14, 266) y 2

y y = x2 7 6 5 4 3 2 y = x2 − 1 1

3.

1 y = – x2 2

1 –2

–1

1

2

x

–1 –2

y 2 1

5. −2

9.

−5 −4 −3 −2 −1−1

y 10 8 6 4 y = x3 + 2 2 −2

y 10 8 6 4 y = |x| 2

7. y = −2x2 + 1

−1 −1 1 2 −2 −3 −4 −5 −6 −7 y = −2x2 −8

x

−10−8 −6 −4 −2−2

y 3

y = x3 1

2

2

y= x

1

y= x−1 1

1

2 1 1 2 3 4 x y = x 2 – 4x + 3 y 5 4 3 2 1

23.

−2 −1−1

1 2 3 4 5 6 7 8 x

−2 −3 −4 −5

25.

–1 –1 –2 –3 –4

–4–3 –2 –1 –1 –2 –3 –4

1 y=— x2 –2 –1

y = – x 2 + 2x – 1 1 2 3 4 x

80,000 x P 150,000

P = 1000 t5/4 + 14,000

120,000

30,000 10

1 2 x

y = 2.7x + 63.8

1. (a) 2 y = – –x

(b) 93.5; (c)

96 x

1 2 3 4

y 2

3

y = 

x

–2

1 2 3 4 x

20

30

40

t

101. 16 cities; 30 cities 103. 105. - 1.831, - 0.856, 3.188 107. 1.489, 5.673 109. - 2, 3 111. 3- 1, 24 113. Approximately (75.11, 7893); produce 7893 units at a price of $75.11 each.

Technology Connection, p. 72

y 4 3 2 1

1

x

(b)

60,000

y 7 g(x) = −3x2 − 4x + 5 6 5 4 3 2 1

–4 –3 –2 –1 –1 1 2

40,000

99. (a) 99,130 particles>cm3, 108,347 particles>cm3, 127,322 particles>cm3

90,000

1 –2 –3 –4

31.

1 2 3 4 5 x

−2 −3 −4 −5

140 120 100 80 60 40 20

5 x

15. Not a parabola

– 4 – 3 – 2––1

1 2 3 4 x

4 3 2 1

4

−2 −3

27.

y

29.

3

−8 −7 −6 −5 −4 −3 −2 −1−1

f(x) = 2x2 − 6x + 1

y 4 3 2 y = –x 2 1

2

−5 −4 −3 −2 −1−1

1 2 3 x

37. 1 ; 23, or - 0.732, 2.732 39. - 3 ; 210, or - 6.162, 1 ; 22 - 4 ; 210 0.162 41. , or - 0.207, 1.207 43. , or 2 3 - 7 ; 213 , or - 5.303, - 1.697 - 2.387, - 0.279 45. 2 47. x 3>2 49. a 3>5 51. t 1>7 53. x 3 55. t -5>2 1 1 57. 1x 2 + 72-1>2 59. 2 65. 5 x 61. 2 3 y 2 63. 2 2 5 t 2 3 b 1 1 1 67. 69. 71. 73. 27 75. 16 2 6 e 17 2x 2 - 3 2 3 t2 77. 8 79. 5x ƒ x Z 56 81. 5x ƒ x Z 2, x Z 36 83. E x ƒ x Ú - 45 F 85. 5x ƒ x … 76 87. (50, 500); x = $50, 89. 15, 12; price is $500, and quantity is 1000. q = 500 items 91. 11, 42; price is $1, and quantity is 400. 93. 12, 32; price is $2000, and quantity is 3000. 95. $140.90 per share 97. (a) 166 mi, (b) R(x) R(x) = 11.74 x0.25 200 176 mi, 180 160 184 mi

x

13. Parabola with vertex at 1- 2, - 112 17. 3 y 19. y

21.

y = |x − 3| 2 4 6 8 10 x

−4 −6 −8 −10

11.

−1 −2 −4 −6 −8 −10

–1–1 –2

1 2 3 4 5 x

−2 −3

y = – 1– x 2 2

y 5 4 x2 − 1 3 f(x) = 2 x−1 1

−2 −3 −4 −5

Exercise Set R.5, p. 65 1.

35.

6

11 80

ANSWERS

Technology Connection, p. 74 1. (a)

−0.00005368295x 4

0.037566680x 3

+ y= − 3.4791715x 2 + 105.81080x − 916.68952

(b)

130

(b) 18, 30; (c) [15, 45]; this interval covers typical human child-bearing ages. 16. $1340.24 17. $5017.60 18. Not a function. One input, Richard, has three outputs. 19. (a) f 132 = - 6; (b) f 1- 52 = - 30; (c) f 1a2 = - a 2 + a; (d) f 1x + h2 = - x 2 - 2xh - h2 + x + h y y 20. 21. y = | x + 1|

10

45

62500

y 10 8 6 4 2

22.

75

−10−8 −6 −4 −2−2

0

62500

75 0

(c)

1 2 3 4 5 x

y = −0.0771x 4 + 11.3952x 3 − 639.2276x 2 + 17037.1915x − 135483.9938

9 8 7 6 5 4 3 2 1

−5 −4 −3 −2 −1−1 f(x) =

x2 − 16 x+4

y 5 4 3 2 1

23.

1 2 3 4 5 x

g(x) = x + 1

−1−1 −2 −3 −4 −5

2 4 6 8 10 x

−4 −6 −8 −10

y = −1.6519x 3 + 145.6606x 2 − 2658.3088x + 36491.7730

10

f(x) = (x − 2)2

−2 −3 −4 −5

2. (a) y = −62.8327x 2 + 5417.8404x − 57264.7856

(b)

5 4 3 2 1

−5 −4 −3 −2 −1−1

0

10

A-5

1 2 3 4 5 6 7 8 9 x

24. Not a function 25. Function 26. Function 27. Not a function 28. (a) f 122 = 1; (b) 3- 4, 44; (c) x = - 3; (d) 3- 1, 34 y 29. (a) f 1- 12 = 1, (b) 5 f 11.52 = 4, 4 3 f 162 = 3 y = f(x) 2 1

62500

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

10

75 0

(d) quartic function; (e) $38,853, $58,887 3. y = 93.2857x 2 − 1336x + 5460.8286 1400

5 200

9

30. (a) 3- 2, 54; (b) 1- 1, 34; (c) 1- q , a2 31. (a) 3- 4, 52; (b) 12, q 2;

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5 −4 −3 −2 −1 0 1 2 3 4 5

(b) 5- 3, - 2, - 1, 0, 1, 2, 36; (c) - 1, 3; (d) 5- 2, 1, 2, 3, 46 33. (a) 1- q , 52 ´ 15, q 2; (b) 3- 6, q 2 34. Slope, - 3; y-intercept, 2 35. y + 5 = 141x - 82, or y = 14 x - 7 36. - 3 7 37. - $350 per year 38. 75 pages per day 39. A = V 200 40. (a) C1x2 = 0.50x + 4000; (b) R1x2 = 10x; y (c) P1x2 = 9.5x - 4000 (d) 422 CDs R $8000

Exercise Set R.6, p. 76

6000

1. Linear 3. Quadratic, a 6 0 5. Linear 7. Polynomial, neither linear nor quadratic 9. Linear 11. (a) y = 29 x + 3.6; (b) $6.3 million; $8.0 million; (c) 2024

4000

13. (a) y = 0.144x 2 - 4.63x + 60; (b) 188.5 ft; (c) 15. Answers will vary. 17. 19. 21. (a) y = - 0.224x + 6.5414; (b) 2.51% (c) The regression answer seems more plausible; it uses all the data. (d) y = - 0.009856x 3 + 0.1993x 2 - 1.3563x + 8.103; - 9.217%. (e)

−2000

Chapter Review Exercises, p. 85 1. (d) 2. (b) 3. (f) 4. (a) 5. (e) 6. (c) 7. True 8. False 9. True 10. True 11. False 12. False 13. True 14. True 15. (a) About 56 per 1000 women;

32. (a) f 1- 32 = - 2;

C P

2000 200 400 600

x

−4000

41. (a)

y 5 4 3 2 1 −1−1 −2 −3 −4 −5

y 5 4 3 2 1

(b) y=

x

1 2 3 4 5 6 7 8 9 x

−5 −4 −3 −2 −1−1

y= x−3 y = x3

−2 −3 −4 −5

y = (x − 1)3 1 2 3 4 5 x

A-6

ANSWERS y 9 f(x) = x2 − 6x + 8 8 7 6 5 4 3 2 1

42. (a)

−2 −1−1

y 5 4 3 2 1

(b)

y=−

1 x

y 5 4 3 2 1 −2 −3 −4 −5

y3

y1 60 0

y=

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

y4

0

y 5 4 3 2 1

(d)

(c)

y2

2 4 6 8 10 x

−2 −3 −4 −5

1 2 3 4 5 6 7 8 x

−5 −4 −3 −2 −1−1

2400

−10−8 −6 −4 −2−1

Vertex at 13, - 12 (c)

y1 = 37.58x + 294.48 y2 = −0.59x 2 + 74.61x − 117.72 y3 = 0.02x 3 − 2.60x 2 + 125.71x − 439.65 y4 = 0.0028x 4 − 0.324x 3 + 11.46x 2 − 88.51x + 507.84

(b) 3

g(x) = x + 2

Chapter R Test, p. 88

x2 + x − 6 x−2

1 2 3 4 5 x

−2 −3 −4 −5

43. (a) x = 1, x = 3; (b) x =

2 ; 210 2

(c) m -2>3; (d) 1x 2 - 92-1>2

45. (a) 2 5 x 2 (b)

44. (a) x 4>5; (b) t 4; 1

; 2 5 m3 (c) 2x 2 - 5; (d) 2 47. 13, 162; price = $3, 3 t 46. C 92, q B quantity = 1600 units 48. About 3.3 hr 49. (a) y = 0.2x + 160 (b) M 177 M = 0.2r + 160 (c) 173.4 beats> min 176 175 174 173 172 171 170

1. [R.1] $750 2. [R.2] (a) f 1- 32 = - 4; (b) f 1a + h2 = - a 2 - 2ah - h2 + 5 3. [R.4] Slope, 45; y-intercept, - 23 4. [R.4] y - 7 = 141x + 32, or y = 14 x + 31 4 1 5. [R.4] - 2 6. [R.4] - $700>yr 7. [R.4] 12 lb> bag 8. [R.4] F = 23W 9. [R.4] (a) C1x2 = 0.08x + 8000; (b) R1x2 = 0.50x; (c) P1x2 = 0.42x - 8000; (d) 19,048 cards 10. [R.5] 13, 252; x = $3, q = 25 thousand units 11. [R.2] Yes 12. [R.2] No 13. [R.3] (a) f 112 = - 4; (b)  (c) x = ;3; (d) 3- 5, q 2 y 14. [R.5] 15. [R.5] t -1>2 12 1 16. [R.5] 8 2 5 t3 f(x) = 8 x 4

–12

–8

–4

4

8

12 x

–4 –8 –12 50

50. (a)

60

70

80

r

(b) Data fit a quadratic function. (c) y = 2.0x 2 - 89.84x + 870; (d) About - $25.33; (e)

P 1000 800

y

17. [R.5]

2–1 f(x) = x——– x +1

–1 600 400 −200

x

–2 10

20

30

40

18. [R.5] 1- q , - 72 ´ 1- 7, 22 ´ 12, q 2 20. [R.3] 3c, d2 21. [R.2] f(x) =  x 2 + 2, for x  0 22. [R.6] (a)

x

51. (a) 525,375 lb; (b) $4.31> lb 52.

1 –1

53.

f(x) = x 3 − 9x 2 + 27x + 50 400

−10

y=

3

x 2 – 2, for x  0

|4 − x 2| + 1 6

−10 −400

10 −1

Zero: none; domain: ; Zero: x = - 1.25; domain: ; range:  range: 31, q 2 54. 1- 1.21, 2.362 55. (a) y = 0.2x + 160; (b) 173.4 beats> min; (c)

56. (a) y = 1.86x 2 - 84.18x +

57. (a) y = 37.58x + 294.48; 943.86; (b) $95.46; (c) y = - 0.59x 2 + 74.61x - 117.72; y = 0.02x 3 - 2.60x 2 + 125.71x - 439.65; y = 0.003x 4 - 0.324x 3 + 11.46x 2 88.51x + 507.84

–3 –2 –1 –1 –2 –3

y

2600 2500 2400 2300 2200 2100 2000 1900 1800 1700

5 4 3 2 1

10

19. [R.5] 1- 2, q 2

1 2 3

10

20

30

40

x

(b) yes; (c) y = - 1.94x 2 + 102.74x + 1253.49; (d) 2589.9 1 calories; (e) 23. [R.5] 16 24. [R.5] Domain: A - q , 53 D ; 2 zero: x = - 798 3 25. [R.5] Answers will vary. One possibility is 1x + 321x - 121x - 42 = 0. 26. [R.4] 51 7 27. [R.5] Zeros: ; 28 L ;2.828,

y=

; 210 L ;3.162; domain: ; range: 3- 1, q 2

3

|9 − x 2| − 1 12

−10

10

−12

A-7

ANSWERS

28. [R.6] (a) y = - 1.51x 2 + 79.98x + 1436.93; (b) 2480.4 calories; (c)

Chapter 1 Technology Connection, p. 96

Extended Technology Application, p. 90 1. (a) y = 0.1205093525x - 0.4578553957 (b) 10 (c) $7.01, $7.98. Not reasonable estimates. The data are increasing, but these prices seem lower than the trend of the curve indicates. (d) 2120; seems too far in the 0 70 future. 0

2. (a) y = 0.0014052879x + 0.0323240014x + 0.3756147625 (b) 10 (c) $7.78, $9.52. Yes, the curve seems to better follow the trend in the data. (d) 2057; seems more reasonable. 2

0

1. 5 2. - 4 3. g1x2 = 327, 456.95, 475.24, 492.1, 493.81, 5. - 1 494.19, 495.9, 513.24, 573.9, 685.17 4. 494

Exercise Set 1.1, p. 106

1. 11 3. - 2 5. The limit, as x approaches 4, of f 1x2 7. The limit, as x approaches 5 from the left, of F1x2 9. lim+ 11. 2 13. - 3 15. Does not exist 17. 3 x:2

19. 31. 43. 55. 63.

70

0

f(x) = | x |

−2 −3 −4 −5

x:0

x : -2 y 5 4 3 2 1

x : -1

lim f 1x2 = 2

67.

−2 −1−1

F(x) =

1 x−3

1 2 3 4 5 6 7 8 x

x: q

x:4

x:0 y 9 8 7 6 5 4 3 2 1

−7 −6 −5 −4 −3 −2 −1−1

y 7 6 5 4 3 2 1 −2 −1−1

1 2 3 4 5 x

x:1

lim F1x2 does not exist

x:1

y 4 3 2

g

G 1

1 2 3 4 5 6 7 8 x

−2

−1

1 −1

lim g1x2 = 1; lim+ g1x2 = 0; x:3

lim g1x2 does not exist

x:3

F

lim F1x2 = 3; lim+ F1x2 = 1;

x : 1-

77.

−2 −3

x : 3-

1 −2 x

−2 −3 −4 −5

1 2 3 x

lim g1x2 does not exist

75.

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

lim g1x2 = 4;

f(x) =

lim f 1x2 does not exist

73.

x: q

1 2 3 4 5 x

lim f 1x2 = - 2;

x:3

x : -2

70

−2 −3 −4 −5 −6 −7

lim F1x2 = 1

0 y2, y3 y1

−5 −4 −3 −2 −1−1

lim F1x2 does not exist;

70 y4

y 3 2 1

69.

1 +4 x+2

g(x) = x2 − 5

lim g1x2 = - 4

−2 −3 −4 −5

g(x) =

1 2 3 4 5 x

lim g1x2 = - 5;

x:0

71.

(c) Look at the leading coefficient 10 of the cubic function. Note that it is virtually 0, so the term can be deleted, making a quadratic. Plus, the graphs of the quadratic and cubic functions are virtually identical. 0 0 Using a higher-order polynomial such as a quartic allows results to show very erratic increases, and since the quadratic does just as good a job as the cubic, the researcher might reject the quartic.

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

lim f 1x2 = 0;

5. (a) If the scatterplot and the linear function are graphed on the same axes, the predicted prices seem lower than the trend of the curve indicates. 10 (b) Graphing the scatterplot, the y2 linear function, and the quadratic y1 function, we see that the quadratic function seems to better follow the trend of the data. 0

5 4 3 2 1

−2 −3 −4 −5

70 0

5 4 3 2 1

−5 −4 −3 −2 −1−1

0

3. (a) y = 0.000000758225x 3 + 0.0013427851x 2 + 0.0337649942x + 0.3698599855 (b) 10 (c) $7.79, $9.55. Yes, the curve seems to follow the trend in the data, but the resulting ticket prices are virtually the same as those found with the quadratic 0 70 function; (d) 2056; seems 0 reasonable. 4. (a) y = 0.0000015883525x 4 - 0.0001885529x 3 + 0.0084799832x 2 - 0.0537491459x + 0.539914429 y4 (b) 10 (c) $8.33, $11.79. Yes, but the estimates are higher than those found using the quadratic or cubic function. (d) 2031; seems too soon.

4 21. - 1 23. - 1 25. 0 27. 5 29. Does not exist 2 33. 2 35. 1 37. 4 39. Does not exist 41. 0 0 45. 1 47. 4 49. Does not exist 51. 1 53. 1 Does not exist 57. 0 59. 3 61. 1 y y 65.

lim G1x2 = 1

x : -1

2

x

A-8

ANSWERS y 5 4 3 2 1

79.

−5 −4 −3 −2 −1−1

lim H1x2 does not exist;

x:0

lim H1x2 = 2

x:1 H

1 2 3 4 5 x

−2 −3 −4 −5

81. $3.30; $3.30; $3.30 83. $3.70; $4.10; limit does not exist 85. $1.05; $1.22; limit does not exist 87. Limit does not exist. 89. 1100 deer; 1200 deer; limit does not exist 91. 93. 35 bears; 34 bears; limit does not exist 95. 3 97. - 1 99. Limit does not exist; 2 101. Limit does not exist; limit does not exist.

Technology Connection, p. 113 1. 53

2. 2.8284

3. 0.25

4. 0.16667

Exercise Set 1.2, p. 117 1. True 3. False 5. True 7. False 9. 5 11. - 3 13. 4 15. 15 17. 1 19. 6 21. 72 23. 13 4 1 25. 3 27. 10 29. Limit does not exist. 31. 27 33. Limit does not exist. 35. 0 37. Not continuous 39. Not continuous 41. Not continuous 43. (a) - 2, - 2, - 2; (b) - 2 (c) yes, lim g1x2 = g112; (d) limit does not exist; (e) - 3 x:1

(f ) no, the limit does not exist 45. (a) 2; (b) 2; (c) yes, lim h1x2 = h112; (d) 0; (e) 0; x:1

(f) yes, lim h1x2 = h1 - 22

47. (a) 3; (b) 1; (c) limit does

x : -2

about 2.6% per yr; about 3.25% per yr 23. About 0.97% per yr; about 3% per yr; about 1.9% per yr 25. 1.045 quadrillion BTUs> yr; 0.638 quadrillion BTUs> yr; - 0.337 quadrillion BTUs> yr; 27. (a) 70 pleasure units> unit of product, 39 pleasure units/unit of product, 29 pleasure units> unit of product, 23 pleasure units> unit of product; (b) 29. (a) $11.35; (b) $26.82; (c) $15.47; (d) $1.19, the average price of a ticket increases by $1.19> yr. 31. $909.72 is the annual increase in the debt from the 2nd to the 3rd year. 33. $19.95 is the cost to produce the 301st unit. 35. (a) 1.0 lb> month; (b) 0.54 lb> month; (c) 0.77 lb> month; (d) 0.67 lb> month; (e) growth rate is greatest in the first 3 months. 37. (a) Approximately 1.49 hectares> g; (b) 1.09 represents the average growth rate, in hectares> g, of home range with respect to body weight when the mammal grows from 200 to 300 g 39. (a) 1.25 words> min, 1.25 words> min, 0.625 words> min, 0 words> min, 0 words> min; (b) 41. (a) 256 ft; (b) 128 ft> sec 43. (a) 125 million people> yr for both countries; (b) (c) A: 290 million people> yr, - 40 million people> yr, - 50 million people> yr, 300 million people> yr, B: 125 million people> yr in all intervals; (d) 45. (a) 1985–86; (b) 1975–76, 2003–04, and 2004–05; (c) about $2472 for public and about $5356 for private 47. 2ax + b + ah 49. 4x 3 + 6x 2h + 4xh2 + h3 51. 5ax 4 + 10ax 3h + 10ax 2h2 + 5axh3 + ah4 + 4bx 3 + 6bx 2h + 1 53. 4bxh2 + bh3 11 - x - h211 - x2 2 55. 22x + 2h + 1 + 22x + 1

not exist; (d) 1; (e) no, lim G1x2 does not exist; (f ) yes,

Technology Connection, p. 138

lim G1x2 = G102; (g) yes, lim G1x2 = G12.92

1. f ¿1x2 = -

x:3

x:0

49. Yes; lim f 1x2 = f 152 x:5

G102 does not exist.

x : 2.9

51. No; lim G1x2 does not exist and x:0

53. Yes; lim g1x2 = g132

55. No; lim F1x2 does not exist. x:3

x:3

57. Yes; lim f 1x2 = f 132

59. No; lim G1x2 does not equal G122. x:2

x:3

61. Yes; lim f 1x2 = f 152 x:5

63. No; g152 does not exist and lim g1x2 does not exist. 65. Yes; lim F1x2 = F142 x:4

x:5

67. Yes; g1x2 is continuous at each

point on 1 - 4, 42. 69. No; f 1x2 is not continuous at x = 0. 71. Yes, since g1x2 is not continuous at each real number 73. (a) k = 5; (b) so that the Candy Factory does not lose revenue 1 75. Limit does not exist. 77. 6 79. - 0.2887, or 223 81. 0.75 83. 0.25

Technology Connection, p. 125 1–2. Left to the student

Exercise Set 1.3, p. 128 1. (a) 8x + 4h; (b) 48, 44, 40.4, 40.04 3. (a) - 8x - 4h; (b) - 48, - 44, - 40.4, - 40.04 5. (a) 2x + h + 1; -2 (b) 13, 12, 11.1, 11.01 7. (a) # ; x 1x + h2 2 1 4 40 (b) - 35 9. (a) - 2; (b) - 2, - 2, - 2, - 2 , - 15 , - 51 , - 501 11. (a) - 3x 2 - 3xh - h2; (b) - 109, - 91, - 76.51, 75.1501 13. (a) 2x + h - 3; (b) 9, 8, 7.1, 7.01 15. (a) 2x + h + 4; (b) 16, 15, 14.1, 14.01 17. About 0.3% per yr; about - 0.5% per yr; about - 0.07% per yr 19. About 0.35% per yr; about - 0.56% per yr; about - 0.05% per yr 21. About 3.7% per yr;

3 ; f ¿1 - 22 = - 34; f ¿ A - 12 B = - 12 x2 2. y = - 34 x - 3; y = - 12x - 12 3. Left to the student

Exercise Set 1.4, p. 141 1. (a) and (b)

3. (a) and (b)

y 9 8 (−2, 6) 7 6 5 4 3 2 1 −2

x-axis is tangent to curve at (0, 0). y 1

f

−2

(1, 1.5)

−1 −1

1

2

x

x-axis is tangent to curve at (0, 0).

(c) f ¿1x2 = 3x; (d) - 6, 0, 3 5. (a) and (b) y 10 8 6 4 2 −5 −4 −3 −2 −1−2 −4 −6 (−2, −8) −8 −10

f

(1, 1) 1 2 3 4 5 x x-axis is tangent to curve at (0, 0).

(c) f ¿1x2 = 3x 2; (d) 12, 0, 3

−1 −1 −2 −3 −4 −5 −6 −7 −8 (−2, −8)−9

1

2

x

(1, −2)

f

(c) f ¿1x2 = - 4x; (d) 8, 0, - 4 7. (a) and (b) All tangent lines are identical to the graph of the original function. y 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1

f

1 2 3 4 5 x

−2

(c) f ¿1x2 = 2; (d) 2, 2, 2

A-9

ANSWERS

9. (a) and (b) All tangent lines are identical to the graph of the original function. y 4 3 2 1 −2 −1−1

(−2, 2) 1 2 3 4 5 6 7 8 x

−2 −3 −4 −5 −6

f

21 23. 54 25. x4 29. - 10 3 x 2 31. 10x 3 2 12 5 37. - 4 39. - 2 7x x

11. (a) and (b) y 8 7 6 5 4 3 2 1

21. -

f

(1, 2)

45. - 0.02x - 0.5

−3 −2 −1 −1 −2

1 2 (0, 0)

3 x

(c) f ¿1x2 = 2x + 1; (d) - 3, 1, 3

1 1 1 1 ; (d) , , 2 2 2 2 13. (a) and (b) (c) f ¿1x2 = f

y 10 8 6 4 (0, −2) 2

−3 −2 (−2, 0)

−1 −2 −4 −6 −8 −10

15. (a) and (b) There is no tangent line for x = 0. y 5 4 3 2 1

(1, 3) 1

x −5 −4 −3

−1−1 −2 1 (−2, − 2 ) −3 −4 −5

(c) f ¿1x2 = 4x + 3; (d) - 5, 3, 7

f (1, 1) 1 2 3 4 5 x

1 ; x2 1 (d) - 4, does not exist, - 1 17. (a) y = 6x - 9; (b) y = - 2x - 1; (c) y = 20x - 100 19. (a) y = - 2x + 4; (b) y = - 2x - 4; (c) y = - 0.0002x + 0.04 21. (a) y = 2x + 5; (b) y = 4; (c) y = - 10x + 29 23. f ¿1x2 = m 25. x 0, x 3, x 4, x 6, x 12 27. x 1, x 2, x 3, x 4 29–34. Answers will vary. 35. x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 1 37. x … 0 39. No 41. 43. f ¿1x2 = 11 - x22 2 1 45. f ¿1x2 = - 3 47. f ¿1x2 = x 22x + 1 (c) f ¿1x2 = -

49. (a) f ¿ is not defined at x = - 3; (b) 51. (a) x = 3; (b) h¿102 = - 1, h¿112 = - 1, h¿142 = 1, h¿1102 = 1 53. f ¿1x2 is not defined at x = - 1. 55. (a) lim F1x2 = 5,

1 3 1 1 + 2 27. + x 42 4 x3 22x x2x - 7 33. 0.9x 0.5 35. 23 2 -1>3 41. 4 43. 13 x 1>3 x 3 3 6 24 47. - 2x -5>3 + x -1>4 + x 1>5 - 4 4 5 x

2 1 51. 24 53. 1 55. 14 57. 43 2 x2 59. (a) y = 10x - 15; (b) y = x + 3; (c) y = - 2x + 1 2 61. (a) y = - 2x + 3; (b) y = - 27 x + 13; (c) y = 14 x + 34 49. -

63. 10, - 32 65. 10, 12 67. A 56, 23 69. 1- 25, 76.252 12 B 71. None 73. The tangent line is horizontal at all points of the graph. 75. 1- 1, - 42, A 53, 5 13 77. A 23, 2 - 223 B , 27 B or approximately 11.73, - 1.462; A - 23, 2 + 223 B , or approximately 1 - 1.73, 5.462 79. 10, - 22, A - 1, - 11 6 B

81. 19.5, 99.752

83. 160, 1502

85. A - 2 + 23, 43 - 23 B ,

or approximately 1- 0.27, - 0.402; A - 2 - 23, 43 + 23 B ,

or approximately 1- 3.73, 3.072 87. (a) A¿1r2 = 6.28r; (b) 89. (a) w¿1t2 = 1.82 - 0.1192t + 0.002274t 2; (b) about 21 lb; 6000 (c) about 0.86 lb>month 91. (a) R¿1v2 = - 2 ; v (b) 75 beats>min; (c) - 0.94 beat>min per mL dP 93. (a) = 4000t; (b) 300,000 people; (c) 40,000 people>yr; dt 0.61 (d) 95. (a) V¿ = ; (b) 244 mi; (c) 0.0031 mi>ft; 2h (d) 97. 12, q 2 99. 1- q , - 12 and 13, q 2

1 55 101. 10, - 22, A 213, - 55 27 B , A - 23 , - 27 B 103. Always increasing 105. Always increasing on 10, q 2 1 107. 2x + 1 109. 3x 2 - 1 111. 3x 2 - 2 113. - 192x 2 x 2 1 115. 117. 1 - 2 119. 3x 2 + 6x + 3 121. 2 x 32 3 x 123. y = x 4 − 3x 2 + 1 125. y = 10.2x 4 − 6.9x 3 1

2

x:2

F122 = 5; therefore, lim F1x2 = F122 (b) no, the graph has a x:2

corner there 57. m = 11, b = - 18 58–63. Left to the student 65. f ¿1x2 does not exist for x = 5.

Technology Connection, p. 141 1–8. Left to the student

Technology Connection, p. 147 1. 152, - 76, - 100, - 180 3. - 2.31, 3.69, 0.81

−2

f

1

−1

−2

10, 02, 10.507, - 0.2252

10, 12, 1- 1.225, - 1.252, 11.225, - 1.252 100 127.

2. 36, 0, 12, 0, 43.47

Technology Connection, p. 153

−1

2

2

129. f´

f´

f

−2 −6

2

6

1. Tangent line is horizontal at A 2, 83 B .

Exercise Set 1.5, p. 154 1. 7x 6

3. - 3

13. 3x 2 + 6x

5. 0 7. 30x 14 9. - 6x -7 11. - 8x -3 4 15. 17. 0.9x -0.1 19. 25 x -1>5 2x

−100

f ¿112 = 45

−4

f ¿112 = 1

A-10

ANSWERS 5

131.

133. (a)

(d) factoring; (e) the limit of a product is the product of the limits and lim f 1x + h2 = f 1x2; (f) definition of a derivative;

4 f'

f´

h:0

f −5

f

5

−2

2

f´ −3

f ¿ 112 = 0

125. $2.58>jacket; $0.014>

(g) using Leibniz notation jacket at x = 184 20 127.

There are no points at which the tangent line is horizontal.

f´ f

−2

(b) f ¿1x2 = 3x - x; x = 0, x = (c) 0 6 x 6 23; (d) no, all x such that f ¿1x2 = 0 were found in part (b). 2

2 3;

−4

4 f −20

129.

131.

1

Technology Connection, p. 161

f

f´

1. (c)

f´

–1

1

3. 12x + 7

9. 24x 2 + 12x + 11

5. 20x 4 + 60x 2 7. 1 92t 11. + 2 2 22t

15 3>2 2 x

39.

1x

1>2

+ 32 A

1 -2>3 3x

A 2x + 3 B 2

6 - 2x + 21x 1>6 6x 43. 47.

2>3

A 2x + 3 B

B - 1x

1>3

2

41.

- 72 A

1 -1>2 2x

B

x1x + 22 2x -1 + 1 , or , for x Z 0 -1 2 1x + 12 1x + 122

- 21x 2 - 3x - 12 -1 45. 1t - 422 1x 2 + 122 2 1t - 2t + 421- 2t + 32 - 1- t 2 + 3t + 5212t - 22 1t 2 - 2t + 422

–1

−4

1 - 0.2, - 0.752, 10.2, 0.752

, or

,

- t 2 - 18t + 22 49–96. Left to the student 1t 2 - 2t + 422 97. (a) y = 2; (b) y = 12 x + 2 99. (a) y = x + 5; 21 (b) y = 21 101. - $0.006875>jacket 4 x - 4 103. - $0.0053>jacket 105. About $0.0016>jacket 107. $1.64>vase 109. (a) P¿1t2 = 57.6t 0.6 - 104; - 41t 2 - 12 (b) $461.4 billion>yr; (c) 111. (a) T¿1t2 = 1t 2 + 122 (b) 100.2°F; (c) - 0.48°F>hr 113. 30t 2 + 10t - 15 - 4x x2 + 1 115. 3x 2 a 2 b + 2 1x 3 - 82, or x - 1 1x - 122 3x 6 - 4x 4 - 3x 2 + 32x - x 6 + 4x 3 - 24x 2 117. 1x 2 - 122 1x 4 - 3x 3 - 522 - 2x - 2x 119. (a) f ¿1x2 = 2 ; (b) g¿1x2 = 2 ; (c) 2 1x - 12 1x - 122

or

121. 123. (a) Definition of a derivative; (b) adding and subtracting the same quantity is the same as adding 0; (c) the limit of a sum is the sum of the limits;

3

f´

+ 4x

13. 4x 3, for x Z 0 15. 8x 3 + 1, for x Z 0 17. 8x + 2, for x Z 12 19. 1, for t Z - 4 - 2x15x 3 - 3x - 152 21. 48x 3 + 3x 2 + 22x + 17 23. 12x 3 + 322 3 3>2 25. 120x 2 + 25 + 32 x -1>2 + 24 27. + 15t 2 2 x 13 - t22 29. 2x + 6 31. 2x1x 2 - 4213x 2 - 42 6 33. 5 - 100x -3 + 30x -4 35. 3t 2 - 1 + 2 t - x 4 - 3x 2 - 2x 37. - 10x 1x 3 - 122

f

−3

Exercise Set 1.6, p. 163 1. 11x 10

5

1- 1, - 22, 11, 22

Exercise Set 1.7, p. 173 1. 8x + 4 7.

3. - 5517 - x254

3x 23x 2 - 4

5.

9. - 640x18x - 62-41

4x15x + 142

13.

-6 13x + 823

17. 9x 211 + x 322 - 32x 712 + x 823

17 - 5x24

19. 4x - 400

21 + 8x

2

11. 41x - 42712x + 32517x - 62 15.

4

1

+ 31x - 322 22x 23. - 512x - 323110x - 32 25. 13x - 12612x + 124172x + 112 2x15x - 12 5x 4 + 6 5x 4 + 6 27. 29. 3 , or 2>3 4 3x 1x + 622>3 24x - 1 321x 5 + 6x22 3 4413x - 12 -7 31. 33. 15x + 225 21x + 323>214 - x21>2 35. 20012x 3 - 3x 2 + 4x + 129913x 2 - 3x + 22 6815x - 123 -1 37. 39. 12x + 325 1x - 123>21x + 121>2 - 12x + 32316x + 612 13x - 421>41138x - 192 41. 43. 13x - 226 12x + 121>3 1 x 45. , 2x, 2 22u 2x - 1 47. 50u49, 12x 2 - 4x, 5014x 3 - 2x 2249112x 2 - 4x2 49. 2u + 1, 3x 2 - 2, 12x 3 - 4x + 1213x 2 - 22 212x - 12 51. 3x 2110x 3 + 132 53. 312x 2 - 2x + 522>3 31- 6t - 112 - 6t - 11 55. , or 2 2 15 + 3t2 16 + 3t2 31t + 22213t + 522 5 3 2x - 3x 2 57. y = x + 59. y = 4x - 3 61. (a) ; 4 4 11 + x26 2x - 3x 2 (b) ; (c) They are the same. 63. f 1x2 = x 5, 11 + x26 x + 1 g1x2 = 3x 2 - 7 65. f 1x2 = , g1x2 = x 3 67. - 216 x - 1 69. 41132-2>3, or about 0.72 21.

A-11

ANSWERS

71. f ¿1x2 = 632x 3 + 14x - 52245 36x 2 + 814x - 524 1 3 73. f ¿1x2 = b a 2x 2 221 - 3x 23x + 21 - 3x 50012x - 0.12 4000x 75. $1,000,000>item 77. P¿1x2 = 2 31x 2 + 222>3 2x - 0.1x 3 2 79. (a) 0.84x - 17.76x + 101.06x - 18.92; (b) ; 2 (c) $336 billion>yr 81. (a) dA>di = 300011 + i2 ; (b) 80,000 83. (a) D1t2 = ; (b) - 4.482 units>day 1.6t + 9 95w 85. (a) D1c2 = 4.25c + 106.25, c1w2 = L 2.199w 43.2 (b) 4.25 mg>unit of creatine clearance; (c) 2.199 units of creatine clearance> kg; (d) 9.35 mg>kg; (e) 1 1 # a1 + 1 b 89. 271 x -26>27 87. 1 + + 22x 22x 23x + 2x 3x 21x - 22 6x 7 + 32x 5 + 5x 4 91. 93. 1x 3 + 6x + 122>3 21x - 125>2 31x 2 - x - 1221x 2 + 4x - 12 1 95. 97. 1x 2 + 124 11 - x2 21 - x 2 62t + 1

99.

1- 2.14476, - 7.7282, 12.14476, 7.7282

f´ −3

3 -6 18 , f –1x2 = , f ‡1x2 = , 1x + 222 1x + 223 1x + 224 - 72 f 41x2 = 69. 2.29 sec 71. (a) 3.24 m; 1x + 225 (b) 3.24 m>sec; (c) 1.62 m>sec2; (d) It is the gravitational constant for the moon. 73. 42.33 ft>sec 3 3 75. 77. 67. f ¿1x2 =

s

4 - 2x 2 24 - x 2

107.

22x - 1

5. 8

7. 0

9.

6 x4

11.

,

-1 3>2

4x 48 6 4 13. 12x 2 + 3 15. 17. 5 x x 25x 9>5 19. 141x 2 + 3x25113x 2 + 39x + 272 21. 1012x 2 - 3x + 1281152x 2 - 228x + 852 31x 2 + 22 -2 45x 4 - 54x 2 - 3 23. 25. 27. 41x 2 + 125>4 9x 4>3 161x 3 - x25>4 5 -3>4 1 -3>2 -5 -4 - 4x 29. 8 x 31. 24x + 6x 44 33. 24x - 2 35. 37. 24 39. 720x 12x - 323 15 -7>2 x 41. 120x -6 + 43. 0 45. (a) v1t2 = 3t 2 + 1; 16 (b) a1t2 = 6t; (c) v142 = 49 ft> sec, a142 = 24 ft>sec2 47. (a) v1t2 = 3; (b) a1t2 = 0; (c) v122 = 3 mi> hr,

a122 = 0 mi/hr2; (d) 49. (a) 144 ft; (b) 96 ft> sec; (c) 32 ft>sec2 51. v122 = 19.62 m>sec, a122 = 9.81 m>sec2 53. (a) The velocity at t = 20 sec is greater, since the slope of a tangent line is greater there. (b) The acceleration is positive, since the velocity (slope of a tangent line) is increasing over time. 55. (a) $146,000>month, $84,000>month, - $4000>month; (b) - $68,000>month2, - $56,000>month2, - $32,000>month2;

−2

4

v1t2 switches at t = 1.

Chapter Review Exercises, p. 190

Exercise Set 1.8, p. 182 3. 24x 2

s

−3

v1t2 switches at t = 0.

2. False 3. True 4. False 5. True 7. False 8. True 9. (e) 10. (c) 12. (f) 13. (b) 14. (d) f 1x2

x : - 7-

5 A 22x - 1 + x 3 B 4 A 3x 2 22x - 1 + 1 B

v

a

or 5 A 22x - 1 + x 3 B 4312x - 12-1>2 + 3x 24

1. 20x 3

3

−3

f

105.

a

−3

3

−8

v

1. False 6. True 11. (a) 15. (a)

101.

42t23t + 1t 8 103.

57. (a) 11.34, 1.98, 0.665; (b) - 0.789, - 0.0577, 6 - 15 59. 61. - 0.0112; (c) 11 - x24 12x - 127>2 3x 1>2 - 1 63. 3>2 1>2 65. k1k - 121k - 221k - 321k - 42x k - 5 2x 1x - 123

(c)

f 1x2

x : - 7+

-8

- 11

-6

-9

- 7.5

- 10.5

- 6.5

- 9.5

- 7.1

- 10.1

- 6.9

- 9.9

- 7.01

- 10.01

- 6.99

- 9.99

- 7.001

- 10.001

- 6.999

- 9.999

- 7.0001

- 10.0001

- 6.9999

- 9.9999

(b) lim - f 1x2 = - 10; lim + f 1x2 = - 10; lim f 1x2 = - 10 16.

x : -7

y 10 8 6 4 2

−10 −8 −6 −4 −2−2

x : -7

x : -7

2 4 6 8 10 x

−4 f −6 −8 −10

1x + 721x - 32 x 2 + 4x - 21 = lim = x : -7 x : -7 x + 7 x + 7 lim 1x - 32 = - 10 18. - 4 19. 10 20. - 12

17. lim x : -7

21. 3

22. Not continuous, since lim g1x2 does not exist x : -2

23. Continuous 24. - 4 25. - 4 26. Continuous, since lim g1x2 = g112 28. - 2 30. 2

x:1

27. Does not exist

29. Not continuous, since lim g1x2 does not exist 31. - 3

32. 4x + 2h

x : -2

33. y = x - 1

34. 14, 52

A-12

ANSWERS

35. 15, - 1082

36. 45x 4

8 -2>3 24 x 38. 9 39. 6x -3>5 3 x 41. 52 x 5 + 32x 3 - 2 42. 2x, x Z 0 37.

40. 0.7x 6 - 12x 3 - 3x 2 - x 2 + 16x + 8 43. 44. 215 - x212x - 1241- 7x + 262 18 - x22 x111x + 42 45. 35x 41x 5 - 326 46. 47. - 48x -5 14x + 221>4 48. 3x 5 - 60x + 26 49. (a) v1t2 = 1 + 4t 3; (b) a1t2 = 12t 2; (c) v122 = 33 ft>sec, a122 = 48 ft>sec2 50. (a) AC1x2 = 5x -1>2 + 100x -1, AR1x2 = 40, AP1x2 = 40 - 5x -1>2 - 100x -1; (b) average cost is dropping at approximately $1.33 per item. 51. (a) P¿1t2 = 100t; (b) 30,000; (c) 2000>yr 52. 1 f  g21x2 = 4x 2 - 4x + 6; - 9x 4 - 4x 3 + 9x + 2 1g  f 21x2 = - 2x 2 - 9 53. 221 + 3x11 + x 322 50 54. - 0.25 55. 16 56. f´

f

−3

3

1 1 23. [1.3] 4x + 3 + 2h 24. [1.4] y = 34 x + 2 Z lim+ x:0 x x 25. [1.5] 10, 02, 12, - 42 26. [1.5] 23x 22 10 27. [1.5] 34 x -2>3 + 52 x -1>2 28. [1.5] 2 29. [1.5] 54 x 1>4 x 30. [1.5] - 1.0x + 0.61 31. [1.5] x 2 - 2x + 2 - 61x - 22 5 32. [1.6] 33. [1.6] x4 15 - x22 34. [1.7] 1x + 32317 - x241- 9x + 132 35. [1.7] - 51x 5 - 4x 3 + x2-615x 4 - 12x 2 + 12 2x 2 + 5 36. [1.6, 1.7] 37. [1.8] 24x 38. [1.6] (a) AR = 50, 2x 2 + 5 AC = x -1>3 + 750x -1, AP = 50 - x -1>3 - 750x -1; (b) average cost is dropping at approximately $11.74 per item. 39. [1.5] (a) M¿1t2 = - 0.003t 2 + 0.2t; (b) 9; (c) 1.7 words>min 40. [1.7] 1 f  g21x2 = 4x 6 - 2x 3; - 1 - 9x 1g  f 21x2 = 21x 2 - x23 41. [1.6, 1.7] 211 - 3x22>311 + 3x25>6 42. [1.2] 27 43. [1.5] 50 lim

x : 0-

f

f´

−50

1- 1.7137, 37.4452, 10, 02, 11.7137, - 37.4452

0

Chapter 1 Test, p. 192

5

−30

11.0836, 25.10292 and 12.9503, 8.62472

1. [1.1] (a) f 1x2

x : 6-

x : 6+

f 1x2

5

11

7

13

5.7

11.7

6.5

12.5

5.9

11.9

6.1

12.1

5.99

11.99

6.01

12.01

5.999

11.999

6.001

12.001

5.9999

11.9999

6.0001

12.0001

(b) lim- f 1x2 = 12; lim+ f 1x2 = 12; lim f 1x2 = 12 x:6

x:6

2. [1.1]

y 18 16 14 12 10 8 6 4 2 −8 −6 −4 −2−2

f(x) =

x:6

x 2 − 36 x−6

2 4 6 8 10 12 x

1x + 621x - 62 x 2 - 36 = lim = lim 1x + 62 = 12 x:6 x - 6 x:6 x:6 x - 6 4. [1.1] Does not exist 5. [1.1] 0 6. [1.1] Does not exist 7. [1.1] 2 8. [1.1] 4 9. [1.1] 1 10. [1.1] 1 11. [1.1] 1 12. [1.2] Continuous 13. [1.2] Not continuous, since lim f 1x2 does not exist 14. [1.1, 1.2] Does not exist

44. [1.5] 0.5

Extended Technology Application p. 195 1.

160

160

0

500 0

0

500 0

Xscl = 50, Yscl = 20 3

Xscl = 50, Yscl = 20

2. (a) y = - 0.0000045x + 0.000204x 2 + 0.7806x + 4.6048; (b) 160 (c) acceptable fit; (d) about 441 ft;

0

500 0

Xscl = 50, Yscl = 20 2

(e) dy>dx = - 0.0000135x + 0.000408x + 0.7806; (f ) approximately 1256, 1422; at about 256 ft from home plate, the ball reached its maximum height of approximately 142 ft. 3. (a) y = - 0.0000000024x 4 - 0.0000026x 3 - 0.00026x 2 + 0.8150x + 4.3026; (b) 160 (c) acceptable fit; (d) about 440 ft;

3. [1.2] lim

x:3

15. [1.1, 1.2] 1 18. [1.1, 1.2] 3 21. [1.1, 1.2] 18

16. [1.1, 1.2] No 17. [1.1, 1.2] 3 19. [1.1, 1.2] Yes 20. [1.1, 1.2] 6 22. [1.1, 1.2] Does not exist, since

0

500 0

Xscl = 50, Yscl = 20

(e) dy>dx = - 0.0000000096x 3 - 0.0000078x 2 0.00053x + 0.815; (f) approximately (257, 142); at about 257 ft from home plate, the ball reached its maximum height of approximately 142 ft.

A-13

ANSWERS

4. (a)

0

500 0

9. Relative minimum at 11, 42; relative maximum at 1- 1, 82

(b) 450 ft; dy 303.75 - 0.003x 2 (c) = ; dx 2202,500 - x 2

160

Xscl = 50, Yscl = 20

(d) approximately 1318, 1522; at about 318 ft from home plate, the ball reached its maximum height of approximately 152 ft.

5. The two models are very similar. The main difference seems to be that the maximum height is reached further from home plate with the model in Exercise 4. 6. 466 ft, 442 ft, 430 ft 7. The estimate of the reporters is way off. Even with a low trajectory, the ball would at best have traveled a horizontal distance of about 526 ft. 8. (a) 523 ft; (b) 430 ft; (c) 464 ft (rounded)

−5 −4 −3 −2 −1−1 −2 −3

f(x) = 2 − (x − 1)2/3

f´(x) = − 2 (x − 1)−1/3 3 4

2.

4

y 5 4 3 2 f(x) = 3x2 + 2x3 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

1 2 3 4 5 x

y 15 10 5 −10 −15 −20 −25 −30 −35

Technology Connection, p. 208

f(x) = x 3 − 3x + 6

13. No relative extrema exist.

−5 −4 −3 −2 −1−5

Chapter 2 1.

y 9 8 7 6 5 4 3 2 1

11. Relative minimum at 10, 02; relative maximum at 1- 1, 12

15. Relative minimum at 14, - 222; relative maximum at 10, 102 y

1 2 3 4 5 x

20

g(x) = 2x 3 − 16

10 −3 −2 −1

1 2 3 4 5 6 7 x

−10 −20 G(x) = x 3 − 6x2 + 10

−4

6

−4

6

−2

−2

The derivative is not defined at 11, 22.

Technology Connection, p. 210

1. Relative maximum at 1 - 1, 192; relative minimum at 12, - 82

Technology Connection, p. 212 Exercise Set 2.1, p. 212

f(x) = x 2 + 4x + 5

y 10 9 8 7 6 5 4 3 2 1

−6 −5 −4 −3 −2 −1−1

5. Relative minimum at 1- 1, - 22 y 5 4 3 g(x) = 1 + 6x + 3x 2 2 1 −5 −4 −3 −2 −1−1 −2 −3 −4 −5

3. Relative maximum at A - 12, 214 B y 7 6 f(x) = 5 − x − x 2 5 4 3 2 1

1 2 x

1 2 3 4 5 x

y 0.5 0.4 0.3 0.2 0.1

g(x) = x 3 − x4

−1.0 −0.5−0.1 −0.2 −0.3 −0.4 −0.5

0.5

1.0

19. No relative extrema exist. y 10 8 6 4 2 −5 −4 −3 −2 −1−2

1 2 3 4 5 x

−4 −6 −8 −10

x

f(x) =

1 3 x − 2x 2 + 4x − 1 3

21. Relative minima at A - 25, - 32 B and A 25, - 32 B ; relative maximum at 10, 182

1–8. Left to the student 1. Relative minimum at 1- 2, 12

17. Relative maximum at 27 A 34, 256 B

−5 −4 −3−2 −1−1 −2 −3 −4 −5

1 2 3 4 5 x

20 10 −3 −2 −1 −10

1

2

3 x

−20 −30 g(x) = 2x 4 − 20x2 + 18

23. No relative extrema exist.

7. Relative minimum at 11, 12; relative maximum at A - 13, 59 27 B

25. Relative maximum at 10, 12 y

y

1

2 1 −10 −8 −6 −4 −2

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

y 30

−1 −2

3

F(x) = x − 1 2 4 6 8 10 x

−2

−1

1 −0.5 −1

1 2 3 4 5 x

−2 −3 −4 −5

G(x) = x 3 − x 2 − x + 2

f (x) = 1 – x 2/3

0.5

2 x

A-14

ANSWERS

27. Relative minimum at 10, - 82

29. Relative minimum at 1- 1, - 22; relative maximum at 11, 22

y 1 −5 −4 −3 −2 −1−1

y 3 2 1

1 2 3 4 5 x

−2 −3 −4 −5 −6 −7 −8 −9

−8 G(x) = 2 x +1

−4 −3−2 −1−1 −2 −3

Relative minimum at 12, 02; increasing on 12, q 2; decreasing on 1- q , 22; f ¿ does not exist at x = 2

|

4

–4 1 2 3 4 x 4x g(x) = ——– x2 + 1

6

–2

|

103.

33. Relative minimum at 1- 1, 22

y 2 3

|

f(x) = x − 2

f(x) = x 2 − 1

|

3

31. No relative extrema exist.

f (x) = x

101.

y 10 9 8 7 6 5 4 3 2 1

1

−5 −4 −3−2 −1

1 2 3 4 5 x

−1 −2

−12 −8 −4 −1

2

–2

–2

|

f(x) = 9 − x 2

105.

Relative maximum at 10, 9); relative minima at 1- 3, 02 and 13, 02; increasing on 1- 3, 02 and 13, q 2; decreasing on 1- q , - 32 and 10, 32; f ¿ does not exist at x = - 3 and x = 3

|

10

4

8

12 x

g(x) = √x 2 + 2x + 5

35–83. Left to the student. 87. Relative minimum at 11.94, 15,8822; relative maximum at 17.05, 17,7732 E(t) 18,000

89. Relative maximum at 16, 102.22 105 104 103 102 101 100 99 98 97 96 95

17,000 16,000 15,000 4

6

8

–1

107.

|

f(x) = x 3 − 1

Relative minimum at 11, 02; increasing on 11, q 2; decreasing on 1- q , 12; f ¿ does not exist at x = 1

|

10

T(t)

E(t) = −28.31t3 + 381.86t2 − 1162.07t + 16,905.87

2

–4

85.

10 t

Relative maximum at 10, 12; relative minima at 1- 1, 02 and 11, 02; increasing on 1- 1, 02 and 11, q 2; decreasing on 1- q , - 12 and 10, 12; f ¿ does not exist at x = - 1 and x = 1

T(t) = −0.1t2 + 1.2t + 98.6 –2

3 –1

109. 1 2 3 4 5 6 7 8 9 10 11 12 t

91. Increasing on 1- 1, q 2, decreasing on 1- q , - 12; relative minimum at x = - 1. 93. Increasing on 1- q , 12, decreasing on 11, q 2; relative maximum at x = 1. 95. Increasing on 1- 4, 22, decreasing on 1- q , - 42 and 12, q 2; relative minimum at x = - 4, relative maximum at x = 2. 97. f (x) = −x 6 − 4x 5 + 54x 4 + 160x 3 Relative minima at − 641x 2 − 828x + 1200 1- 3.683, - 2288.032 and 7000 12.116, - 1083.082; relative maxima at 1 - 6.262, 3213.82 and 1- 0.559, 1440.062 and 15.054, 6674.122 −8 8

111.

Technology Connection, p. 224 Left to the student

Technology Connection, p. 226

1. Relative minimum at 11, - 12; inflection points at 10, 02, 10.553, - 0.5122, 11.447, - 0.5122, and 12, 02 6

6

f´

f

−1

3

6

− 3000 f(x) =

3

Relative minima at 1 - 2, 12 and 12, 12; relative maximum at 10, 2.5872

|4 − x 2| + 1 6

−1

3 −2

Technology Connection, p. 230 1–8. Left to the student −10

10 0

3 −2

−2 f˝

99.

−1

ANSWERS

17.

Technology Connection, p. 231 f´ (x) = 3x 5 − 5x 3

f´ (x) = 15x 4 − 15x 2

10

10

y 6 −1

1

2

3

4 x

−6 −12 −18

−3

3

−3

3

−24 −30

−10

−10

f ˝(x) = 60x 3 − 30x

1. Critical values: - 1, 0, and 1 2. Inflection points at - 0.707, 0, and 0.707

10

−3

3

−10

Exercise Set 2.2, p. 231

1. Relative maximum is f 102 = 3. Relative minimum is f A 12 B = 5. Relative maximum is f A 45 B = 7. Relative minimum is f A 12 B = f A - 12 B = 3. y 16 12 8 4

f(x) = x 3 − 12x

−5 −4 −3 −2 −1 −4 −8 −12 −16

1 2 3 4 5 x

9.

Relative minimum at 12, - 162, relative maximum at 1- 2, 162; inflection point at 10, 02; increasing on 1- q , - 22 and 12, q 2, decreasing on 1- 2, 22, concave down on 1- q , 02, concave up on 10, q 2

y 50 40 30 20 10 −5 −4 −3 −2 −1 −10 −20 −30 −40 −50

1 2 3 4 5 x

Relative minimum at 12, - 512, relative maximum at 1 - 2, 452; inflection point at 10, - 32; increasing on 1- q , - 22 and 12, q 2, decreasing on 1- 2, 22; concave down on 1- q , 02, concave up on 10, q 2

f(x) = 3x 3 − 36x − 3 y f(x) = 8– x 3 – 2x + 1– 2 3

3

1 −1

1 x −1

15.

y 10 5 −2−1−5

−10 −15 −20

f(x) = – x 3 + 3x 2 – 4 1 2 3 4 x

Relative minimum at A 12, - 13 B , relative maximum at A - 12, 1 B ; inflection point at A 0, 13 B ; increasing on A - q , - 12 B and A 12, q B , decreasing on A - 12, 12 B ; concave down on 1- q , 02, concave up on 10, q 2

Relative minimum at 10, - 42, relative maximum at 12, 02; inflection point at 11, - 22; increasing on 10, 22, decreasing on 1- q , 02 and 12, q 2; concave up on 1- q , 12, concave down on 11, q 2

Relative minima at 10, 02 and 13, - 272, relative maximum at 11, 52; inflection points at 10.451, 2.321) and 12.215, - 13.3582; increasing on 10, 12 and 13, q 2, decreasing on 1- q , 02 and 11, 32; concave up on 1- q , 0.4512 and 12.215, q 2, concave down on 10.451, 2.2152

Relative minima at A - 23, - 9 B and A 23, - 9 B , relative maximum −3 −2 −1 1 2 3 x −1 at 10, 02; inflection points at −2 1- 1, - 52 and 11, - 52; increasing −3 −4 on A - 23, 0 B and A 23, q B , −5 −6 decreasing on A - q , - 23 B and −7 −8 A 0, 23 B ; concave up on −9 1- q , - 12 and 11, q 2, concave −10 down on 1- 1, 12 21. f(x) = x 3 – 2x 2 – 4x + 3 Relative minimum at 12, - 52, y relative maximum at A - 23, 121 10 27 B ; 8 7 inflection point at A 23, - 27 B ; in6 4 creasing on A - q , - 23 B and 2 12, q 2, decreasing on A - 23, 2 B ; 1 2 3 4 x −3 −2 −1 −2 −4 concave down on A - q , 23 B , −6 concave up on A 23, q B −8 19.

f(x) = x4 – 6x 2 y 1

−10

23.

y 3 2 1

11.

13.

5. - 14. - 19 5. - 1; relative maximum is

f(x) = 3x 4 − 16x 3 + 18x 2

A-15

−2 −1.5 −1 −0.5 −1 f(x) = 3x 4 + 4x 3 −2

0.5

1 x

−3

25.

y 600 450 300 150 −15 −10 −5 −150 −300 −450 −600 −750 −900

5

10 15 x

f(x) = x 3 − 6x 2 − 135x y 40

27.

30 20 10 −2 −1 −10 −20

1

2

3

4 x

2 −1

1

2

3

4 x

−2 −4 −6 −8

Relative minimum at 19, - 9722, relative maximum at 1- 5, 4002; inflection point at 12, - 2862; increasing on 1- q , - 52 and 19, q 2, decreasing on 1 - 5, 92; concave down on 1- q , 22, concave up on 12, q 2 Relative minimum at 13, - 172; inflection points at 10, 102 and 12, - 62; increasing on 13, q 2, decreasing on 1- q , 32; concave down on 10, 22, concave up on 1- q , 02 and 12, q 2

f(x) = x 4 − 4x 3 + 10

y 4

29.

Relative minimum at 1- 1, - 12; inflection points at A - 23, - 16 27 B and 10, 02; increasing on 1- 1, q 2, decreasing on 1- q , - 12; concave up on A - q , - 23 B and 10, q 2, concave down on A - 23, 0 B

f(x) = x 3 − 6x 2 + 12x − 6

No relative extrema; inflection point at 12, 22; increasing on 1- q , q 2; concave down on 1- q , 22, concave up on 12, q 2

A-16

ANSWERS

31.

y 3

y

43.

x f(x) = ——– x2 + 1

0.5

f(x) = 5x 3 − 3x 5

2 1 −2

−1

1

2

x

−3−2 −1

−1

1 2 3 x

−2 −3

Relative minimum at 1- 1, - 22, relative maximum at 11, 22; inflection points at 1- 0.707, - 1.2372, 10, 02, and 10.707, 1.2372; increasing on 1- 1, 12, decreasing on 1- q , - 12 and 11, q 2; concave down on 1- 0.707, 02 and 10.707, q 2, concave up on 1- q , - 0.7072 and 10, 0.7072 33.

y 7 6 5 4 3 2 1 −1

35.

f(x) = x 2 (3 − x)2

1

−1

2

f(x) = (x + 1) 2/3 2

3

4 x

Relative minimum at 1- 1, 02; no inflection points; increasing on 1 - 1, q 2, decreasing on 1- q , - 12; concave down on 1- q , - 12 and 1- 1, q 2

y

1 −2

−1 y 2

37.

f(x) = (x − 3)1/3 − 1

1 −2

1 x

−0.5

2

4

6

8

Relative minima at 10, 02 and 13, 02, relative maximum at A 32, 81 16 B ; inflection points at 10.634, 2.252 and 12.366, 2.252; increasing on A 0, 32 B and 13, q 2, decreasing on 1- q , 02 and A 32, 3 B ; concave down on 10.634, 2.3662, concave up on 1- q , 0.6342 and 12.366, q 2

10 x

−1

No relative extrema; inflection point at 13, - 12; increasing on 1- q , q 2; concave up on 1- q , 32, concave down on 13, q 2

−2 −3 −4

39.

y 6 5 4 3 2 1 −2 −1

f(x) = −2(x − 4)2/3 + 5

2

4

6

8 10 12 x

y 3 2

f(x) = x√4 − x2

1 −3 −2 −1

1 1 Relative minimum at a - 1, - b, relative maximum at a1, b; 2 2 23 23 b and 10, 02 and a 23, b; inflection points at a - 23, 4 4 increasing on 1- 1, 12, decreasing on 1- q , - 12 and 11, q 2; concave up on A - 23, 0 B and A 23, q B ; concave down on A - q , - 23 B and A 0, 23 B 45. y 4 3 2

f(x) =

3 x2 + 1

1 −6 −4 −2

2

4

6 x

−1

Relative maximum at 10, 32; inflection points at a -

1 9 , b and A3 4

1 9 , b; increasing on 1- q , 02, decreasing on 10, q 2; concave up A3 4 1 1 1 1 on a - q , b and a , q b, concave down on a , b A3 A3 A3 A3 47–101. Left to the student 103. 105. a

y 1300 1200 1100 1000 900 800 700 600 500 400 300 200 100

p(x) R(x)

13 12 11 10 9 8 7

P(x)

p

10

C(x)

20

30

40

x

10 20 30 40 50 60 70 80 90100 x

−2 −3 −4

41.

Relative maximum at 14, 52; no inflection points; increasing on 1 - q , 42, decreasing on 14, q 2; concave up on 1- q , 42 and 14, q 2

−0.5

1 −1 −2 −3

2

3 x

Relative minimum at A - 22, - 2 B , relative maximum at A 22, 2 B ; inflection point at 10, 02; increasing on A - 22, 22 B , decreasing on A - 2, - 22 B and A 22, 2 B ; concave up on 1- 2, 02, concave down on 10, 22

40 1 , or 13 mm 109. 111. 3 3 113. Left to the student 115. False 117. True 119. True 121. True f (x) = x 2 (1 − x )3 f (x) = 4x − 6x 2/3 123. 125. 107. Radius =

0.2

4 −1

6 −0.5

−8

Relative maximum at 10, 02; relative minimum at 11, - 22

1.5

−0.2

Relative minimum at 10, 02; relative maximum at 10.4, 0.0352

ANSWERS y 5 4 3 2 1

29.

f (x) = (x − 1) 2/3 − (x + 1) 2/3

127.

2

−6 −4 −2−1 −2 −3 −4 −5

10

−10

f(x) =

−2

Relative maximum at 1- 1, 1.5872; relative minimum at 11, - 1.5872

y 8 7 6 5 4 3 2 1

Technology Connection, p. 237 1. Vertical asymptotes: x = - 7 and x = 4 asymptotes: x = 0, x = 3, and x = - 2

2. Vertical

−5 −4 −3 −2 −1−1

Technology Connection, p. 238 1. and 2. Left to the student

3. 2

f(x) =

Technology Connection, p. 239 (top)

33. f(x) = x +

1. and 2. Left to the student

Technology Connection, p. 239 (bottom)

1 2 3 4 5 x

1. Horizontal asymptote: y = 0 2. Horizontal asymptote: y = 3 3. Horizontal asymptote: y = 0 4. Horizontal asymptote: y = 12

Increasing on A - q , - 22 B and A 22, q B ;

y 2 10 x 8 6 4 2

−10 −8 −6 −4 −2 −2

decreasing on A - 22, 0 B and A 0, 22 B Relative minimum at A 22, 222 B ;

2 4 6 8 10 x

−4 −6 −8 −10

Technology Connection, p. 240 (top) 1. y = 3x - 1

2. y = 5x

35.

Technology Connection, p. 240 (bottom)

−4 −6

Exercise Set 2.3, p. 247 1. x = 5 3. x = - 3 and x = 3 5. x = 0, x = 2, and x = 4 7. x = - 1 9. No vertical asymptotes 11. y = 34 13. y = 0 15. y = 5 17. No horizontal asymptotes 19. y = 0 21. y = 12 y Increasing on 1- q , 02 and 10, q 2 23. 10 No relative extrema 8 6 Asymptotes: x = 0 and y = 0 5 f(x) = − 4 x Concave up on 1- q , 02; 2 concave down on 10, q 2 −10 −8 −6 −4 −2 −2 2 4 6 8 10 x No intercepts −4 −6 −8 −10

25.

−4 −2−1 −2 −3 −4 −5

27.

f(x) =

1 x−5

2 4 6 8 10 12 x

y 5 4 3 2 1 −6 −5 −4 −3 −2 −1−1 f(x) =

1 x+2

−2 −3 −4 −5

1 2 3 4 x

Decreasing on 1- q , 52 and 15, q 2 No relative extrema Asymptotes: x = 5 and y = 0 Concave down on 1- q , 52; concave up on 15, q 2 y-intercept: A 0, - 15 B Decreasing on 1- q , - 22 and 1- 2, q 2 No relative extrema Asymptotes: x = - 2 and y = 0 Concave down on 1- q , - 22; concave up on 1- 2, q 2 y-intercept: A 0, 12 B

1 2 3 4 5 f(x) = −12 x

−8

relative maximum at A - 22, - 222 B ; Asymptotes: x = 0 and y = x Concave down on 1- q , 02; concave up on 10, q 2 No intercepts Decreasing on 1- q , 02; increasing on 10, q 2 No relative extrema x Asymptotes: x = 0 and y = 0 Concave down on 1- q , 02 and 10, q 2 No intercepts

−10

37.

y 10 8 6 4 2 −10 −8 −6 −4 −2 −2

f(x) =

x x+2

2 4 6 8 10 x

−4 −6 −8 −10

39.

y 5 4 3 2 1

y 2 −5 −4 −3 −2 −1 −2

1. x-intercepts: 10, 02, 13, 02, and 1 - 5, 02; y-intercept: 10, 02 2. x-intercepts: 10, 02, 11, 02, and 1- 3, 02; y-intercept: 10, 02

Increasing on 1- q , 02 and 10, q 2 No relative extrema Asymptotes: x = 0 and y = 3 Concave up on 1- q , 02; concave down on 10, q 2 x-intercept: A 13, 0 B

3x − 1 x

−2

4. Left to the student

Increasing on 1- q , 32 and 13, q 2 No relative extrema Asymptotes: x = 3 and y = 0 Concave up on 1- q , 32; concave down on 13, q 2 y-intercept: 10, 12

−3 x−3

2 4 6 8 10 x

31.

A-17

y 0.25 −5 −4 −3 −2 −1 −0.25 −0.5 −0.75

1 2 3 4 5 x f(x) =

−1 x2 + 2

Increasing on 1- q , - 22 and 1- 2, q 2 No relative extrema Asymptotes: x = - 2 and y = 1 Concave up on 1- q , - 22; concave down on 1- 2, q 2 x- and y-intercept: 10, 02 Decreasing on 1 - q , 02; increasing on 10, q 2 Relative minimum at A 0, - 12 B Asymptote: y = 0 Concave up on A - 223, 223 B ; concave down on A - q , - 223 B and

A 223, q B

Inflection points: A - 223, - 38 B

A 223, - 38 B y-intercept: A 0, - 12 B and

A-18 41.

ANSWERS y 3 2 1

−6 −4 −2 −1

Decreasing on 1- q , - 32, 1- 3, 32, and 13, q 2 No relative extrema Asymptotes: x = 3 and y = 0 8 10 x Concave down on 1- q , - 32 and 1- 3, 32; concave up on 13, q 2 y-intercept: A 0, - 13 B Increasing on 1- q , - 22 and 1- 2, q 2 No relative extrema x – 1 f(x) = ——– x+2 Asymptotes: x = - 2 and y = 1 Concave up on 1- q , - 22; concave down on 1- 2, q 2 x-intercept: 11, 02; y-intercept: 2 4 6 8 x A 0, - 12 B

x+3 f(x) = x2 − 9

2 4 6

−2 −3

43.

y 8 6 4 2 1

−8 −6 −4 −2

Increasing on 1- q , - 42 and 1- 4, 02; decreasing on 10, 42 and 14, q 2 Relative maximum at 10, 02 Asymptotes: x = - 4, x = 4, and y = 2 Concave up on 1- q , - 42 and 14, q 2; concave down on 1- 4, 42 x- and y-intercept: 10, 02 y 51. Increasing on 1- q , - 12 and 1 - 1, 02; 3 decreasing on 10, 12 and 11, q 2 2 Relative maximum at 10, - 12 1 Asymptotes: x = - 1, x = 1, and y = 0 1 2 3 x −3 −2 −1 Concave up on 1- q , - 12 and 11, q 2; −1 concave down on 1- 1, 12 1 f(x) = ——– −2 x2 – 1 y-intercept: 10, - 12 −3

2

x +1 f(x) = ——–– x

−4

−6 −5 −4 −3 −2 −1

1 2

x

−10 −20

Increasing on A - q , - 3 - 25 B and A - 3 + 25, q B , or approximately 1- q , - 5.2362 and 1- 0.764, q 2; decreasing on A - 3 - 25, 3 B and A - 3, - 3 + 25 B , or approximately 1- 5.236, - 32 and 1- 3, - 0.7642 Relative maximum at A - 3 - 25, - 6 - 225 B or approximately 1- 5.236, - 10.4722; relative minimum at A - 3 + 25, - 6 + 225 B , or approximately 1- 0.764, - 1.5282 Asymptotes: x = - 3 and y = x - 3 Concave down on 1- q , - 32; concave up on 1- 3, q 2 x-intercepts: 1- 2, 02, 12, 02; y-intercept: A 0, - 43 B y 5 4 3 2 1 −4 −3 −2 −1 −2 −3 −4 −5

1 2 3 4 5 6 x

Decreasing on 1- q , - 12, 1- 1, 32, and 13, q 2 No relative extrema Asymptotes: x = 3 and y = 0 Concave down on 1- q , - 12 and 1- 1, 32; concave up on 13, q 2 y-intercept: A 0, - 13 B

x+1 f(x) = 2 x − 2x − 3

49.

2

4

6 x

−4

10

47.

2

−2

20

x2 – 4 f(x) = ——– x+3

4

−6 −4 −2

y

45.

y

53.

−2

y 7 6 5 4 3 2 1 −7−6 −5 −4 −3−2 −1 −1 −2 −3 −4 −5 −6 −7

2x 2 f(x) = ——–— x 2 – 16

1 2 3 4 5 6 7 x

Increasing on 1- q , - 12 and 11, q 2; decreasing on 1- 1, 02 and 10, 12 Relative maximum at 1- 1, - 22; relative minimum at 11, 22 Asymptotes: x = 0 and y = x Concave down on 1- q , 02; concave up on 10, q 2 No intercepts 55.

y 10 8 6 4 2 −10 −8 −6 −4 −2−2

f(x) =

x2 − 9 x−3

2 4 6 8 10 x

−4 −6 −8 −10

Increasing on 1- q , 32 and 13, q 2 No relative extrema No asymptotes No concavity x-intercept: 1- 3, 02; y-intercept: 10, 32 - 2x x2 - 2 57. f 1x2 = 59. g1x2 = 2 x - 2 x - 1 x - 9 61. h1x2 = 2 x + x - 6 63. (a) $50, $37.24, $32.64, $26.37; (b) maximum = 50 at t = 0; (c) V(t) 2 50 40 30 20 10

25t V(t) = 50 – ——— (t + 2)2

5 10 15 20 25 30 35 40 45 50 55 60 65 70 t

65. (a) $480, $600, $2400, $4800;

(d)

(b) 30, 1002

A-19

ANSWERS

(c)

C(p) $20,000 18,000 16,000 14,000 12,000 10,000 8,000 6,000 4,000 2,000

(d) 48,000 C(p) = ——–– 100 – p

20

40

60

80

100

67. (a) $1.22, $0.79, $0.47; 69. (a)

p

(b) 36.8 yr after 1970;

(c) 0

n

9

6

3

1

2>3

1>3

E

4.00

6.00

12.00

36.00

54.00

108.00

(b) lim E1n2 = q . The pitcher gives up one or more runs but n:0 gets no one out (0 innings pitched). (c) E = 2.00; pitcher gave up an average of 2 earned runs per game (9 innings). 71. 73. Does not exist 75. - q 77. 32 79. - q 81. x f (x) = 83. x 3 + 2x 2 − 15x 2

f (x) =

x2 + 1

x 2 − 5x − 14

35

−8

8 −10

85.

−2 1 −2 x

f (x) =

|

5

20 −10

|

87.

4

−4 −1

89. (a)

(b) f ¿1x2 =

x 2 - 10x + 1 ; critical 1x 2 + x - 622

values: ' 0.101 and ' 9.899; (c)

; (d)

; (e)

Technology Connection, p. 253

1. On 3- 2, 14, absolute minimum is - 8 at x = - 2, and absolute maximum is 2.185 at x = - 0.333; on 3- 1, 24, absolute minimum is 1 at x = - 1 and x = 1, and absolute maximum is 4 at x = 2

Technology Connection, p. 254 1. Absolute minimum: - 4 at x = 2; no absolute maximum

15. Absolute maximum: 4 at x = - 1; absolute minimum: - 12 1 at x = 3 17. Absolute maximum: 16 5 at x = - 5 ; absolute minimum: - 48 at x = 3 19. Absolute maximum: 50 at x = 5; absolute minimum: - 4 at x = 2 21. Absolute maximum: 2 at 23. Absolute x = - 1; absolute minimum: - 110 at x = - 5 maximum: 513 at x = - 8; absolute minimum: - 511 at x = 8 25. Absolute maximum: 17 at x = 1; absolute minimum: - 15 at x = - 3 27. Absolute maximum: 32 at x = - 2; absolute 3 minimum: - 27 29. Absolute maximum: 13 at x = - 2 16 at x = 2 and x = 2; absolute minimum: 4 at x = - 1 and x = 1 31. Absolute maximum: - 1 at x = 5; absolute minimum: - 5 at 1 at x = 20; absolute x = - 3 33. Absolute maximum: 20 20 minimum: 2 at x = 1 35. Absolute maximum: 45 at x = - 2 and x = 2; absolute minimum: 0 at x = 0 37. Absolute maximum: 3 at x = 26; absolute minimum: - 1 at x = - 2 39–47. Left to the student 49. Absolute maximum: 36 at x = 6 51. Absolute minimum: 70 at x = 10 53. Absolute maximum: 31 at x = 12 55. Absolute maximum: 900 at x = 30 57. Absolute maximum: 223 at x = - 23; absolute minimum: - 223 at x = 23 59. Absolute maximum: 5700 at x = 2400 61. Absolute minimum: - 55 13 at x = 1 63. Absolute maximum: 2000 at x = 20; absolute minimum: 0 at x = 0 and x = 30 65. Absolute minimum: 24 at x = 6 67. Absolute minimum: 108 at x = 6 69. Absolute maximum: 3 at 71. Absolute x = - 1; absolute minimum: - 38 at x = 12 maximum: 2 at x = 8; absolute minimum: 0 at x = 0 73. No absolute maximum or minimum 75. Absolute maximum: - 1 at x = 1; absolute minimum: - 5 at x = - 1 77. No absolute maximum: absolute minimum: - 5 at x = - 1 79. Absolute maximum: 1 at x = - 1 and x = 1: absolute minimum: 0 at x = 0 81. No absolute maximum or minimum 83. Absolute maximum: - 10 3 + 223 at x = 2 - 23; absolute 10 minimum: - 3 - 223 at x = 2 + 23 85. No absolute maximum; absolute minimum: - 1 at x = - 1 and x = 1 87–95. Left to the student 97. 1430 units; 25 yr of service 99. 1986 101. 1999; 37.4 billion barrels 103. (a) P1x2 = - 12 x 2 + 400x - 5000; (b) 400 items 105. About 1.26 at x = 19 cc, or about 0.11 cc 109. Absolute maxima: 1 at 107. Absolute maximum: x = 0 and 1 at x = 2; 3 at x = 1; absolute minimum: - 15 at absolute minimum: x = -4 - 5 at x = - 3 y

y

4

−4

2 −4

2

4

x

x

−2

−10

−4 −15

1. No absolute maximum; absolute minimum: 6.325 at x = 0.316 111. (a)

y 6

1. (a) 55 mph (b) 5 mph (c) 25 mpg 3. Absolute maximum: 5 14 at x = 12; absolute minimum: 3 at x = 2 5. Absolute maximum: 4 at x = 2; absolute minimum: 1 at 1 7. Absolute maximum: 86 x = - 1 and x = 1 27 at x = - 3 ; absolute minimum: 2 at x = - 1 9. Absolute maximum: 8 at x = 3; absolute minimum: - 17 at x = - 2 11. Absolute maximum: 15 at x = - 2; absolute minimum: - 13 at x = 5 13. Absolute maximum: - 5 for - 1 … x … 1; absolute minimum: - 5 for - 1 … x … 1

2 −5

−2

Technology Connection, p. 256 Exercise Set 2.4, p. 257

−2

4 2 −2

2 −2

4

6

8 x

(b) Absolute maximum: 6 at x = - 2; (c) f 1x2 = 2 over [0, 4] 113. Absolute maximum: 326 at x = 3; absolute minimum: - 2 at x = - 2 115. Minimum: $20,000 at x = 7 “quality units”

A-20

ANSWERS

117. 119. (a) 2.755 billion barrels in 1981; (b) 0.0765 billion barrels> yr; 0.1014 billion barrels> yr 121. No absolute maximum; absolute minimum: 0 at x = 1 123. (a) P1t2 = t + 8.857; P172 = 15.857 mm Hg; (b) P1t2 = 0.117t 4 - 1.520t 3 + 6.193t 2 - 7.018t + 10.009; P172 = 24.86 mm Hg; P10.7652 = 7.62 mm Hg is the smallest contraction

Technology Connection, p. 263 1.

x

y = 20 - x

A = x120 - x2

0

20

0

4

16

64

6.5

13.5

87.75

8

12

96

10

10

100

12

8

96

13.2

6.8

89.76

20

0

2. Left to the student

0 3. Maximum: 100 at x = 10

maximum area = 182.25 ft2 17. Dimensions: 33 13 cm by 7 33 13 cm by 8 13 cm; maximum volume = 9259 27 cm3 19. Dimensions: 5 in. by 5 in. by 2.5 in.; minimum surface area = 75 in2 21. Dimensions: 2.08 yd by 4.16 yd by 1.387 yd 23. $1048; 46 units 25. $19; 70 units 27. $5481; 1667 units 29. (a) R1x2 = x1150 - 0.5x2; (b) P1x2 = - 0.75x 2 + 150x - 4000; (c) 100 suits; (d) $3500; (e) $100> suit 31. $12.75>ticket; 57,500 people 33. 25 trees>acre 35. (a) q1x2 = 3.13 - 0.04x; (b) $39.13 37. 4 ft by 4 ft by 20 ft 39. Order 5 times>yr; lot size is 20. 41. Order 12 times>yr; lot size is 60. 43. Order 8 times>yr; lot size is 32. 45. r L 3.414 in., h L 6.828 in. 47. r L 2.879 in., h L 9.598 in. 49. 14 in. by 14 in. by 28 in., 3 51. x L 3.36 ft, y L 3.36 ft 53. 20.1, or approximately 0.4642 55. 9% 57. S is 3.25 mi downshore from A. x¿ # C¿1x2 - C1x2 bp 59. x = 61. (a) A¿1x2 = ; a + b x2 # x 0 C¿1x 02 - C1x 02 (b) A¿1x 02 = 0 = ; solving for C¿1x 02, x 20 C1x 02 we get C¿1x 02 = 63. x = - 22, y = 0, = A1x 02. x0 Q = - 322 L - 4.24

65. Order 25 times; lot size: 100 units

Technology Connection, p. 279 (top) 1. P1x2 = - 3 + 40x - 0.5x 2; R1402 = 1200, C1402 = 403, P1402 = 797, R¿1402 = 10, C¿1402 = 10, P¿1402 = 0; marginal cost is constant.

Technology Connection, p. 264

Technology Connection, p. 279 (bottom)

1.

1. P¿1502 = $140>unit; P1512 - P1502 = $217>unit

8 - 2x

4x 3 - 32x 2 + 64x

0

8

0

0.5

7

24.5

1.0

6

36

1.5

5

37.5

2.0

4

32

2.5

3

22.5

3.0

2

12

3.5

1

3.5

4.0

0

0

x

Exercise Set 2.6, p. 285

2. Left to the student

3. Maximum: about 37.9 at x L 1.33

Technology Connection, p. 271 1. Left to the student

2. Minimum: $23,500 at x = 100, yes

Exercise Set 2.5, p. 273 1. Maximum Q = 625; x = 25, y = 25 3. 5. Minimum product = - 4; x = 2, y = - 2 7. Maximum Q = 14; x = 12, y = 212 9. Minimum Q = 30; x = 3, y = 2 11. Maximum Q = 2113; x = 2, y = 10 23 13. Maximum area = 4050 yd2; width is 45 yd, and length (parallel to shoreline) is 90 yd 15. x = 13.5 ft, y = 13.5 ft;

1. (a) P1x2 = - 0.001x 2 + 3.8x - 60; (b) R11002 = $500, C11002 = $190, P11002 = $310; (c) R¿1x2 = 5; C¿1x2 = 0.002x + 1.2; P¿1x2 = - 0.002x + 3.8; (d) R¿11002 = $5, C¿11002 = $1.40, P¿11002 = $3.60; (e) 3. (a) $1234.38; (b) $24.52; (c) $24.38; (d) $48.75; (e) $1283.13 5. (a) $1799; (b) $235.88; (c) $75.40; (d) R1712 = $1874.40, R1722 = $1949.80, R1732 = $2025.20 7. (a) $4572.78; (b) $594.03; (c) $593.63; (d) $5166.41 9. If the price increases from $1000 to $1001, sales will decrease by 100 units. 11. $2.01; $2.00 13. $2; $2 15. (a) P1x2 = - 0.01x 2 + 1.4x - 30; (b) - $0.01; $0 17. (a) dD>dp = 0.021p 2 - p + 150; (b) 3547 units; (c) (d) 19. - $0.01 21. $491.03 billion 23. 25. About $0.21 paid in taxes per dollar earned 27. 0.0401; 0.04 29. 0.2816; 0.28 31. - 0.556; - 1 33. 6; 6 35. 5.1 37. 10.1 39. 10.017 1 1 41. dx 43. 9x 2 22x 3 + 1 dx 45. dx 51x + 2724>5 22x + 1 47. 14x 3 - 6x 2 + 10x + 32 dx 49. 3.1 51. 7.2 53. 657.00 55. - 0.01345 m2 57. The concentration 10 changes more from 1 hr to 1.1 hr 59. = 1.59 ft 2p 2 61. (a) dA = 628 ft (b) 3 extra cans (c) $90 32x 63. R¿1x2 = 100 65. R¿1x2 = 500 - 2x 2 67. R¿1x2 = 5 69.

ANSWERS

Exercise Set 2.7, p. 292 -y 1 4x 8 -x x 3 ; - 2 3. 2 ; 5. ; 7. ; y A2 9 2x 4 2y 2 9y 6x 2 - 2xy 36 3x - 2y 2 1 - y 1 1 ;;;9. 11. 13. 2 2xy 12 x + 2 9 23 x - 3y 2 2 2 -y - 3xy - 2y 3x x 15. 17. 19. 4 21. x y 5y 4x 2y + 3x -p 2 - p -p - 4 3 23. 2 25. 27. 29. 31. - 34 3x x - 2 x + 3 3p + 1 33. $400>day, $80>day, $320>day 35. $16>day, $8>day, $8>day 37. - 1.18 sales>day 39. - 21,830 mi2>yr 41. Decreasing by dV dR 43. (a) 0.0256 m2>month = 952.38R ; dt dt 1 2 (b) 0.0143 mm>sec 45. - 2 ft>sec 47. 494.8 cm3>week 12 x11 - y 22 - y3 2x 49. 3 51. , or x y1x 2 + 122 y11 + x 22 5x 4 - 31x - y22 - 31x + y22 6y 2 - 5x 4 + 6x 2 53. , or 2 2 4 31x + y2 - 31x - y2 - 5y y15y 3 - 12x2 2 2 3 - 61y - xy + x 2 2x1y - x 32 55. 57. 59. 12y - x23 y5 x 3 = y 2 (2 − x ) x4 = y2 + x6 63. 61. 2

16. Relative minimum: - 4 at x = 1; relative maximum: 4 at x = - 1 y

1.

8

0.5

−1

−2

1

4

4 3 2 1

–8x f(x) = ——– x2 + 1

−5−4−3−2−1 −1 −2 −3 −4

1 2 3 4 5 x

17. No relative extrema y 8 7 6 5 4 3 2 1 −1 −1 −2

−8

1

−1 −1

1 2 3 x

f(x) = 4 − 3x − x2

15. Relative minima: 2 at x = - 1 and 2 at x = 1; relative maximum: 3 at x = 0 f(x) = x4 – 2x 2 + 3 y 8 7 6 5 4 3 2 1 −2 −1

1

2

2 x

f(x) = 3x 2/3

y 8 7 6 5 4 3 2 1 −4 −3−2 −1

−2 −3 −4

1

19. Relative minimum: 0 at x = 0

1. (g) 2. (e) 3. (f) 4. (a) 5. (b) 6. (d) 7. (c) 8. False 9. False 10. True 11. False 12. False 3 13. False 14. Relative maximum: 25 4 at x = - 2

−7 −6 −5 −4 −3−2 −1−1

3 x

y 10 9 8 f(x) = x 3 + x 2 – x + 3 7 6 5 4 3 2 1

Chapter Review Exercises, p. 301

y 6 5 4 3 2 1

2

1 18. Relative minimum: 76 27 at x = 3 ; relative maximum: 4 at x = - 1

−2

−0.5

f(x) = 4 + (x – 1)3

1 2 3 4 x

20. Relative maximum: 17 at x = - 1; relative minimum: - 10 at x = 2 y 40 30 20 10 −3 −2 −1 −10 −20 −30 −40

f(x) = 2x 3 – 3x 2 – 12x + 10

1 2 3 4 x

21. Relative maximum: 4 at x = - 1; relative minimum: 0 at x = 1 f(x) = x 3 – 3x + 2 x

y 5 4 3 2 1

−3−2 −1 −1

1 2 3 x

A-21

A-22

ANSWERS

22.

No relative extrema Inflection point at 1- 3, - 72 Increasing on 1- q , q 2 Concave down on 1- q , - 32; concave up on 1- 3, q 2

y 8 6 4 2 −7 −6 −5 −4 −3−2 −1−2

1 2 3 x

−4 −6 −8 −10 −12

f(x) =

23.

1 3 x + 3x 2 + 9x + 2 3

f(x) =

Relative minimum: - 17 at x = 5 Decreasing on 1- q , 52; increasing on 15, q 2 Concave up on 1- q , q 2 2 4 6 8 10 12 14 16 x

−8 −12 −16 −20

f(x) = x2 − 10x + 8

Relative minimum: - 35 at x = 2; relative maximum: 19 at x = - 1 Inflection point at A 12, - 8 B Increasing on 1- q , - 12 and 12, q 2; decreasing on 1- 1, 22 Concave down on A - q , 12 B ; concave up on A 12, q B

y 24 18 12 6

24.

−3 −2 −1 −6 −12 −18 −24 −30 −36

1

2

3

4 x

f(x) = 4x3 − 6x2 − 24x + 5

25.

y 5 4 3 1 −1

−1 −2

1

2 x

f(x) = x4 − 2x2

1 5 3 4 4 3 x + x − x +8 5 4 3

Relative minimum: 457 60 at x = 1; relative maximum: 1208 15 at x = - 4 Inflection points at 1- 2.932, 53.7012, 10, 82, and 10.682, 7.7692 Increasing on 1- q , - 42 and 11, q 2; decreasing on 1- 4, 12 Concave down on 1- q , - 2.9322 and 10, 0.6822; concave up on 1- 2.932, 02 and 10.682, q 2 y 28. 10 8 6 4 2 −5 −4 −3 −2 −1 −2 −4 −6

Concave up on A - q , - 213 B and

23 B

A 213, q B ;

1

f(x) =

2

5 4 3 2 1

−10 −8 −6 −4 −2−1

1 1

2 x

−2 f(x) = 3x4 + 2x3 − 3x2 + 1

1 Relative minima: - 1 at x = - 1 and 11 16 at x = 2 ; relative maximum: 1 at x = 0 Inflection points at 1- 0.608, - 0.1472 and 10.274, 0.8332 Increasing on 1- 1, 02 and A 12, q B ; decreasing on 1- q , - 12 and A 0, 12 B Concave down on 1- 0.608, 0.2742; concave up on 1- q , - 0.6082 and 10.274, q 2

x x−2

No relative extrema Decreasing on 1- q , 22 and 12, q 2 Concave down on 1- q , 22; concave up on 12, q 2 Asymptotes: x = 2 and y = 1 x-intercept: 10, 02; y-intercept: 10, 02 y 30.

3

−1

2x + 5 x+1

1 2 3 4 5 6 x

4

−1

f(x) =

−2 −3 −4 −5

5

−2

1 2 3 4 5 x

No relative extrema Decreasing on 1- q , - 12 and 1- 1, q 2 Concave down on 1- q , - 12; concave up on 1- 1, q 2 Asymptotes: x = - 1 and y = 2 x-intercept: A - 52, 0 B ; y-intercept: 10, 52 y 29.

−4 −3 −2 −1−1

Relative minima: - 1 at x = - 1 and - 1 at x = 1; relative maximum: 0 at x = 0 Inflection points at A - 213, - 59 B and A 213, - 59 B Increasing on 1- 1, 02 and 11, q 2; decreasing on 1- q , - 12 and 10, 12 concave down on A - 213, y 26.

1 2 3 4 x

5 4 3 2 1

2

−2

y 80 70 60 50 40 30 20 10 −6 −5 −4 −3 −2 −1 −10 −20

y 20 16 12 8 4 −4 −2−4

27.

2 4 6 8 10 x

−2 −3 −4 −5

f(x) =

5 x2 − 16

5 Relative maximum at A 0, - 16 B Decreasing on 10, 42 and 14, q 2; increasing on 1- q , - 42 and 1- 4, 02 Concave down on1- 4, 42; concave up on 1- q ,- 42 and 14, q 2 Asymptotes: x = - 4, x = 4, and y = 0 5 y-intercept: A 0, - 16 B

ANSWERS

31.

y 5 4 3 2 1 −4 −3 −2 −1−1

1 2 3 4 5 6 x

−2 −3 −4 −5

f(x) = −

x+1 x2 − x − 2

No relative extrema Increasing on 1- q , - 12, 1- 1, 22, and 12, q 2 Concave up on 1- q , - 12 and 1- 1, 22; concave down on 12, q 2 Asymptotes; x = 2 and y = 0 y-intercept: A 0, 12 B y 32. 2 x – 2x + 2 f(x) = ——–——– x–1

6 5 4 3 2 1

−4 −3−2 −1 −1 −2 −3 −4 −5 −6

1 2 3 4 x

A-23

minimum: 0 at x = 3 52. Absolute maxima: 4 at x = 2 and x = 6; absolute minimum: - 2 at x = - 2 3x 5 - 21x - y23 - 21x + y23 53. 54. Relative maximum 21x + y23 - 21x - y23 - 3y 5 at 10, 02; relative minima at 1- 9, - 94772 and 115, - 37,1252 3x + 3 55. f 1x2 = (answers may vary) 56. Relative maxima at x + 2 1- 1.714, 37.4452; relative minimum at 11.714, - 37.4452 57. Relative maximum at 10, 1.082; relative minima at 1- 3, - 12 and 13, - 12 58. (a) Linear: y = 6.998187602x - 124.6183581 Quadratic: y = 0.0439274846x 2 + 2.881202838x - 53.51475166 Cubic: y = - 0.0033441547x 3 + 0.4795643605x 2 - 11.35931622x + 5.276985809 Quartic: y = - 0.00005539834x 4 + 0.0067192294x 3 - 0.0996735857x 2 - 0.8409991942x - 0.246072967 (b) The quartic function best fits the data. (c) The domain is [26, 102]. Very few women outside of the age range from 26 to 102 years old develop breast cancer. (d) Maximum: 466 per 100,000 women at x = 79.0 years old

Chapter 2 Test, p. 303

Relative minimum at 12, 22; relative maximum at 10, - 22 Decreasing on 10, 12 and 11, 22; increasing on 1- q , 02 and 12, q 2 Concave down on 1- q , 12; concave up on 11, q 2 Asymptotes: x = 1 and y = x - 1 y-intercept: 10, - 22 33. y 10 8 6 4 2 −10 −8 −6 −4 −2−2 −4 −6 −8 −10

2 4 6 8 10 x f(x) =

x2 + 3 x

Relative minimum at A 23, 223 B ; relative maximum at A - 23, - 223 B Decreasing on A - 23, 0 B and A 0, 23 B ; increasing on A - q , - 23 B and A 23, q B Concave down on 1- q , 02; concave up on 10, q 2 Asymptotes: x = 0 and y = x No intercepts 34. Absolute maximum: 66 at x = 3; absolute minimum: 2 at 16 x = 1 35. Absolute maximum: 75 23 27 at x = 3 ; absolute minima: 0 at x = 0 and x = 8 36. No absolute maxima; absolute minimum: 1022 at x = 522 37. No absolute maxima; absolute minima: 0 at x = - 1 and x = 1 38. 30 and 30 39. Q = - 1 when x = - 1 and y = - 1 40. Maximum profit is $451 when 30 units are produced and sold. 41. 10 ft by 10 ft by 25 ft 42. Order 12 times per year with a lot size of 30 43. (a) $108; (b) $1>dinner; (c) $109 44. ¢y = - 0.335, 45. (a) 16x 2 + 12 dx; (b) 0.25 46. 9.111 dy = - 0.35 2 3y 2x 4 47. dV = ;240,000 ft3 48. 49. - 1.75 ft>sec ; 2y 2 + 3x 5 50. $600>day, $450>day, $150>day 51. No maximum; absolute

1. [2.1, 2.2] Relative minimum: - 9 at x = 2 Decreasing on 1- q , 22 increasing on 12, q 2 f(x) = x 2 – 4x – 5 y 1 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

1 2 3 4 5 x

2. [2.1, 2.2] Relative minimum: 2 at x = - 1; relative maximum: 6 at x = 1 Decreasing on 1- q , - 12 and 11, q 2; increasing on 1- 1, 12 y 8 7 f(x) = 4 + 3x − x3 6 5 4 3 2 1 −3 −2 −1 −1 −2 −3

1

2

3 x

3. [2.1, 2.2] Relative minimum: - 4 at x = 2 Decreasing on 1- q , 22; increasing on 12, q 2 y 3 f(x) = (x − 2)2/3 − 4 2 1 −8 −6 −4 −2 −1 −2 −3 −4 −5

2 4 6 8 10 12 x

A-24

ANSWERS

4. [2.1, 2.2] Relative maximum: 4 at x = 0 Increasing on 1- q , 02; decreasing on 10, q 2 y 4 3 2 1

16 f(x) = ——– x2 + 4

−4−3−2 −1

5. [2.3]

10. [2.3]

−5 −4 −3 −2 −1−1

1 2 3 4 x

f(x) =

y 5 4 3 2 1 1

2 x

2

–6

4 x

3

No relative extrema Inflection point: 12, 32 y 8. [2.3] f(x) = x 9 – x2

5 4 3 2 1

−3 −2 −1 −1 −2 −3 −4 −5

1 2 3 x

Relative maximum: 92 at x = at x = - 292 Inflection point: 10, 02 y 9. [2.3]

22; relative minimum: - 2 9

20 2 f(x) = ——– 15 x – 1 10 5 −1

−5 −10 −15 −20

1

−3

−2

−1

1

−1 −2 −3 −4 −5 −6 −7

15 10 5

10 7.5 5 2.5

1

1 3

2

3

x

No relative extrema Asymptotes: x = 0 and y = x y 12. [2.3]

Relative maximum: 1 at x = 0; relative minima: - 1 at x = - 1 and x = 1 Inflection points: A - 213, - 19 B and A 213, - 19 B y 7. [2.3] f(x) = (x – 2)3 + 3

–1 –2.5 –5 –7.5 –10

7 6 5 4 3 2 1

x2 – 1 f(x) = ——–– x

Relative maximum: 2 at x = - 1; relative minimum: 22 27 at x = Inflection point: A - 13, 38 27 B 6. [2.3] f(x) = 2x4 – 4x 2 + 1

−1 −2

−8 x2 − 4

Relative minimum: 2 at x = 0 Asymptotes: x = - 2, x = 2, and y = 0 y 11. [2.3]

1 2 3 x −3 −2−1 −1 −2 −3

−1

1 2 3 4 5 x

−2 −3 −4 −5

y 8 7 6 5 4 f(x) = x 3 + x 2 – x + 1 3 2 1

−2

y 5 4 3 2 1

2

No relative extrema Asymptotes: x = 1 and y = 0

3 x

9

–4

–2

–5 –10 –15

f(x) = x – 3 x+2

2

4 x

No relative extrema Asymptotes: x = - 2 and y = 1 13. [2.4] Absolute maximum: 9 at x = 3; no absolute minimum 14. [2.4] Absolute maximum: 2 at x = - 1; absolute minimum: - 1 at x = - 2 15. [2.4] Absolute maximum: 28.49 at x = 4.3; no absolute minimum 16. [2.4] Absolute maximum: 7 at x = - 1; absolute minimum: 3 at x = 1 17. [2.4] There are no absolute extrema. 18. [2.4] Absolute minimum: - 13 12 at 1 19. [2.4] Absolute minimum: 48 at x = 4 x = 6 20. [2.5] 4 and - 4 21. [2.5] Q = 50 for x = 5 and y = - 5 22. [2.5] Maximum profit: $24,980; 500 units 23. [2.5] Dimensions: 40 in. by 40 in. by 10 in; maximum volume: 16,000 in3 24. [2.5] Order 35 times per year; lot size, 35 25. [2.6] ¢y = 1.01; f ¿1x2 ¢x = 1 26. [2.6] 7.0714 - x2 1 x 27. [2.6] (a) dx; (b) 0.00756 28. [2.7] 2 ; 2 4 y 2x + 3 29. [2.6] dV = ;1413 cm3 30. [2.7] - 0.96 ft>sec 2 2>3 3 31. [2.4] Absolute maximum: L 0.529 at x = 22; absolute 3 minimum: 0 at x = 0 32. [2.5] 10,000 units 33. [2.4] Absolute minimum: 0 at x = 0; relative maximum: 25.103 at x = 1.084; relative minimum: 8.625 at x = 2.95 34. [2.4] Relative minimum: - 0.186 at x = 0.775; relative maximum: 0.186 at x = - 0.775 35. [2.1, 2.2] (a) Linear: y = - 0.7707142857x + 12691.60714 Quadratic: y = - 0.9998904762x 2 + 299.1964286x + 192.9761905 Cubic: y = 0.000084x 3 - 1.037690476x 2 + 303.3964286x + 129.9761905

A-25

ANSWERS

Quartic: y = - 0.000001966061x 4 + 0.0012636364x 3 - 1.256063636x 2 + 315.8247403x + 66.78138528 (b) Since the number of bowling balls sold cannot be negative, the domain is 30, 3004. This is supported by both the quadratic model and the raw data. The cubic and quartic models can also be used but are more complicated. (c) Based on the quadratic function, the maximum value is 22,575 bowling balls. The company should spend $150,000 on advertising.

Exercise Set 3.1, p. 319 y

1.

3.

16 y = (0.25)

12 y = 4x

8 4 –1

1

2

x

Extended Technology Application, p. 306 1. (a)

(b) 4500; (c) 20,250

25

0

2. (a)

−5−4−3−2−1

5.

y 10 9 8 7 6 5 4 3 2 1

12 0

Xscl = 1, Yscl = 5

(b) 60,000; (c) 90,000

200

−5−4−3−2−1

9.

0

3. (a)

Xscl = 25, Yscl = 25

(b) 50,000; (c) 25,000

33. 0

250 0

4. (a)

Xscl = 50, Yscl = 50

5. (a)

2000

(b) 30,513; (c) 205,923

250

50 0

Xscl = 10, Yscl = 50 3

6. (a) y = - 0.0011P + 0.0715P 2 - 0.0338P + 4 (b) 50 (c) 33,841

0 Xscl = 5, Yscl = 5

Chapter 3 Technology Connection, p. 308 1. 156.993

2. 16.242

3. 0.064

Technology Connection, p. 312 Left to the student

1 2 3 4 5 x

3x

35. e x1x 2 + 5x - 62 + 7x

41. - xe -x >2

ex

2

-2x

10 9 8 7 6 5 4 3 2 1 −5−4−3−2−1

4. 0.000114

f(x) = e2x

1 2 3 4 5 x

No critical values No inflection points Increasing on 1- q , q 2 Concave up on 1- q , q 2 y 59. 10 9 8 7 6 5 4 3 2 1

70 0

−5−4−3−2−1

-x

1 2 3 4 5 x

x5

()

−5−4−3−2−1

-2x

37. 43.

e x1x - 42 x5 e

2x - 7

22x - 7

47. - 2xe + e - e + 3x 2 49. e -x 22e x - 1 2 2 51. ke -kx 53. 14x 2 + 3x2e x - 7x12x - 72 + 18x + 32e x - 7x, 3 2 x2 - 7x or 18x - 22x - 13x + 32e y y 55. 57.

Xscl = 200, Yscl = 200

0

x

2e 1x - 22

g(x) = 23 x

11. - e 13. 3e 15. 6e x 17. - 7e -7x 19. 8e 4x 21. 3e -x 23. - 52 e -5x 4x # x2 25. 27. 15e 5x e 3 29. 5x 4 - 12e 6x 31. 5x 4 # e 2x + 2x 5 # e 2x

f(x) = (2.5)

39. 1- 2x + 72e -x 45.

0

()

2

(b) 400,000; (c) 400,000

2000

0

f(x) = 32 x

1 2 3 4 5 x

−5−4−3−2−1 2x

1 2 3 4 5 x

y 10 9 8 7 6 5 4 3 2 1

7.

y 10 9 8 7 6 5 4 3 2 1

200

0 150

x

y 10 9 8 7 6 5 4 3 2 1

-x

10 9 8 7 6 5 4 3 2 1 −5−4−3−2−1

1 2 3 4 5 x

No critical values No inflection points Decreasing on 1- q , q 2 Concave up on 1- q , q 2

(1/2)x

1 2 3 4 5 x

No critical values No inflection points Increasing on 1- q , q 2 Concave up on 1- q , q 2 y 61. 1

−4 −3 −2 −1−1

f(x) = 12 e−x

g(x) = e

−2 −3 −4 −5 −6 −7 −8 −9

1 2 3 4 5 6 x

F(x) = −e(1/3)x

No critical values No inflection points Decreasing on 1- q , q 2 Concave down on 1- q , q 2

A-26

ANSWERS

63.

y 3 2 1 −5 −4 −3 −2 −1−1

g(x) = 2(1 − e−x) x≥0 1 2 3 4 5 x

−2 −3

No critical values on 30, q 2 No inflection points Increasing on 30, q 2 Concave down on 30, q 2

Technology Connection, p. 323 Left to the student

Technology Connection, p. 325 1. Graph is left to the student; function values are 1000, 5, 0.699, 3. 2. Left to the student; graph is obtained by entering y = log x>log 2.

65–73. Left to the student 75. 1 77. y = - x + 1 79. Left to the student 81. (a) $1.6 billion, $2.7 billion; (b) 15 yr 83. (a) C¿1t2 = 50e -t; (b) $50 million> yr; (c) $916,000> yr; (d) 85. (a) 113,000; (b) q (c) q¿1x2 = - 0.72e -0.003x; 240 220 200 180 160 140 120 100 80 60 40 20

(d)

Technology Connection, p. 327 1. t = 6.9 2. x = - 4.1 5. x = 38.7 Left to the student

q = 240e−0.003x

Exercise Set 3.2, p. 334

80

C(t) = 10t 2 e –t

160 240 320

x

(e)

1 2 3 4 5 6 7 8 9 10 t

89. 15e 3x1e 3x + 124 e 2x

91. - e -t - 3e 3t

93.

1x 2 - 2x + 12e x 1x 2 + 122

1 x>2 x 4 e 97. e x>2 a 99. x b 2 1e + e -x22 22x 22x - 1 101. 2; 2.25; 2.48832; 2.59374; 2.71692 103. 4e -2 L 0.5413, for x = 2 105. Left to the student 109. 107. f (x) = f '(x) = f "(x) = e x f(x) = x 2 e − x 95.

+

4

1

−3 −2

3

7

0

−1

Relative minimum at 10, 02; relative maximum at 12, 0.54132 111. f(x) = 2e 0.3 x

f '(x) = 0.6e 0.3x 10

10

−10

10

−10

10 −2

−2

113.

f "(x) = 0.18e 0.3 x

(

1 f (x) = 1 + −− x

10

)

x

3

−10

10 −2

4. x = 46.2

Technology Connection, p. 328

87. (a) 0 ppm, 3.7 ppm, 5.4 ppm, 4.5 ppm, 0.05 ppm; (b) C (c) C¿1t2 = 10te -t12 - t2; 6 (d) 5.4 ppm at t = 2 hr 5 4 3 2 1

3. t = 74.9

0

0

40

1. 2 3 = 8 3. 81>3 = 2 5. a J = K 7. 10-p = h 9. ln b = M 11. log 10 100 = 2 13. log 10 0.1 = - 1 15. log M V = p 17. 0.51 19. 2.708 21. 2.609 23. 2.9957 25. 0.2231 27. 2.6094 29. 3 31. - 1.3863 33. - 0.6094 35. 8.681690 37. - 4.006334 39. 8.999619 41. t L 4.382 43. t L 3.454 45. t L 2.303 8 47. t L 140.671 49. 51. x 3 + 41ln x2x 3 - x x 1 1 - 4 ln x 2 53. 55. x + 2x ln 17x2 57. 59. x x x5 213x + 12 x2 + 7 2e x x 61. 2 63. 65. + 2e ln x x 3x + 2x - 1 x1x 2 - 72 41ln x23 ex 1 67. x 69. 71. x e + 1 x ln 18x2 ln 15x2 + ln 13x2 73. 75. y = 8.455x - 11.94 x 77. y = 0.732x - 0.990 79. (a) 2000 units; 500 (b) N¿1a2 = , N¿1102 = 50 units per $1000 spent on a advertising; (c) minimum is 2000 units; (d) 81. 58 days (d) 83. (a) $58.69, $78.00; (b) V¿1t2 = 63.8e -1.1t; (c) 2.7 months; (d) 85. (a) P¿1x2 = 1.7 - 0.3 ln x;

(b)

(c) 289.069

87. (a) 68%; (b) 35.8%; (c) 3.6%; - 20 (d) 5.3%; (e) S¿1t2 = ; (f) maximum = 68%, and t + 1 minimum approaches 0%; (g) 89. (a) 2.4 ft>s; (b) 3.4 ft>s; ln 1P>P02 712t - 12 0.37 (c) v¿1p2 = ; (d) 91. t = 93. p k t1t - 12 1 1 -1 95. 97. , x ln 13x2 # ln 1ln 13x22 1 - t 1 + t 1 -2 x or 99. 101. 2 11 - t211 + t2 x ln 5 x + 5 1 1 4 103. x ln x 105. , or 2x A 1 - 2x B A 1 + 2x B 2x A 1 - x B 107. Definition of logarithm; Product Rule for exponents; definition of logarithm; substitution 109. Definition of logarithm; if a = b, then a c = b c; Power Rule for exponents; definition of logarithm; substitution and the commutative law for multiplication 111. 1 113. e p 115. 0 117. and 119. Left to the student 121. Minimum: - e -1 L - 0.368

ANSWERS

Technology Connection, p. 341 1. 8.26 billion 2. 13.22 billion 3. 15.46 billion 4. 21.15 billion 5. y = 9689.99111.03476232x = 6. $16,741; $18,548; $39,336 9689.991e 0.0341717x

Exercise Set 3.3, p. 347

1. f 1x2 = ce 4x 3. A1t2 = ce -9t 5. Q1t2 = ce kt 0.046t 7. (a) N1t2 = 112,000e ; (b) N1402 = 705,212; (c) 15.1 yr 9. (a) P1t2 = P0e 0.059t; (b) $1060.78, $1125.24; (c) 11.7 yr 11. (a) G1t2 = 4.7e 0.093t; (b) 48.07 billion gallons; (c) 7.5 yr 13. 4.62% 15. 6.9 yr after 2006 17. 11.2 yr; $102,256.88 19. $7,500; 8.3 yr 21. (a) k = 0.151, or 15.1%, V1t2 = 30,000e 0.151t; (b) $549,188,702; (c) 4.6 yr; (d) 69 yr 23. (a) E1t2 = 1.031e 0.047101t; (b) $3.347 billion; (c) after about 48.3 yr, or in 2038 25. (a) y = 136.3939183 # 1.071842825x, y = 136.3939183e 0.0693794334x, and exponential growth rate = 0.069, or 6.9%; (b) 444 million, 628 million; (c) 18.7 yr; (d) 10 yr 27. Approximately $8.6 billion 29. (a) S1t2 = 4e 0.05t; (b) 5% per yr; (c) 51¢, 60¢, 69¢ (d) For the years 2010–2020, the total cost of Forever Stamps is 11 * $4400, or $48,400. For the years 2010–2012, the cost of regular first-class stamps is 3 * $4400, or $13,200. If the price of a regular postage stamp increases to 51¢ in 2013, the cost of postage for the years 2013–2015 would be 3 * $0.51 * 10,000, or $15,300. If the price increases to 60¢ in 2016, the cost for the years 2016–2018 would be 3 * $0.60 * 10,000, or $18,000. If the price increases to 69¢ in 2019, the cost for the years 2019–2020 would be 2 * $0.69 * 10,000, or $13,800. Thus, the total cost of regular first-class stamps for the years 2010–2020 would be $13,200 + $15,300 + $18,000 + $13,800, for a total of $60,300. Thus, by buying Forever Stamps, the firm would save $60,300 - $48,400, or $11,900. (e) 31. (a) 2%; (b) 3.8%, 7%, (d) P(x) 100 21.6%, 50.2%, 93.1%, 98%; 90 80 637e -0.13x 70 (c) P¿1x2 = -0.13x 2 100 60 11 + 49e 2 P(x) = ——–———– –0.13 50 40 30 20 10

1 + 49e

x

10 20 30 40 50 60 x

33. (a) V1t2 = 0.10e ; (b) $9,486,828; (c) 3.09 yr; (d) after 87.1 yr, or in 2025 35. 2019 37. 1%>yr 39. 4%>yr 41. B1t2 = 190e 0.035t; 886 bears 43. (a) 1000, 1375, 1836, (c) P(t) 6000 3510, 5315, 5771; 5500 -0.4t 5000 11,051.36e 4500 (b) P¿1t2 = 4000 11 + 4.78e -0.4t22 0.224t

3500 3000 2500 2000 1500 1000 500

P(t) =

5780 1 + 4.78e−0.4t

2 4 6 8 10 12 14 16 18 20 t

45. N1t2 = 48,869e growth rate = 3.78%

, where t0 = 1930; exponential

0.0378t

47. (a) 0%, 33%, 55%, 70%, 86%, 99.2%, 99.8%; (b) at 7 months, the percentage of doctors who are prescribing the medication is growing by 2.4% per month

(c)

P(t) 100 90 80 70 60 50 40 30 20 10

A-27

P(t) = 100(1 – e –0.4t)

2 4 6 8 10 12 14 16 t

49. (a) N1t2 =

29.47232081 ; 1 + 79.56767122e -0.809743969t

(b) 29 students; (c)

N(t) =

29.47232081 1 + 79.56767122e−0.809743969t

30

0

15 0

(d) N¿1t2 =

1898.885181e -0.809743969t ; (e) 11 + 79.56767122e -0.809743969t22

57. ln 4 = kT4 59. 2 yr 61. 7.57% 63. 9% ln 1y2>y12 65. k = 67. 69. (a) R102 = $2 million; this t2 - t1 represents the initial revenue of the corporation at its inception. (b) lim R1t2 = $4000 million = Rmax; this represents the upper

51–55.

t: q

limit of the revenue of the company over all time. It is never actually attained. (c) t = 24

Exercise Set 3.4, p. 360 1. (a) N1t2 = N0e -0.096t; (b) 341 g; (c) 7.2 days 3. (a) A1t2 = A0e -kt; (b) 11 hr 5. 23.1%>min

7. 22 yr 9. 42.9 g 11. 4223 yr 13. 25 days 15. 3965 yr 17. $13,858.23 19. $6,393,134 21. $42,863.76 23. (a) $40,000; (b) $5413.41; (c) 25. (a) 0.022, 0.031, 0.069; (b) Q(t) 0.10 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01

Q(t) = (Q0 − 0.00055)e 0.163 t + 0.00055 1 2 3 4 5 6 7 8 9 10 t

27. (a) N1t2 = 5,650,000e -0.018t; (b) 1,953,564 farms, 1,753,573 farms; (c) about 2046 29. (a) B1t2 = 64.6e -0.0068t; (b) 58.3 lb; (c) 2172 31. (a) P1t2 = 51.9e -0.0091t; (b) 43.3 million; (c) after 434 yr, or in 2429 33. (a) 27; (b) 0.05878; (c) 83°; (d) 28.7 min; (e) 35. The murder was committed at 7 P.M. 37. (a) 145 lb; (b) - 1.2 lb>day 39. (a) 11.2 W; (b) 173 days; (c) 402 days; (d) 50 W; (e) 41. (c) 43. (e) 45. (f) 47. (d) 49. (a) 51. x = $166.16, q = 292 printers 53. and 55.

A-28

ANSWERS

Exercise Set 3.5, p. 368

# ln 5.4 # 15.42x + 3x 215.42x 11. 1ln 32 # 3x + 1 # 14x 32

1. 1ln 727 3. 1ln 828 5. x 4 4 7. 1ln 72 # 7x + 2 # 4x 3 9. 8e 8x 1 1 5 13. # 15. # 17. x ln 4 x ln 17 15x + 12 ln 6 6 3x 2 + 1 2 19. 21. 3 23. 16x - 72 ln 10 1x + x2 ln 8 A x - 22x B ln 7 6x 1 25. # 27. 51log 12 x24 a + 6x # ln 6 # log 7 x b x ln 7 x ln 12 x# x 14x + 127 ln 7 - 4 # 7 29. 14x + 122 x

x

3

6 # 52x - 1 3 + 1ln 5252x - 1 # 6x 2 # log 16x + 52 16x + 521ln 102 7x # 9 # 1log 4 x28 33. 1ln 727x # 1log 4 x29 + x # ln 4 13x 5 + x25 35. 513x 5 + x24115x 4 + 12 # 1log 3 x2 + ln 3 # x 37. (a) V¿1t2 = 52001ln 0.80210.802t; (b) 39. (a) $19.84 trillion; (b) ; (c) 41. (a) 0.82; (b) - 0.015>yr; (c) 43. 8.8 45. (a) I = I01010; (b) I = I010; (c) the power mower is 109 times louder than a just audible sound; (d) dI>dL = I0100.1L1ln 10210.12; (e) dL 10 47. (a) 49. ln 3 L 1.0986 = ; (b) dI 1ln 102I 1 4 51. 1ln 222 x # 4x 3 53. ln 3 # log x # ln 10 # x 3

31.

Chapter Review Exercises, p. 382 1. (b) 2. (e) 3. (f) 4. (c) 5. (a) 7. False 8. True 9. True 10. False

6. (d) 11. True 1 12. False 13. True 14. False 15. True 16. x 3 22x 4x e 1 17. e x 18. 4 19. 20. 21. 3x 4e 3x + 4x 3e 3x 2x x + 5 2x 2 ex 1 1 - 3 ln x x2 22. 23. 24. 4e 4x 1ln 4x2 + 2xe x x x4 1 - x 8 1 7 x 25. 8x 26. 27. 1ln 929 28. x ex 1ln 22x 13x22 29. 3x1ln 321log 4 12x + 122 + 12x + 121ln 42 y 30. 31. y 12

f(x) = 4 x

g(x) = 13 x

()

8 4 −1

1

2

x

−5−4−3−2−1

g1x2 # f ¿1x2 f 1x2

+ g¿1x2 # ln 1 f 1x22b # 3 f 1x24g1x2

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

59.

Technology Connection, p. 373 1. E1x2 =

x ; R1x2 = 300x - x 2 300 - x

Exercise Set 3.6, p. 376 x 5 ; (b) , inelastic; (c) $200 400 - x 11 x 3. (a) E1x2 = ; (b) 11.5, elastic; (c) $25 50 - x 5. (a) E1x2 = 1; (b) 1, unit elasticity; (c) total revenue is independent of x. x 7. (a) E1x2 = ; (b) 0.10, inelastic; (c) $400 21600 - x2 9. (a) E1x2 = 0.25x; (b) 2.5, elastic; (c) $4 2x 11. (a) E1x2 = ; (b) 0.5, inelastic; (c) $3 x + 3 25x 13. (a) E1x2 = ; (b) approximately 19¢; 967 - 25x (c) prices greater than 19¢; (d) prices less than 19¢; (e) approximately 19¢; (f) decrease 81 3x 3 15. (a) E1x2 = ; (b) ; (c) increase 346 21200 - x 32 17. (a) E1x2 = n; (b) no; (c) yes, at n = 1 19. E1x2 = 1- x2L¿1x2

2 4 6 8 10 12 14 16 t

2x ; (b) 0.4, inelastic; x + 4 -8 (c) 1.5, elastic; (d) decrease; (e) $4 49. 2x 1e - e -2x22 1 50. 52. 0 L 0 51. e1/x f (x) = 1024e 1/x 2 47. $186,373.98

1. (a) E1x2 =

21.

p(t) = 1 – e –0.7t

2. Left to the student

3. $150

1 2 3 4 5 x

32. 6.93 33. - 3.2698 34. 8.7601 35. 3.2698 36. 2.54995 37. - 3.6602 38. Q1t2 = 25e 7t 39. 4.3% 40. 10.2 yr 41. (a) C1t2 = 15.81e 0.024t; (b) $29.51, $35.75 42. (a) N1t2 = 60e 0.12t; (b) 123 franchises; (c) 5.8 yr after 2007 43. 5.3 yr 44. 18.2% 45. (a) A1t2 = 800e -0.07t; (b) 197 g; (c) 9.9 days 46. (a) 0.50, 0.75, 0.97, 0.999, 0.9999; (b) p¿1t2 = 0.7e -0.7t; (c) (d) p

55. ln a # a f 1x2 # f ¿1x2 57. a

10 9 8 7 6 5 4 3 2 1

16

48. (a) E1x2 =

(1 + e ) 0.5

−4

4 −0.1

53. (a) y = 9.033880177 # 1.431864118x, y = 9.033880177e 0.3589771744x, 0.3589771744; (b) $671.0 billion, $24.3 trillion; (c) 10.56 yr; (d) 1.93 yr

Chapter 3 Test, p. 384 1. [3.1] 6e 3x 4. [3.2]

1 x

2. [3.2]

41ln x23 x

5. [3.1] e x - 15x 2

7. [3.5] 1ln 727x + 1ln 323x

3. [3.1] - 2xe -x

2

6. [3.1, 3.2]

8. [3.5]

1 1ln 142x

3e x + 3e x # ln x x 9. [3.2] 1.0674

ANSWERS

10. [3.2] 0.5554 11. [3.2] 0.4057 12. [3.3] M1t2 = 2e 6t 13. [3.3] 23.1% 14. [3.3] 10.0 yr 15. [3.3] (a) C1t2 = 3.22e 0.021t; (b) $3.65, $4.14 16. [3.4] (a) A1t2 = 3e -0.1t; (b) 1.1 cc; (c) 6.9 hr 17. [3.4] About 16.47 centuries, or 1647 yr 18. [3.4] 4.0773%>sec 19. [3.3] (a) 4%; (b) 5.2%, 14.5%, 40.7%, 73.5%, 91.8%, 672e -0.28t 99.5%, 99.9%; (c) P¿1t2 = ; (d) ; 11 + 24e -0.28t22

(e)

P(t) 100 90 80 70 60 50 40 30 20 10

100 P(t) = ——–———– 1 + 24e –0.28t

10

20

30

40

A-29

eventually the total revenue does not change. From the logistic function, it would be about $251.1 million, but the table shows about $254 million. 6. From the logistic function, it would be about $745 million, but the table shows about $762 million.

Chapter 4 Technology Connection, p. 395 1. (a) Left to the student; (b) 400; (c) the area is the square of x; (d) A1x2 = x 2; (e) A1x2 is the antiderivative of f 1x2. 2. (a) Left to the student; (b) 60; (c) the area is 3 times x; (d) A1x2 = 3x; (e) A1x2 is the antiderivative of f 1x2 . 3. (a) Left to the student; (b) 800; (c) the area is the cube of x; (d) A1x2 = x 3; (e) A1x2 is the antiderivative of f 1x2. Exercise Set 4.1, p. 396

t

20. [3.4] $10,043,738 21. [3.6] (a) E1x2 = 0.2x; (b) 0.6, inelastic; (c) 3.6, elastic; (d) increase; (e) $5 256 22. [3.2] 1ln x22 23. [3.1] Maximum is 4 L 4.689; e minimum is 0 24. [3.1] e x − e −x f(x) =

e x + e −x 2

−4

4

−2

25. [3.1] 0 26. [3.3] (a) y = 686996.582 # 1.086747476x, y = 686996.582 # e 0.0831892684x; (b) $4.28 million, $5.50 million; (c) 87.5 yr; (d) 8.3 yr; (e)

Extended Technology Application, p. 387 1. Linear: R1t2 = - 5.2444t + 52.3333 Quadratic: R1t2 = 0.756x 2 - 13.5603x + 68.965 Cubic: R1t2 = - 0.2107x 3 + 4.2327x 2 - 29.5954x + 87.044 Exponential: R1t2 = 68.455210.78682x The linear and cubic functions both reach R = 0 too fast. The value from the quadratic function decreases as we might expect over time, but then it makes a dramatic rise. Most movies do not have this revenue pattern. The exponential function shows a steady decrease and approaches 0 as a limit, but never reaches it. It is a reasonable assumption that G gets smaller and smaller over time. Eventually, box office revenue reaches 0. Gross revenue thereafter comes from DVD rentals, TV rights, and electronic outlets, such as iTunes. 2. G = $4.90, $3.85, $3.03, $2.39, 3. R = $239.79, $1.88, $1.48, $1.16, $0.91, all in millions $243.64, $246.67, $249.06, $250.94, $252.43, $253.58, $254.49, all in millions. There are costs, such as marketing and shipping costs, associated with distributing a movie to theaters. Eventually, movie executives want the jump in revenue that comes with DVD and electronic rentals. 251.1 4. R1t2 = 1 + 3.4687e -0.4183t 364.3354 5. R¿1t2 = , which represents the rate of 11 + 3.4687e -0.4183t22 change of the total revenue; lim R¿1t2 = 0, which means that t: q

1.

x7 + C 7

3. 2x + C

9. 23 t 3 + 52 t 2 - 3t + C

5. 45 x 5>4 + C 11. -

7. 13 x 3 + 12 x 2 - x + C

x -2 + C 2

13. 34 x 4>3 + C

x -3 + C 19. ln x + C 3 -1 21. 3 ln x - 5x + C 23. - 21x 1>3 + C 25. e 2x + C 27. 13 e 3x + C 29. 17 e 7x + C 31. 53 e 3x + C 2 -9x 33. 34 e 8x + C 35. - 27 e + C 37. 53 x 3 - 27 e 7x + C 3 x 39. - x 3>2 - 3x -1>3 + C 41. 3x 3 + 6x 2 + 4x + C 3 2 5x 43. 3 ln x - 52 e 2x + 29 x 9>2 + C 45. 14x 1>2 - 15 e - 8 ln x + C 1 2 1 3 47. f 1x2 = 2 x - 3x + 13 49. f 1x2 = 3 x - 4x + 7 51. f 1x2 = 53 x 3 + 32 x 2 - 7x + 9 53. f 1x2 = x 3 - 52 x 2 + x + 4 55. f 1x2 = 52 e 2x - 2 57. f 1x2 = 8x 1>2 - 13 59. D1t2 = - 270.1t 3 + 865.15t 2 + 3648t + 41,267 x4 x3 61. C1x2 = - x 2 + 7000 63. (a) R1x2 = - 3x; (b) 4 3 4000 65. D1x2 = + 3 67. (a) E1t2 = 32 + 30t - 5t 2; x (b) E132 = 77%, E152 = 57% 69. (a) I1t2 = - 3.17t 2 + 141.6t + 1408; (b) 930 people; (c) 1522 people; (d) 348 people 71. (a) h1t2 = - 16t 2 + 75t + 30; (b) h122 = 116 ft, h¿122 = 75 11 ft>sec; (c) t = L 2.344 sec; (d) h12.3442 L 117.89 ft; 32 (e) 5.06 sec; (f) h¿15.062 = - 86.92 ft>sec 20 6 16 5 7 73. f 1t2 = 23 t 3>2 + 2t 1>2 - 28 75. 25 3 7 t + 3 t + 5 t + C 2 x 77. 23 t 3>2 - 25 t 5>2 + C 79. - 6 ln x + 72 x -2 + C 2 1 # 2 81. ln x + C, or log x + C 83. 3x 4 - 83 x 3 - 17 2 x ln 10 x2 5x + C 85. - x + C 87. 2 15. 27 x 7>2 + C

17. -

Exercise Set 4.2, p. 407 1. $1060

3. 46,800¢, or $468

5. - $255,000

7. $8400

6

9. 23,302.4¢, or $233.02 4

15. a f 1x i2 i=1

15

11. $471.96

17. a G1x i2 i=1

13. a 3i i=1

19. 2 1 + 2 2 + 2 3 + 2 4 , or 30

21. f 1x 12 + f 1x 22 + f 1x 32 + f 1x 42 + f 1x 52

23. (a) 1.4914;

A-30

ANSWERS

(b) 1.1418 25. 3,166,250¢, or $31,662.50 27. 247.68 29. 124 31. 4 33. 12 35. 92 37. 25 39. 8 41. 1.0016 43. 37.96 A Exact area is 12 # 25p. B Technology Connection, p. 413 32 9 2. 3 4 5. 313.24 1.

3.

5 - ln 6 L 0.535 6

4. ' 1.59359

2. 13.75

3. 0.535

4. 27.972

41 23

1 4

10 23

1. 8 3. 8 5. 7. 9. 11. e - 1 L 19.086 13. 3 ln 6 L 5.375 15. Total cost, in dollars, for t days 17. Total number of kilowatts used in t hours 19. Total revenue, in dollars, for x units produced 21. Total amount of the drug, in milligrams, in v cubic centimeters of blood 23. Total number of words memorized in t minutes 25. 4 27. 9 56 29. 12 31. e 5 - e -1 , or approximately 148.045 33. 35. 0; the area above the x-axis is the same as the area below the x-axis. 37. 0; the area above the x-axis is the same as the area below it. 39–42. Left to the student 43. 40 45. 53 47. 637 6 3 b - a3 2 -5 49. e - e , or approximately 7.382 51. 6 e 2b - e 2a e2 + 1 53. 55. , or approximately 4.195 2 2 8 57. 3 59. $628.56 61. $29.13 63. (a) $2948.26; (b) $2913.90 65. $7627.28 billion 67. 18.69 hr; 20.12 hr 69. 7 words 71. About 5 words 73. s1t2 = t 3 + 4 3 t 75. v1t2 = 2t 2 + 20 77. s1t2 = - + 3t 2 + 6t + 10 3 79. (a) 104.17 m; (b) 229.17 m 81. (a) 60 mph; (b) 18 mi 83. (a) 16.67 km> hr; (b) 0.1875 km 85. s1t2 = - 16t 2 + v0t + s0 87. 41 mi 89. 148 mi 7 91. On the 10th day 93. 3.5 95. 359 15 97. 6.75 1 2 99. 30 101. 5 3 103. 14 3 105. 107. 4068.789 109. 7.571 111. 9.524 113. 10.987 3

Technology Connection, p. 429 2. Left to the student

Technology Connection, p. 432 1. 16

65. 16

67. 6

69. 4

1. 4.673 5. - 260

Exercise Set 4.3, p. 421

1. 43

8 (b) - 10°; (c) 46.25° 63. 40 15 71. 4 73. 5.886 75. 0.237

Technology Connection, p. 441

Technology Connection, p. 418 1. 0

(b) $220,155.66 51. $26,534.37 53. $32,781.35 55. (a) Ben; (b) 2 more words; (c) 0.7 word per minute; (d) 0.9 word per minute 57. (a) 90 words per minute; (b) 96 words per minute, at t = 1 min; (c) 70 words per minute 59. (a) 42.03 g>mL; (b) 22.44 g>mL 61. (a) 31.7°;

Over the interval 30, 24, the areas under f 1x2 = x 4 and yav = 3.2 are equal.

f (x) = x 4 and yav = 3.2

Exercise Set 4.5, p. 443

1 1 1. 16 18 + x 326 + C 3. 16 1x 2 - 628 + C 5. 24 13t 4 + 222 + C 1 1 3x 4 7. ln 12x + 12 + C 9. 4 1ln x2 + C 11. 3 e + C 5 2 13. 3e x>3 + C 15. 15 e x + C 17. - 12 e -t + C 19. 12 ln 15 + 2x2 + C 21. 13 ln 112 + 3x2 + C 1 23. - ln 11 - x2 + C 25. 12 1t 2 - 126 + C 1 4 3 2 8 27. 8 1x + x + x 2 + C 29. ln 14 + e x2 + C 1 31. 1ln x22 + C 33. ln 1ln x2 + C 35. 1ax 2 + b23>2 + C 3a P0 1 37. e kt + C 39. + C 41. 56 11 + 6x 226>5 + C k 2412 - x 426 43. e - 1 45. 21 47. ln 5 49. ln 19 51. 1 - e -b 4 1640 53. 1 - e -mb 55. 208 57. 6561 59. 315 3 8 61. Left to the student 63. 32 x - 34 ln 12x + 12 + C 65. x + 5 ln 1x - 22 + C 1 1 67. 13 1x + 1213 - 16 1x + 1212 + 11 1x + 1211 + C 69. 27 1x - 227>2 + 85 1x - 225>2 + 83 1x - 223>2 + C 1500 71. D1x2 = 2000225 - x 2 + 5000 73. P1x2 = 2 x - 6x + 10 1 1 2t 75. 5 3 77. ln 1ax + b2 + C 79. 2e + C a 1 81. 100 1ln x2100 + C 83. 12 1e t + 222 + C 85. 4912 + t 323>4 + C 87. 13 1ln x23 + 32 1ln x22 + 4 ln x + C 9 89. 18 3ln 1t 4 + 8242 + C 91. x + + C x + 3 93. t - 4 - ln 1t - 42 + C , or t - ln 1t - 42 + K , where 1ln x2n + 1 + C K = - 4 + C 95. - ln 11 + e2-x + C 97. n + 1 1 5 99. ln 11 - ae -mx2 + C 101. 12x 3 - 72n + 1 + C am 61n + 12

Technology Connection, p. 449 1. 1.941

Exercise Set 4.6, p. 452 0

2

−1

Exercise Set 4.4, p. 433 1. 22

3. 18 16

5. 89 11 12

7. 5

9. 72

11. x = - 3 and x = 3

3 ; 229 , or approximately x = - 1.193 and x = 4.193 2 15. x = - 3 and x = 5 17. 32 19. 62.5 21. 41 23. 4 12 3 5 1 2 2 25. 20 6 27. 4 2 29. 10 31. 41 3 33. 10 3 35. 3 37. 85 13 39. 83 41. - e -1 + 1, or approximately 0.632 2n + 1 - 1 43. 16 45. 2a + 5 47. 49. (a) $2,201,556.58; 3 n + 1 13. x =

x6 1 5x + C 5. 15 xe 5x - 25 e + C 2 3 3 x x ln x 7. - 12 xe -2x - 14 e -2x + C 9. + C 3 9 11. 14 x 2 ln x - 18 x 2 + C 13. 1x + 52 ln 1x + 52 - x + C x2 x2 15. a + 2xb ln x - 2x + C 2 4 x2 x2 17. a - xb ln x + x + C 2 4 2 4 3>2 19. 3 x1x + 22 - 15 1x + 225>2 + C x 4 ln 2 x 4 ln x x4 x4 x4 21. ln 12x2 + C , or + + C 4 16 4 4 16 1 2 2x 1 2 x x x 2x 23. x e - 2xe + 2e + C 25. 2 x e - 2 xe + 14 e 2x + C 27. - 12 x 3e -2x - 34 x 2e -2x - 34 xe -2x - 38 e -2x + C 1. xe 4x - 14 e 4x + C

3.

ANSWERS 8 8 3x 29. 31 1x 4 + 42 e 3x - 49 x 3e 3x + 49 x 2e 3x - 27 xe 3x + 81 e + C 8 7 31. 3 ln 2 - 9 33. 14 ln 14 - 10 ln 10 - 4 35. 1 5>2 37. 1192 39. C1x2 = 83 x1x + 323>2 - 16 15 15 1x + 32 41. (a) - 10Te -T - 10e -T + 10; (b) about 9.084 kW-h 2 2 43. 125 15x + 125>2 - 75 15x + 123>2 + C; they are the same. x x 45. 21xe 1 - 2e 1 + C 47. 23 x 3>2 ln x - 49 x 3>2 + C 49. 21x 1ln x2 - 41x + C 4 51. 27 127x 3 + 83x - 2213x + 827>6 - 91 181x 2 + 832 1296 2592 13>6 19>6 13x + 82 + 1729 x13x + 82 - 43,22513x + 8225>6 + C n+1 n+1 x 2x 2x n + 1 53. 1ln x22 ln x + + C n + 1 1n + 122 1n + 123 n x n-1 55. Let u = x and dv = e dx. Then du = nx dx and v = e x . Next, use integration by parts. 57. 59. About 355,986

Exercise Set 4.7, p. 457 1.

- 19 e -3x13x

+ 12 + C

1 10

ln `

Chapter Review Exercises, p. 466 1. True 2. False 3. True 4. False 5. (e) 6. (d) 7. (a) 8. (f) 9. (b) 10. (c) 11. $77,000 12. 4x 5 + C 13. 3e x + 2x + C 14. t 3 + 52 t 2 + ln t + C 15. 9 16. 21 17. Total number of words keyboarded in t minutes 18. Total sales in t days b 6 - a6 19. 20. - 25 21. e - 12 22. 2 ln 4, or 4 ln 2 6 23. 22 24. Zero 25. Negative 26. Positive 3 27. 13 12 28. 14 e x + C 29. ln 14t 6 + 32 + C 1 2 -3x 2 30. 41ln 4x2 + C 31. - 3 e 32. xe 3x + C 2x x3 33. 34. x 3 ln x + x ln x 2>3 + C + C 3 3 4

35. e 3x a 13 x 4 1 ln 14 1 38. + 49 36.

`

7 + 7 x 7

4 3 9x

+

4 2 9x

-

54. 56. 57.

59.

5 + x ` + C 7. 3 - x - 3 ln ƒ 3 - x ƒ + C 5 - x 1 x 1 9. + ln ` ` + C 818 - x2 64 8 - x x5 x5 11. 1ln 32 x + x ln x - x + C 13. 1ln x2 + C 5 25 x4 x4 15. 1ln x2 + C 17. ln ƒ x + 2x 2 + 7 ƒ + C 4 16 x - 1>2 2 2 x 5 19. + ln ` ` + C 21. - ln ` ` + C 5 - 7x 5 5 - 7x 4 x + 1>2 23. m 2m 2 + 4 + 4 ln ƒ m + 2m 2 + 4 ƒ + C 5 5 25. 2 1ln x2 + + C 27. x 3e x - 3x 2e x + 6xe x 2x 4x 2 1 - 6e x + C 29. 15 13x - 1211 + 2x23>2 + C 20 31. S1x2 = 100c + ln 120 - x2 d 20 - x x 1 x -1 33. - 4 ln ` ` + C 35. + ln ` ` + C 3x - 2 21x - 22 4 x - 2 -3 37. -x + ln ƒ e -x - 3 ƒ + C e - 3 5.

52.

58.

6x 3. + C ln 6

8 27 x

+

8 81 b

x ` + C 37. 15 x 2e 5x x 1 ln ƒ 7x + 1 ƒ + C 49

1 3x 3e

+ C

+ C 2 5x 25 xe

+

2 5x 125 e

+ C

1 ln x b + C 7 49 1 8x 42. About $70,666.67 64 e 18x - 12 + C 1 -2 or approximately 0.297 44. 80 mi 11 3e 2, 2 About $162,753.79 10x 3e 0.1x - 300x 2e 0.1x + 6000xe 0.1x - 60,000e 0.1x + C 2 ln ƒ 4t 3 + 7 ƒ + C 48. 75 15x - 8224 + 5x + C 1 x5 e + C 50. ln 1x + 92 + C 51. 96 1t 8 + 3212 + C 2 x2 x x ln 17x2 - x + C 53. ln 18x2 + C 2 4 1 5 2 55. - 12 ln 11 + 2e -x2 + C 10 C ln ƒ t + 3 ƒ D + C 1ln 1x22 + C, or 14 1ln x22 + C 1 ln x b + C x 92 a 92 8464 1x - 32 ln 1x - 32 - 1x - 42 ln 1x - 42 + C 1 + C 60. 37 1x + 327>3 - 94 1x + 324>3 + C 31ln x23 1 1 1 2 16 12x + 12 - 4 12x + 12 + 8 ln 12x + 12 + C 1.343

39. ln ƒ x + 2x 2 - 36 ƒ + C 41. 43. 45. 46. 47. 49.

61. 62.

A-31

40. x 7 a

Chapter 4 Test, p. 468 1. 34.24 95

2. 34.14

213 3>2 6 x + C 3. 34.14 500 3 x + C 3 8 11>8 4. 34.14 e x + ln x + 11 6. 34.34 4 ln 3 x + C 5. 34.34 16 7. 34.34 Total miles run in t hours 8. 34.34 12 1 - e -2 9. 34.34 10. [4.3] 1 11. [4.4] 61 12. 34.34 Positive 6 2 -0.5x 13. 34.54 ln 1x + 122 + C 14. 34.54 - 2e + C 1t 4 + 3210 1 15. 34.54 + C 16. 34.64 15 xe 5x - 25 + C 40 x4 x4 x4 17. 34.64 ln x 4 + C, or x 4 ln ƒ x ƒ + C 4 4 4 x 2 x 1 18. 34.74 + C 19. 34.74 ln ` ` + C ln 2 7 7 - x 1 20. 34.44 6 21. 34.44 3 22. 34.44 $49,000 23. 34.34 94 words 24. 34.34 5.4 km 25. 34.54 67 ln 15 + 7x2 + C 26. 34.64 x 5e x - 5x 4e x + 20x 3e x - 60x 2e x + 120xe x - 120e x + C 6 27. 34.54 16 e x + C 28. 34.6, 4.74 23 x 3>21ln x2 - 49 x 3>2 + C 8 + x 1 29. 34.74 ln a b + C 16 8 - x 4 -0.1x 30. 34.6, 4.74 - 10x e - 400x 3e -0.1x - 12,000x 2e -0.1x -0.1x - 2,400,000e -0.1x + C - 240,000xe 2 x2 x 31. 34.64 ln 113x2 + C 2 4 1 32. 34.64 15 13x 2 - 821x 2 + 423>2 + C 1ln x24 4 33. 34.54 - 1ln x23 + 5 ln x + C 4 3 34. 34.64 1x + 32 ln 1x + 32 - 1x + 52 ln 1x + 52 + C 108 2 3 35. 34.6, 4.74 18x 3 + 10215x - 422>3 x 15x - 425>3 + 10 125 243 81 x15x - 428>3 15x - 4211>3 + C 625 34,375 2 36. [4.6] 27 13x - 223>2 + 49 13x - 221>2 + C 16 37. [4.6] x + 8 ln x + C x 1 1ln 52x 5x 38. 34.54 e + C, or + C 39. 34.44 16 ln 5 ln 5

A-32

ANSWERS

Extended Technology Application, p. 470 1. (a) 36%; (b) 33.3 2. (a) 16.7%; (b) 55.5 3. (a) f 1x2 = x 1.75, where 0 … x … 1; (b) 0.272, 27.2; (c) ' 59% 4. (a) f 1x2 = x 2.34, where 0 … x … 1; (b) 0.2, 20; (c) ' 15.4%; (d) ' 21.9% 5. Left to the student 6. (a) f 1x2 = x 2.64; (b) ' 20.6% 7. (a) f 1x2 = x 1.86; (b) ' 32.9% 8. (a) 34.3%; (b) f 1x2 = 0.0000763111022.22x, where 0 … x … 1; (c) 0.409; (d) 0.819, 81.9; (e) ' 0.8%; (f) ' 79.2%

Chapter 5 Technology Connection, p. 474

The point of intersection is 12, 92; this is the equilibrium point.

30

1.

S

D

0

Exercise Set 5.3, p. 496 1. Convergent; 12 3. Divergent 5. Convergent; 1 7. Convergent; 21 9. Divergent 11. Convergent; 2 13. Divergent 15. Divergent 17. Divergent 1000 19. Convergent; 1 21. Covergent; 0.001 23. Divergent p 1 25. 2 27. 1 29. $51,428.57 31. $6250 33. $4500 35. $62,500 37. $900,000 39. 33,333 13 lb 41. (a) 4.20963; (b) 0.702858 rem; (c) 2.37551 rems 43. Divergent 45. Convergent; 2 47. Convergent; 12 1 49. 2 is the total dose of the drug. 51. k 53. 55. About 1.252 y = xe − 0.1x 5

5

0

0

Exercise Set 5.1, p. 479

1. (a) 16, $42; (b) $15; (c) $9 3. (a) 11, $92; (b) $3.33; (c) $1.67 5. (a) 13, $92; (b) $36; (c) $18 7. (a) 150, $5002; (b) $12,500; (c) $6250 9. (a) 12, $32; (b) $2; (c) $0.35 11. (a) 1100, $102; (b) $1000; (c) $333.33 13. (a) 10.8, $10.242; (b) $2.22; (c) $0.98 15. (a) 15, $0.612; (b) $86.36; (c) $2.45 17. 8 19. (a) 16, $22; (b) (c) $7.62; (d) $7.20 D

Consumer surplus

12

(b) y = - 2.5x + 22.5; (c) $45; (d) $24.20

20

0

8 0

Exercise Set 5.2, p. 489 1. $119,721.74 3. $235,955.31 5. $83,527.02 7. $173,773.94 9. $2,617,560 11. $19,765,160 13. $3,207,800 15. $7,981,030 17. $380,920 19. $216,192 21. $70,408.74 23. (a) $1,321,610; (b) $7,995,280 25. $160,777.75 27. A: $598,884, B: $601,377; B is the better buy. 29. (a) Crunchers: $2,338,910, Radars: $2,364,760; (b) the difference of the accumulated present values of the two offers, or $25,850 31. (a) $62,144.41; (b) $4796.74 33. $379,358.53 35. (a) $6,080,740; (b) $1,179,540 37. (a) $2,182,290; (b) $538,145; (c) 4%: $688,339, 6%: $582,338, 8%: $498,815, 10%: $432,332 (d) 39. 1712.4 trillion cubic feet 41. 2042 43. (a) approximately 0.0284, or 2.84%; (b) approximately 39.76 billion barrels; (c) approximately 27.7 years after 2006 45. 16.031 lb 47. $535,847 51.

3

1. 3. 5. 7. 9.

0

21. (a) Linear

Exercise Set 5.4, p. 506

S

Producer surplus 0

49. $732,121

100 0

1 x2 3 9 1 x dx = c d = = 1 8 1 8 8 L1 4 1>3 1 3 dx = 33x41>3 = 3a - 0b = 1 0 3 L0 4 3 2 x3 4 1 x dx = c d = 14 3 - 032 = 1 64 0 64 L0 64 e 1 dx = 3ln x4e1 = ln e - ln 1 = 1 - 0 = 1 L1 x 1 1- 123 x3 1 13 1 1 3 2 x dx = c d = = + = 1 2 2 2 2 2 2 L-1 -1 q

11.

L0

3e -3x dx = lim

b: q L 0

b

3e -3x dx = lim 3- e -3x4b0 b: q

= lim 3- e -3b - 1- e 024 = lim a -

1 + 1b = 1 e 3b 3 2 2 13. 21; f 1x2 = 21 x 15. 2; f 1x2 17. 16; f 1x2 = 16 1 1 2 - x 1 19. ; f 1x2 = 21. ; f 1x2 = 2 2 ln 3 x ln 3 1 ex 8 23. 3 25. (a) , or 0.32; (b) ; f 1x2 = 3 25 e - 1 e - 1 11 27. 16, or 0.6875 29. 0.3297 31. 0.999955 33. 0.3935 1 35. (a) k L L 0.000017; (b) 0.02404 59,119.3 37. 0.950213 39. 22 41. 0.049787 43–53. Check using the answers to Exercises 1–11. b: q

b: q = 32 x 2

Technology Connection, p. 516 1. (a) 0.6295, or 62.95%; (b) 0.0228, or 2.28% 2. (a) About 19.6%; (b) about 90.1%

Technology Connection , p. 517 1. x = 0 2. x = - 0.674 3. x = 9.935 4. x = 2.477 5. x = 103.201

Exercise Set 5.5, p. 520 1. m = E1x2 = 5, E1x 22 =

79 2 4 2 ,s = ,s = 3 3 23 8 8 222 3. m = E1x2 = , E1x 22 = 8, s2 = , s = 3 9 3

ANSWERS

13 11 211 , E1x 22 = 5, s2 = ,s = 6 36 6 3 2 3 3 2 7. m = E1x2 = 0, E1x 2 = , s = , s = 5 5 A5 6 27 27 ln 5 - 36 2 2 9. m = E1x2 = , E1x 2 = ,s = , ln 5 ln 5 1ln 522 5. m = E1x2 =

227 ln 5 - 36 11. 0.4834 13. 0.4147 15. 0.6442 ln 5 17. 0.0150 19. 0.1790 21. 0.0013 23. (a) 0.6826; (b) 68.26% 25. 0.2898 27. 0.4514 29–46. Check using the answers to Exercises 11–28. 47. (a) -0.52; (b) 0; (c) 1.645 49. (a) -15.04; (b) -14.44 51. 0.62% 53. 0.7910 55. (a) 982; (b) 1055; (c) 1185 57. (a) 84; (b) 80 59. 90.8th b + a b 2 + ab + a 2 61. m = E1x2 = , E1x 22 = , 2 3 2 1b - a2 b - a s2 = ,s = 12 223 ln 2 63. 22 65. 67. 7.801 oz 69. k

A-33

19. y = 22x 2 + C1, y = - 22x 2 + C1, where C1 = 2C = 212x + C1 and y = - 212x + C1, where C1 = 2C 2 3 = 8e x >2 - 3 25. y = 215x - 3 29. P = C1e 2t, where = C1e 3x, where C1 = ;e C ;e C 31. f 1x2 = ln x - 2x 2 + 23 x 3>2 + 9 120 C1x2 = 2.6x - 0.01x 2 + 120, A1x2 = 2.6 - 0.01x + x 400 (a) P1C2 = - 40; (b) $97 2C + 3 (a) R = k ln 1S + 12 + C; (b) R = k ln 1S + 12; (c) C1 q = 200 - x 41. q = n , where C1 = e C x 1 45. y = 1>x R = C1S k, where C1 = e C e - C 49. y = 13x 2 + Cx

21. y 23. y 27. y C1 = 33.

s =

35. 37. 39. 43. 47.

51. y1 = 210x + C1 and y2 = - 210x + C1, where C1 = 2C

20 C1 = 100 −20

C1 = 5

40 C1 = −200

Exercise Set 5.6, p. 525 p 15p , or about 1.05 3. , or about 23.56 3 2 p 2p 5. 1e 10 - e -42, or about 34,599.06 7. , or about 2.09 2 3 9 9. 4p ln 4 , or about 10.19 11. 32p, or about 100.53 32p 13. , or about 20.11 15. 56p, or about 175.93 5 32p 17. , or about 33.51 19. 1,703,703.7p ft3 3 y 21. The graphs are semicircles. Their rotation about the x-axis y = 4 − x2 y = r2 − x2, r > 0 creates spheres of radius 2 r and r, respectively. 2 23. 2pe 3 , or about 126.20 25. p x − r −2 2 r 1.

Exercise Set 5.7, p. 533 1. y = x 5 + C; y = x 5, y = x 5 - 1, y = x 5 + p (answers may vary) 3. y = 12 e 2x + 12 x 2 + C; y = 12 e 2x + 12 x 2, y = 12 e 2x + 12 x 2 - 3, y = 12 e 2x + 12 x 2 + 3 (answers may vary) 5. y = 8 ln x - 13 x 3 + 16 x 6 + C; y = 8 ln x - 13 x 3 + 16 x 6, y = 8 ln x - 13 x 3 + 16 x 6 + 5, y = 8 ln x - 13 x 3 + 16 x 6 - 17 (answers may vary) 7. y = 13 x 3 + x 2 - 3x + 4 3 5>3 1 2 9. f 1x2 = 5 x 11. y¿ = ln x + 3; y– = 1>x - 2 x - 61 10 1 0 y– x 1 1 0 x x 0 0 13. y¿ = 4e x + 3xe x; y– = 7e x + 3xe x y– 2y¿ + y 0 17e x + 3xe x2 - 214e x + 3xe x2 + 1e x + 3xe x2 0 17e x - 8e x + e x2 + 13xe x - 6xe x + 3xe x2 0 0 4 3 15. y = C1e x , where C1 = ;e C 17. y = 24x 2 + C

−20

Chapter Review Exercises, p. 541 1. (d) 2. (e) 3. (f) 4. (b) 5. (c) 6. (a) 7. False 8. True 9. True 10. False 11. False 12. True 13. 12, $162 14. $18.67 15. $5.33 16. $7195.37 17. $6603.40 18. $25,948.85 19. $7919.65 per yr 20. $639,668.38 21. 16.03 billion metric tons 22. 53.0 yr from 2005 23. Convergent; 1 24. Divergent 8 1 8 25. Convergent; 2 26. k = 3; f 1x2 = 27. 0.6 3x 3 1 3 1 28. 10 29. 21 30. 21 31. 20 32. 225 33. 78.4th percentile 34. 0.4678 35. 0.8827 36. 0.1002 37. 0.5000 38. 0.3085 39. $6199.06 127p p 11 40. 41. 42. y = C1e x , where C1 = ;e C 7 6 43. y = 24x + C1 and y = - 24x + C1, where C1 = 2C 3 44. y = 5e 4x 45. v = 215t + 19 46. y = 23x 2 + C1 and y = - 23x 2 + C1, where C1 = 2C 2 47. y = C1e -x >2 + 8, where C1 = ;e -C 48. q = 100 - x 9 49. V = - 6.37e -kt + 36.37 50. 24.5 51. Divergent 52. Convergent; 3 53. 1

Chapter 5 Test, p. 543

1. [5.1] 13, $162 2. [5.1] $45 3. [5.1] $22.50 4. [5.2] $18,081.81 5. [5.2] $55,766.35 6. [5.2] 1149.5 million metric tons 7. [5.2] 28.89 yr after 2004 8. [5.2] $4273.39 per yr 9. [5.2] $1,906,391.71 10. [5.2] $1403,270.10 11. [5.3] Convergent; 14 12. [5.3] Divergent 13. [5.4] 14, f 1x2 = 14 x 3 on 30, 24 14. [5.4] 0.9975 15. [5.5] E1x2 = 13 16. [5.5] E1x 22 = 5 6 211 6 20. [5.5] 37.5th percentile 21. [5.5] 0.4032 22. [5.5] 0.1365 23. [5.5] 0.9071 24. [5.5] 0.3085 25. [5.5] $14.59 per lb

17. [5.5] m =

13 6

18. [5.5] s2 =

11 36

19. [5.5] s =

A-34

ANSWERS

5p 8 28. [5.7] y = C1e x , where 2 C1 = e C 29. [5.7] y = 218x + C1 and y = - 218x + C1, where C1 = 2C 30. [5.7] y = 11e 6t 3 31. [5.7] y = ;C1e -x >3 + 5, where C1 = ;e -C 4 32. [5.7] v = ; 28t + C1, where C1 = 4C 2 33. [5.7] y = C1e 4x + x >2, where C1 = ;e C C1 34. [5.7] q = 4 , where C1 = e -C x 35. [5.7] (a) V1t2 = 3611 - e -kt2; (b) k = 0.12; (c) V1t2 = 3611 - e -0.12t2; (d) V1122 = $27.47; 4 (e) t L 14.9 months 36. [5.4] 24, or 22 1 37. [5.3] Convergent; - 4 38. [5.3] p 26. [5.6] p ln 5

27. [5.6]

Extended Technology Application, p. 545 1. y = 0.00525488582427x 3 - 0.31949926791313x 2 + 5.2617546608767x - 8.994864719578 2. 12,348.287 cm3 3. y = 0.00109000713314x 4 - 0.02219272885861x 3 + 0.11944088382992x 2 - 0.11585042438606x + 1.0466143607372 4. 35.170535 in3, 19.493 fl. oz 5. It seems good since the volume estimate was 19.493 fl. oz. 6. The cubic function y = 0.00149729174531x 3 0.05193696325118x 2 + 0.34763221627001x + 0.68805240705899 yields an estimated volume of 35.4635 in3, or 19.655 fl. oz, which is a better estimate.

Chapter 6 Exercise Set 6.1, p. 554 1. 0; - 14; 250 3. 1; - 125 5. 9; 66; 128 7. 6; 12 9 ; 23 9. 51x, y2 ƒ y Ú 3x6 11. 51x, y2 ƒ y Ú 06 13. 25.62 15. $151,571.66 17. (a) $165.70; (b) $143.70; (c) for (a), approximately $11,930.40; for (b), approximately $12,070.80; she spends less with option (a). 19. 244.7 mph 21. 1.939 m2 23. (a) 65; (b) 62; (c) about 30%; (d) 25. Drops by approximately 10% 27. 29. - 10°F 31. - 64°F z z 33. 35. z = (x4  16x2)ey

2

z = y2

y y x x

37.

z

z = x3 − 3xy2

x y

Exercise Set 6.2, p. 562 1. 2; - 3; 2; - 3

3. 6x - 2y; - 2x + 1; - 6; 1 5. 2 - 5y; y x -2 -2 - 5x; - 18; - 20 7. ; ; ; 2 2 2 2x + y 2x + y 2 25 213 9. 2e 2x - y; - e 2x - y 11. ye xy; xe xy y 2y x 13. ; + ln 1x + 2y2 15. 1 + ln 1xy2; y x + 2y x + 2y y x 1 1 17. + 19. 1212x + y - 52; 612x + y - 52 ;- 2 y 3x 3x 2 y 2 21. 4m + 10m + 2b - 22; 3m 2 + 8mb + 10b + 26m - 56 23. 5y - 2l; 5x - l; - 12x + y - 82 25. 2x - 10l; 2y - 2l; - 110x + 2y - 42 27. fxx = 0; fyy = 0; fxy = 5; fyx = 5 29. fxx = 0; fyy = 14x; fxy = 14y + 5; fyx = 14y + 5 31. fxx = 20x 3y 4 + 6xy 2; fyy = 12x 5y 2 + 2x 3; 33. 0; 0; 0; 0 fxy = 20x 4y 3 + 6x 2y; fyx = 20x 4y 3 + 6x 2y 35. 4y 2e 2xy; 4xye 2xy + 2e 2xy; 4xye 2xy + 2e 2xy; 4x 2e 2xy -y 1 1 37. 0; 0; 0; e y 39. 2 ; ; ; 0 41. (a) 614,400 units x x x y 3>5 0p 0p x 2>5 (b) = 960a b , = 1440a b ; x y 0x 0y 0p 0p (c) ` = 7680, ` = 360; (d) 0x 132, 10242 0y 132, 10242 43. (a) $1.274 million; 0P (b) = - 0.005075w-1.638r1.038s0.873t2.468, 0w 0P = 0.008257w-0.638r0.038s0.873t2.468, 0r 0P = 0.006945w -0.638r 1.038s -0.127t 2.468, 0s 0P = 0.019633w-0.638r1.038s0.873t1.468; (c) 0t 2w 45. 99.6°F 47. 121.3°F 49. 51. (a) ; 1202h 2h (b) ; (c) - 0.0243 m2 53. 78.244 55. - 0.846 1202w - 4xt 2 4x 2t 57. fx = 2 ;f = 2 2 2 t 1x - t 2 1x - t 222 59. fx =

1 2x11 + 22t2

; ft =

- 1 - 22x 2t11 + 22t22

61. fx = 4x -1>3 - 2x -3>4t 1>2 + 6x -3>2t 3>2; ft = - 4x 1>4t -1>2 - 18x -1>2t 1>2 - 6y -2 2 -2 2 63. fxx = ; fxy = 3 + 3 ; fyx = 3 + 3 ; 4 x y x y x 6x 65. fyy = 4 y - 2x 2 + 2y 2 2x 2 - 2y 2 67. fxx = and fyy = 2 , so fxx + fyy = 0 2 2 2 1x + y 2 1x + y 222 2 2 2 y1h - y 2 x1x - h22 69. (a) lim (b) = y; lim = x; h : 0 h2 + y 2 h : 0 x 2 + h2 (c) fyx10, 02 = 1 and fxy10, 02 = - 1; at 10, 02, the mixed partial derivatives are not equal.

ANSWERS

Exercise Set 6.3, p. 570

1. Relative minimum = A B 3. Relative maximum = A B 5. Relative minimum = - 1 at 11, 12 7. Relative minimum = - 7 at 11, - 22 9. Relative minimum = - 5 at 1 - 1, 22 11. No relative extrema 13. Relative minimum = e at 10, 02 15. 6 thousand of the $17 sunglasses and 5 thousand of the $21 sunglasses 17. Maximum value of P = $55 million when 19. The bottom measures a = $10 million and p = $3 8 ft by 8 ft, and the height is 5 ft. 21. (a) R = 64p1 - 4p 12 - 4p1p2 + 56p2 - 4p 22; (b) p1 = 6, or $60, p2 = 4, or $40; (c) q1 = 32, or 3200 units, q2 = 28, or 2800 units; (d) $304,000 23. No relative extrema; saddle point at 10, 02 25. Relative minimum = 0 at 10, 02; saddle - 13 at - 13, 23 4 2 2 27 at 3 , 3

points at 12, 12 and 1- 2, 12 - 5 at 10, 02

29. Relative minimum =

27.

31. No relative extrema

Technology Connection p. 574 1. x = - 2, y = 3 2. x = - 10, y = 37 3. x = - 13, y = 19

Exercise Set 6.4, p. 576 29 1. y = 0.6x + 0.95 3. y = 36 35 x - 35 5. (a) y = 0.15x + 3.95; (b) $7.70, $8.45 7. (a) y = 0.17x - 261.07; (b) 80.6 yr, 81.5 yr 9. (a) y = 1.07x - 1.24; (b) 85% 11. 13. (a) y = - 0.0059379586x + 15.57191398; (b) 3:38.2, 3:36.4; (c) the value predicted is 3:42.1, about a second less than the actual record.

A-35

Exercise Set 6.6, p. 592 1. 9 3. 14 5. - 10 7. 0 4 13. 4 15. 15 17. 1 19. 23. 39

25.

13 240

27.

9. 5 12

3 20

11. 12 21. 18,000 fireflies

29. Left to the student

Chapter Review Exercises, p. 598 1. (e) 2. (h) 3. (f) 4. (g) 5. (b) 6. (c) 7. (d) 8. (a) 9. 1 10. 3y 3 11. e y + 9xy 2 + 2 12. 9y 2 13. 9y 2 14. 0 15. e y + 18xy 16. D = 51x, y2 ƒ x Z 1, y Ú 26 17. 6x 2 ln y + y 2 3 2 2x 6x 6x 2 18. 19. 20. 21. 12x ln y + 2xy + 2y + 2y y y y - 2x 3 27 22. 23. Relative minimum = - 549 + 2x 20 at A 5, 2 B y2 24. Relative minimum = - 4 at 10, - 22 3 25. Relative maximum = 45 4 at A 2 , - 3 B 20 3 26. Relative minimum = 29 at 1- 1, 22 27. (a) y = x + ; 5 3 (b) 9.1 million 28. (a) y = 17.94x + 124.41; (b) $358 3 29. Minimum = 125 30. Maximum = 300 at 4 at A - 2 , - 3 B 31. Absolute maximum at 13, 0, 92; absolute 15, 102 1 minimum at 10, 2, - 42 32. 45 33. 60 34. About 1533 students 35. 0 36. The cylindrical container z 37. z = x2 + 4y2

Exercise Set 6.5, p. 586

5 1. Maximum = 25 3. Maximum = - 16 at 12, 42 3 at A 3 , 5 B 5. Minimum = 20 at 14, 22 7. Minimum = - 96 at 18, - 122 9. Minimum = 32 at A 1, 12, - 12 B 11. 25 and 25 13. 3 and - 3 1 15. A 32, 2, 52 B 17. 9 34 in. by 9 34 in.; 95 16 in2; no 3 27 19. r = L 1.6 ft, h = 2r L 3.2 ft; about 48.3 ft2 A 2p 21. Maximum value of S is 1012.5 at L = 22.5, M = 67.5 23. (a) C1x, y, z2 = 7xy + 6yz + 6xz; (b) x = 60 ft, y = 60 ft, 25. 10,000 units on A, z = 70 ft; minimum cost is $75,600 100 units on B 27. Absolute maxima at 1- 1, 2, 92 and 11, 2, 92; absolute minimum at 10, 0, 02 29. Absolute maximum at 12, 2, 82; absolute minimum at 10, 6, - 122 31. (a) P1x, y2 = 45x + 50y, 180 acres of celery and 120 acres of lettuce, $14,100 profit; (b) 270 acres of lettuce and zero acres of 7 3 celery, $14,100 profit 33. Minimum = - 155 128 at A - 16 , - 4 B

35. Maximum =

8 27

at A ; 223, ; 223, ; 223 B

37. Maximum = 2 at

A 12, 12, 12, 12 B

43–49. Left to the student

39. l =

py px = c1 c2

x y

Chapter 6 Test, p. 599 1. [6.1] e - 1 - 2 2. [6.2] e x + 6x 2y 3. [6.2] 2x 3 + 1 4. [6.2] e x + 12xy 5. [6.2] 6x 2 6. [6.2] 6x 2 7. [6.2] 0 7 8. [6.3] Minimum = - 16 at A 34, 12 B 9. [6.3] None 9 17 10. [6.4] (a) y = 2 x + 3 ; (b) $24 million 11. [6.5] Maximum = - 19 at 14, 52 12. [6.6] 720 13. [6.5] $400,000 for labor, $200,000 for capital - x 4 + 4xt + 6x 2t - 2x 3 - 2x 2 14. [6.2] fx = ; f = t 1x 3 + 2t22 1x 3 + 2t22 15. [6.1]

z 1 1 z = x − − y2 − − x 3 2 3

41. x

y

A-36

ANSWERS

and 11, q 2, decreasing on 1- 1, 12; concave down on 1- q , 02, y concave up on 10, q 2

Extended Technology Application, p. 601 1. Case

Bldg.

n

1

B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2 B1 B2

2 3 2 3 4 5 5 10 5 10 10 17 10 17 17 26 17 26 26 50

2 3 4 5 6 7 8 9 10

k 40 32.66 60 48.99 40 35.777 60 42.426 150 106.066 40 30.679 80 61.357 40 32.344 50 40.43 77 55.525

A

h

t1h, k2

3200 3200 7200 7200 6400 6400 18000 18000 112500 112500 16000 16000 64000 64000 27200 27200 42500 42500 154154 154152

15 30 15 30 45 60 60 135 60 135 135 240 135 240 240 375 240 375 375 735

21.5 19.4 31.5 27.5 24.5 23.9 36 34.7 81 66.5 33.5 39.3 53.5 54.7 44 53.7 49 57.7 76 101.3

h k h 4. A = a 1 + + b k 2 = 40,000 10 2 12 5. About 57.7 ft by 57.7 ft by 132.2 ft (11 floors) 6. Left to the student 2. Yes

3. t1h, k2 =

Cumulative Review, p. 602 1. [R.4] y = - 4x - 27 y 3. [1.2] (a) 6 5 4 3 2 1

–2 –1 –1 –2 –3 –4

2. [R.2] x 2 + 2xh + h2 - 5 (b) 3; (c) - 3; (d) no

f(x)

1 2 3 4 5 6 7 8 x

4. [1.2] - 8 5. [1.2] 3 6. [1.2] Does not exist 7. [2.3] 4 8. [2.3] 0 9. [1.4] f ¿1x2 = 2x 10. [1.1] f 11. [1.2] b, f 12. [1.4] b, d, f 13. [1.5] - 9 14. [1.5] 2x - 7 4 5 -2>3 15. [1.5] 14 x -3>4 16. [1.5] - 6x -7 17. [1.7] 10 3 x 12x - 82 3 2x 20x - 15x 2 - 8 18. [1.6] 19. [3.2] 2 20. [3.2] 1 12x - 122 x + 5 ex e 2x - 3 21. [3.1] 3e 3x + 2x 22. [3.1] 23. [3.2] x e - 4 22x - 3 y 24. [1.8] 2 - 4x -3 25. [2.5] - $0.04> pair 26. [2.7] 3xy 2 + x 1 27. [3.1] y = x - 2 28. [1.5] x = 1, x = 3 29. [2.2] Relative maximum at 1- 1, 32, relative minimum at 11, - 12; point of inflection at 10, 12; increasing on 1 - q , - 12

5 4 f(x) = x3 – 3x + 1 3 2 1

-5 -4 -3 -2 -1-1

1 2 3 4 5 x

-2 -3 -4 -5

30. [2.2] Relative maxima at 1- 1, - 22 and 11, - 22, relative 22 1 minimum at 10, - 32; points of inflection at a ,b and 9 23 a

1

,-

22 b ; increasing on 1- q , - 12 and 10, 12, 9

23 decreasing on 1- 1, 02 and 11, q 2; concave down on 1 1 a - q, b and a , q b , concave up on 23 23 y 1 1 2 a, b f(x) = 2x 2 – x4 – 3 1 23 23 -2 -1 -1

1

2

x

-2 -3 -4 -5 -6

31. [2.2, 2.3] Relative maximum at 11, 42; relative minimum at 1- 1, - 42; points of inflection at A - 23, - 223 B and 10, 02 and A 23, 223 B ; decreasing on 1- q , - 12 and 11, q 2, increasing on 1- 1, 12; concave down on A - q , - 23 B and A 0, 23 B , concave up on A - 23, 0 B and A 23, q B ; horizontal asymptote y at y = 0 4 8x f(x) = ——– 3 x2 + 1 2 1 –5–4–3–2–1 –1 –2 –3 –4

1 2 3 4 5

32. [2.3] Relative maximum at 10, - 22; no points of inflection; vertical asymptotes at x = - 2 and x = 2; horizontal asymptote at y = 0; increasing on 1- q , - 22 and 1- 2, 02, decreasing on 10, 22 and 12, q 2; concave up on 1- q , - 22 and 12, q 2, cony cave down on 1- 2, 22 10 8 6 4 2 −10 −8 −6 −4 −2−2

f(x) =

8 x2 − 4

2 4 6 8 10 x

−4 −6 −8 −10

33. [2.4] Minimum = - 7 at x = 1 34. [2.4] No absolute extrema 35. [2.4] Maximum = 6 23 at x = - 1; minimum = 4 13 at x = - 2 36. [2.5] 15 sweatshirts

ANSWERS

37. [2.5] 30 times; lot size of 15 38. [2.6] ¢y L 0.5; f ¿1x2¢x = 3 39. [3.3] (a) N1t2 = 8000e 0.1t; (b) 17,804; x (c) 6.9 yr 40. [3.6] (a) E1x2 = ; (b) E122 = 15, inelastic; 12 - x (c) E192 = 3, elastic; (d) increase; (e) x = $6 41. [2.6] (a) ¢A L ;50 ft2; (b) 6 squares, $48 2 42. [4.2] 12 x 6 + C 43. [4.3] 3 e 1 7 44. [4.7] + ln ƒ 7 - 3x ƒ + C 917 - 3x2 9 x2 x2 1 x4 45. [4.5] 4 e + C 46. [4.6] a + 3xb ln x - 3x + C 2 4 232 47. [4.2] 75 ln x + C 48. [4.3] 2 49. [4.3] 3 , or 7713 50. [5.2] $41,324.72 51. [5.2] (a) $28,992.08; (b) $3832.91/yr; (c) $21,007.91, $11,670.90 1 52. [5.2] $711,632.48 53. [5.3] Convergent, 4374 54. [5.5] 32 ln 3 55. [5.5] 0.6826 56. [5.5] (a) 0.644; (b) 99.4th; (c) $57,322.49 57. [5.1] 17, $1692; $751.33 p 1 58. [5.6] - a 10 - 1b, or approximately 1.571 2 e 2 59. [5.7] y = C1e x /2, where C1 = e C 60. [6.1] (a) 26; (b) 51x, y2 ƒ x 2 + y 2 … 46 61. [6.4] (a) y = 54 x + 35 12 ; (b) $154,167 62. [6.2] 8xy 3 + 3 63. [6.2] e y + 24x 2y 64. [6.3] No relative extrema 65. [6.5] Maximum = 4 at 67. [6.6]1467 shoppers 13, 32 66. [6.6] 31e 3 - 12

Appendix A Exercise Set A, p. 614

1. 5 # 5 # 5, or 125 2. 7 # 7, or 49 3. 1- 721- 72, or 49 4. 1- 521- 521- 52, or - 125 5. 1.0201 6. 1.030301 1 1 7. 16 8. 64 9. 1 10. 6x 11. t 12. 1 13. 1 1 1 1 1 1 14. 3 15. 2 , or 16. 2 , or 17. 8 18. 4 9 16 3 4 1 1 1 1 19. 0.1 20. 0.0001 21. b 22. k 23. 24. b h e t 1 5 7 -6 6 5 25. x 26. t 27. x , or 6 28. x 29. 35x x 30. 8t 7 31. x 4 32. x 33. 1 34. 1 35. x 3 36. x 4 1 1 37. x -3, or 3 38. x -4, or 4 39. 1 40. 1 41. e t - 4 x x 1 42. e k - 3 43. t 14 44. t 12 45. t 2 46. t -4, or 4 t 1 1 47. a 6b 5 48. x 7y 7 49. t -6, or 6 50. t -12, or 12 t t 51. e 4x 52. e 5x 53. 8x 6y 12 54. 32x 10y 20 8 20 y 21z 15 x y 1 8 20 -16 1 -9 21 15 55. 56. x y z , or x y z , or 16 81 125 81z 125x 9 14 4 z 9y 625x 16 57. 9x -16y 14z 4, or 58. 625x 16y -20z -12, or 20 12 16 x y z 64x 6y 3 c 4d 12 59. 60. 9 9 61. 5x - 35 62. x + xt 16q 8 a b 63. x 2 - 7x + 10 64. x 2 - 7x + 12 65. a 3 - b 3 66. x 3 + y 3 67. 2x 2 + 3x - 5 68. 3x 2 + x - 4 69. a 2 - 4 70. 9x 2 - 1 71. 25x 2 - 4 72. t 2 - 1 2 2 2 73. a - 2ah + h 74. a + 2ah + h2 75. 25x 2 + 10xt + t 2 76. 49a 2 - 14ac + c 2 77. 5x 5 + 30x 3 + 45x 78. - 3x 6 + 48x 2 79. a 3 + 3a 2b + 3ab 2 + b 3

A-37

80. a 3 - 3a 2b + 3ab 2 - b 3 81. x 3 - 15x 2 + 75x - 125 82. 8x 3 + 36x 2 + 54x + 27 83. x 11 - t2 84. x 11 + h2 85. 1x + 3y22 86. 1x - 5y22 87. 1x - 521x + 32 88. 1x + 521x + 32 89. 1x - 521x + 42 90. 1x - 1021x + 12 91. 17x - t217x + t2 92. 13x - b213x + b2 93. 413t - 2m213t + 2m2 94. 15y - 3z215y + 3z2 95. ab1a + 4b21a - 4b2 96. 21x 2 + 421x + 221x - 22 97. 1a 4 + b 421a 2 + b 221a + b21a - b2 98. 16y - 5216y + 72 99. 10x1a + 2b21a - 2b2 100. xy1x + 5y21x - 5y2 101. 211 + 4x 2211 + 2x211 - 2x2 102. 2x1y + 521y - 52 103. 19x - 121x + 22 104. 13x - 4212x - 52 105. 1x + 221x 2 - 2x + 42 106. 1a - 321a 2 + 3a + 92 107. 1y - 4t21y 2 + 4yt + 16t 22 108. 1m + 10p21m 2 - 10mp + 100p 22 109. 13x 2 - 121x - 22 110. 1y 2 - 2215y + 22 111. 1x - 321x + 321x - 52 112. 1t - 521t + 521t + 32 113. 74 114. 79 115. - 8 12 4 116. 5 117. 120 118. 140 119. 200 120. 200 121. 0, - 3, 45 122. 0, 2 - 23 123. 0, 2 124. 3, - 3 125. 0, 3 126. 0, 5 127. 0, 7 128. 0, 3 129. 0, 13, - 31 1 1 130. 0, 4, - 4 131. 1 132. 2 133. No solution 134. No solution 135. - 23 136. - 145 137. 5 138. - 27, 27 139. x Ú - 54 140. x Ú 3 141. x 7 - 121 142. x 7 - 135 143. x 7 - 74 144. x … - 56 145. x … - 3 2 146. x Ú - 2 147. x 7 3 148. x … 53 149. x 6 - 52 150. x … 54 151. 2 6 x 6 4 152. 2 6 x … 4 153. 23 … x … 11 154. 56 … x 6 16 155. - 1 … x … 14 2 5 5 156. - 5 … x 6 - 2 157. $650 158. $800 159. More than 7000 units 160. More than 4200 units 161. 480 lb 162. 340 lb 163. 810,000 164. 720,000 165. 60% … x 6 100% 166. 50% … x 6 90%

Appendix B Exercise Set B, p. 617 1. y = 3.45x + 0.6 2. y = 2.5283x + 13.34 3. y = 0.0732x 2 + 2.3x + 11.813

Diagnostic Test, p. xix Part A The blue bracketed references indicate where worked-out solutions can be found in Appendix A: Review of Basic Algebra. For example, [Ex.1] means that the problem is worked out in Example 1 of the appendix. 1. 64 [Ex. 1] 2. - 32 [Ex. 1] 3. 18 [Ex. 1] 4. - 2x [Ex. 2] 1 1 5. 1 [Ex. 2] 6. 5 [Ex. 3] 7. 16 [Ex. 3] 8. [Ex. 3] t x 10 9. x 11 [Ex. 4] 10. x [Ex. 4] 11. 7 [Ex. 4] 12. a [Ex. 5] x y 15 1 7 13. e [Ex. 5] 14. 6 [Ex. 6] 15. 12 9 [Ex. 6] x 8x z 16. 3x - 15 [Ex. 7] 17. x 2 - 2x - 15 [Ex. 7] 18. a 2 + 2ab + b 2 [Ex. 7] 19. 4x 2 - 4xt + t 2 [Ex. 8] 2 2 20. 9c - d [Ex. 8] 21. h12x + h2 [Ex. 9] 22. 1x - 3y22 [Ex. 9] 23. 1x + 221x - 72 [Ex. 9] 24. 12x - 1213x + 52 [Ex. 9] 25. 1x - 721x + 221x - 22 [Ex. 10] 26. x = 6 [Ex. 11] 27. x = 0, 2, - 45 [Ex. 12] 28. x = 0, 12, - 12 [Ex. 13]

A-38

ANSWERS

q , 21 29. No solution [Ex. 15] 30. x … 21 13 or A 13 D [Ex. 17] 31. 660 lb [Ex. 18] 32. 351 suits [Ex. 19]

Part B The blue bracketed references indicate where worked-out solutions can be found in Chapter R. For example, [Ex. R.2.5] means that the problem is worked out in Example 5 of Section R.2. y y 1. 2. 4

−4

−2

2

4

2

x

−2

−4

−2

4

2

2 2

4

−4

3x + 5y = 10 2

4

x

[Ex. R.5.1] 15.

6 y = x2 − 1

4

−4

6

8

2

4

x

−2

4

f(x) =

for x ≤ 0, 4, 3 − x 2, for 0 < x ≤ 2, 2x − 6, for x > 2.

2 1 −2

−1

1 −1 −2

4

2

3 x

2

x −4

−2

2

4

−2

[Ex. R.5.6] 17. y

[Ex. R.5.8] 18. $1,166.40 [Ex. R.1.6]

2

[Ex. R.1.3] [Ex. R.1.4] 5. f 102 = 8; f 1- 52 = 143; f 17a2 = 147a 2 - 14a + 8 [Ex. R.2.4] 1 f 1x + h2- f 1x22 6. = 1 - 2x - h [Ex. R.2.5] h y 7. [Ex. R.2.9] 3

2

f(x) = | x |

x

−4

2

4

6

−2 −4 −6

[Ex. R.5.9]

x

4

−4 4

4

y 6

1

x

−2

−2

−2

f(x) =

−2

2 2

2

−4

x = y2

2

−4

[Ex. R.5.4] 16.

2

4

−2

f(x) = x2 − 2x − 3

y

[Ex. R.1.2] 4. y

y 8

−4

f(x) = x3

−2

4

[Ex. R.1.1] 3.

6 x

−2

−2

−4

4

−2

4

2

A

9. E x ƒ x is any real number and q B F [Ex. R.3.4]

5 2,

10. Slope m = 12; y-intercept: A 0, - 74 B [Ex. R.4.4] 11. y = 3x - 2 [Ex. R.4.5] 12. m = - 32 [Ex. R.4.7] y y 13. 14.

−4

6

y = 2x + 1

8. 1- 4, 52 [Ex. R.3.1a] x Z 52 F or E x ƒ A - q , 52 B ´

f(x) = −√x

8

x

x

Index of Applications Business and Economics Accumulated future value, 485, 486, 487, 489, 490, 536, 541 Accumulated present value, 485, 489, 490, 542, 543 Accumulated sales, 423, 424, 434 Actuarial science, 362 Advertising, 129, 213, 260, 286, 304, 332, 334, 335, 350, 384, 385 Annual interest rate, 348 Annual sales, 69, 76, 348 Annuities, 12 Area, maximizing, 262, 273, 275, 276, 579 Art value, 348 Average cost, 162, 164, 191, 193, 248, 286, 287, 602 Bakery bonuses, 520 Beverage-can problem, 582 Book value, 46 Bottled water sales, 348 Break-even point, 166 Cable TV subscribers, 71, 72 Capital expansion, 533 Capital outlay, 490 Capitalized cost, 492, 497 Cellphone calling plans, 30 Cobb–Douglas production function, 559, 562, 563, 599 Coffee production, 521, 539 College costs, 47, 131, 342 College enrollment, 598 Comic book value, 307, 342, 350 Compound interest. See Interest Constant elasticity curve, 377 Construction costs, minimizing, 587 employment in, 128 Consumer credit, 175 Consumer demand, 175 Consumer price index, 349 Consumer surplus, 473, 474, 477, 480 Container cost, minimizing, 571, 587 Continuous income stream, 485, 486, 490, 541, 542, 543 Contract buyout, 543, 603 Contract negotiation, 491 Contract value, 536 Cost, 291, 292 average, 162, 164, 191, 193, 248, 286, 287, 602 capitalized, 492, 497 of college, 47, 131, 342 of a first-class postage stamp, 349

of a Hershey bar, 350 marginal, 278, 279, 285, 287, 320 of milk, 384 minimizing, 269, 270, 271, 272, 275, 276, 277, 302, 304, 571, 587 overage, 603 of pollution control, 248 of a prime rib dinner, 383 of storage equipment, 550, 554 and tolerance, 283, 284, 287 total, 43, 129, 175, 232, 248, 261, 302, 397, 399, 402, 407, 408, 423, 452, 466, 467, 468, 496, 533, 550, 587, 594 of a wedding, 47 Credit card debt, 129 Credit market debt, 389, 397, 423 Daily sales, 49 Demand, 63, 66, 67, 88, 397, 444. See also Elasticity of demand; Supply and demand for chocolate chip cookies, 376 for coal, 348 for computer games, 376 consumer, 175 for DVD rentals, 372, 375 for in-ear radios, 373 marginal, 286, 320 for oil, 348, 376, 491 for tomato plants, 376 Depreciation, 46, 49, 248, 369 Diamond mine production, 536 Distance, minimizing, 276 Distribution of wealth, 469 Domar’s capital expansion model, 533 Double declining-balance depreciation, 369 Doubling time, for an investment, 339 Dow Jones Industrial Average, 142 DVD release, and movie revenue, 385 Early retirement, 490, 542 Effective annual yield, 352 Efficiency, of workers, 318, 398 Elasticity of demand, 371, 377, 383, 385, 530, 534, 542, 544, 603 Employment, 128, 213 Enrollment in colleges and universities, 598 Equilibrium point, 64, 66, 68, 87, 89, 477 Equimarginal productivity, 588 Estate planning, 361 Exponential demand curve, 377 Exports, value of, 320 Facebook membership, growth of, 340 Failure, time to, 507

Federal budget, 349 Fencing, minimizing cost of, 275 Forever stamp, 349 Franchise expansion, 347, 383, 603 Future value, 485, 486, 487, 489, 490, 536, 541, 543 Gasoline prices, average, 84 Gini coefficient, 470 Gold price, and stock prices, 59 Gold production, 488 Gross domestic product, 164, 286 Health insurance premiums, 598 Highway tolls, 46 Household income, median, 75 Household liability, 369 Income median, 76 per capita, 348 Industrial learning curve, 423 Inkjet cartridges, 46 Insurance settlement, 483, 490, 491 Interest annual rate, 348 compounded annually, 9, 86, 88, 175 compounded continuously, 339, 345, 347, 348, 352, 383, 384 compounded monthly, 554 compounded n times a year, 10, 11, 12, 23, 29, 32, 86, 129, 175 prime rate, 77, 78 Internet, shopping on, 383 Inventory costs, minimizing, 269, 270, 271, 272, 275, 277, 302, 304, 603 Investment, growth or value of, 144, 339, 444, 481, 603, 615 Juice-can problem, 586 Labor force, 232, 260 Law of Equimarginal Productivity, 588 Life cycle of a product, 198 Light bulbs, life of, 502 Loan payments, 12, 548, 549, 554 Lorenz function, 469 Lottery winnings, 491 Mail orders, 520 Manhattan Island, value of, 349 Manufacturing, 128, 197, 265, 520 Marginal cost, 278, 279, 285, 287, 320 Marginal demand, 286, 320 Marginal productivity, 286, 559 Marginal profit, 278, 279, 285, 335 Marginal revenue, 278, 279, 285, 288, 302, 335, 460 Marginal supply, 320

I-1

I-2

I N D E X O F A P P L I C AT I O N S

Marginal tax rate, 286 Material used, minimizing, 197, 265 Median income, 76 Minimum wage, 547, 577 Mining and logging, 128 Movie revenue, and DVD release, 385 Movie ticket prices, 90 Nitrogen prices, 274 Nursing facilities, profits of, 563 Oil demand for, 348, 376, 491 production of, 260, 261 supply of, 287 Organic food, 320 Paper shredder sales, 349 Payment table, 549 Per capita income, 348 Phone calls, duration of, 507, 543 Pizza sales, 542 Pollution control, cost of, 248 Postage function, 108, 142 Present value, 356, 361, 383, 484, 486, 489, 490, 495, 496, 497, 536, 537, 541, 542, 543, 603 accumulated, 485, 489, 490, 542, 543 Price breaks, 116, 120 Price distribution, 543 Price–earnings ratio, 554 Prime interest rate, 77, 78 and stock prices, 67 Processing time, 520 Producer surplus, 476, 477 Product life cycle, 198 Production, 121, 559 maximizing, 588, 599 rate of, 171 total, 497 Productivity equimarginal, 588 marginal, 286, 559 monthly, 259 Professional services, 128, 213 Profit, 171, 291, 292, 415, 444 average, 162, 164, 166, 191, 193 marginal, 278, 279, 285, 335 maximizing, 260, 274, 276, 302, 304, 569, 570, 571, 587, 603 total, 175, 232, 302, 401, 407, 408, 418, 423, 434, 452, 496, 533 Profit-and-loss analysis, 43, 46, 87, 89 Purchasing power, 248 Quality control, 504, 518 Rates of change average, 122, 128, 129 comparing, 131 Recycling aluminum cans, 369 Reliability, of a machine, 507 Retirement account, 12 Revenue, 187, 291, 292, 385, 542, 572 average, 162, 164, 191 marginal, 278, 279, 285, 288, 302, 335, 460 maximizing, 267, 268, 274, 275, 571, 584

total, 175, 232, 248, 293, 302, 349, 352, 397, 407, 423, 460, 467, 533, 615 Room area, maximizing, 586 Salaries, 603 of actors, 361 of athletes, 76, 77, 349, 361 Sales, 180, 183, 184, 285, 293 accumulated, 423, 424, 434 annual, 69, 76, 348 of bottled water, 348 daily, 49 leveling off of, 93, 172 of paper shredders, 349 total, 586, 587, 599, 613 Salvage value, 361, 362 Satellite radio subscribers, 344 Service area, of supplier, 290 Shipping charges, 87 Shoppers, in a region, 604 Shopping, on the Internet, 383 Small business, 232, 260, 369 Smart phones, production of, 559 Stocks gold price and prices of, 59 growth in value of, 335, 533, 542, 544 prime rate and prices of, 67 yield of, 554 Straight-line depreciation, 46, 49 Supply, 63, 66, 67, 286, 397, 458 marginal, 320 Supply and demand, 361, 364, 604 Surface area, minimizing, 265, 273, 274, 583, 586, 599 Surplus. See Consumer surplus; Producer surplus Taxation, 286 Taxicab fares, 108, 142 Television commercial, on Superbowl, 385 Ticket prices, 268 for Major League baseball, 129, 142, 157, 287 for movies, 90 for NFL football, 577 Ticket profits, 87, 88 Tolerance, and cost, 283, 284, 287 Total cost, 43, 129, 175, 232, 248, 261, 302, 397, 399, 402, 407, 408, 423, 452, 466, 467, 468, 496, 533, 550, 587, 594 Total profit, 175, 232, 302, 401, 407, 408, 418, 423, 434, 452, 496, 533 Total revenue, 175, 232, 248, 293, 302, 349, 352, 397, 407, 423, 460, 467, 533, 615 Total sales, 613 maximizing, 586, 587 predicting, 599 Trade deficit, 76, 78, 129 Transportation planning, 505, 506, 507 Travel time in a building, minimizing, 600 Trust funds, 484, 486, 489, 490, 542 Typing area, maximizing, 586 U.S. farms, decline of, 362 U.S. patents, 347 Utility, 129, 175, 533

Vanity license plates, 274 Vehicle ownership, 591 Vending machine overflow, 521 Waiting time, 542 Water storage, 524 Workers’ efficiency, 318, 398 Yield effective annual, 352 maximizing, 274 of a stock, 554

Environmental Science Arctic ice cap, rate of change of, 293 Cooling tower volume, 525 Daylight hours, 233 Deer population in Maine, 12 Dewpoint, 555 Emissions, 435 control, 430 Energy conservation, 47 Energy consumption, 129, 453 Fuel economy, 257 Grizzly bears, in Yellowstone National Park, 350 Harvest, maximum sustainable, 305 Manatee population, 48 Natural resources, demand for and depletion of, 488, 491, 542, 543 Pollution, 424, 430 control, 67, 248 Population growth, 108, 109, 184, 350 Radioactive buildup, in the atmosphere, 492, 495, 497 Radioactive waste storage, 302 Recycling aluminum cans, 369 Smog, and light intensity, 364 Solar eclipse, 214 Temperature, average, 76, 232, 430, 435, 521 Temperature–humidity heat index, 563 Trail maintenance, 87 Wind chill temperature, 555

Health Care Blood pressure, 260 Breast cancer incidence, 11, 32, 303 Cancer research spending, 42 Cancer treatment, 361 Chemotherapy, 23, 175 Employment in health care, 128 Health insurance premiums, 47, 598 High blood pressure, 77 Lung cancer, 33 Medication absorption of, 77 acceptance of new, 335, 351 concentration of, 10, 184, 248, 320 dosage of, 287, 370, 384, 435, 453, 497 sensitivity to, 165 Mortality rate, 507 Nutrition, 89 Radioactive implants, 497 Spread of infection, 351, 398

I N D E X O F A P P L I C AT I O N S

Temperature during an illness, 155, 165, 213 Tuberculosis, cases of, 363 Tumor, volume of, 149

Life and Physical Sciences Acceleration, 179, 183, 190 Agriculture, 369 Aluminum cans, recycling of, 369 Aluminum ore (bauxite), demand for and depletion of, 491 Atmospheric pressure, 363 Beer–Lambert Law, 364 Births by age of mother, 74, 75, 85 Blood flow, 155, 293, 436, 554 Body fluids, weight of, 89 Body surface area, 287, 293, 555 Brain weight, 48 Braking distance, 77, 420 Caloric intake, 315 Carbon dating, 355, 361 Chemistry, 361 Coffee, cooling of, 358 Contractions, of pregnant women, 262 Cooling, 363 Newton’s Law of, 358, 365 and solving a crime, 359, 363 Coughing velocity, 232 Death rate, 73, 75 Decay rate, 353, 355, 360, 361, 383, 384, 490 Distance traveled, 179, 183, 294, 399, 400 from velocity and acceleration, 419, 420, 424, 460, 468 Drug dosage, 370, 384, 435, 453, 497 Dry lake elevation, 261 Earthquake intensity and magnitude, 370 Eclipse, solar, 214 Fasting, 363 Forensics, 359, 363 Free fall, 180, 183, 184 Gold reserves, depletion of, 488 Growth of a baby, 130, 155 Half-life, 353, 354, 355, 361, 383, 384 Healing wound, 155, 287, 293 Hearing-impaired Americans, 11, 12 Heart rate, 87, 88, 155 Height of a thrown object, 394, 398 Home range of an animal, 62, 67, 130 Homing pigeons, 276 Infant mortality and caloric intake, 215 Iron ore, consumption and depletion of, 542 Kiln temperature, 120 Life expectancy, 3, 215, 577 Light intensity, 364, 365 Memory, 130, 398 Muscle weight, 47 Nerve impulse speed, 47 Newton’s Law of Cooling, 358, 365 Poiseuille’s Law, 293, 436, 554 Population density, 593, 599

Population growth, 131, 192, 308, 340, 341, 350, 351, 383, 395, 398, 534 Potash, production and depletion of, 543 Power plant output, 414 Radar range, 67, 130 Radioactive decay, 353, 355, 360, 361 Radioactive waste storage, 302 Reaction time, 48 Richter scale, 370 Satellite power, 363 Sleep duration, and death rate, 73, 75 Sound intensity and loudness, 370 Speed, 424, 432 Stopping distance, on glare ice, 48 Surface area of body, 287, 293, 555, 563 of Gabriel’s horn, 526 and mass, 67 Temperature, at a point on a plate, 571 Temperature–humidity index, 563 Tornado, wind speed of, 555 Uniform motion, 183 Velocity, 179, 183 average, 130, 131 Weight(s) of brain, 48 distribution of, 515, 516 on Earth vs. the moon, 36 gain, 613, 615 of muscle, 47 Wind speed, of a tornado, 555

Social Sciences Beef consumption, decline in, 363 Brentano–Stevens Law, 534 Daytime accidents, 77 Diffusion of information, 351 Divorce rate, 444 Driver fatalities by age, 69 Ebbinghaus learning model, 321 Education, employment in, 128 Forgetting, 331, 335, 362 Gravity model, for phone calls between growing cities, 550 Heights, estimating (anthropology), 48 Home computer use, 88, 234 Hullian learning model, 335, 351, 383 Internet use, by young adults, 48 Keyboarder’s speed, 435 Learning curve, 468 Learning rate, 458 Love, three aspects of, 233 Marriage age, median, 49, 156 Maze, time in, 507 Memory, 193, 398, 423, 434 Population average, 435 decrease, 363 growth, 131, 156, 165, 192, 615 Power Law of Stimulus–Response, 534 Practice, results of, 435 Reading ease, 564 Rumor, spread of, 351

Stimulus–response model, 534 Stress factors, scaling of, 23 Studying, results of, 73, 435 Test score distribution, 518, 519, 520, 521 Urban population, 49 Walking speed, 335 Weber–Fechner Law, 531 Women college graduates, 351 Zipf’s Law, 67

Sports Baseball, 194 earned-run average, 248 Major League salaries, 349 ticket prices in, 129, 142, 157, 287 Basketball heights of players, 521 NBA players’ average salary, 76, 77 Bowling scores, 521 Bungee jumping, 473, 480 Distance to the goal line, 94 Earned-run average, 248 Female participation in high school athletics, 8 Football, volume of, 525 Goals against average, 555 Hang time, 184 High jump, world record in, 578 Mile run, world record in, 578 Running records, 10 Skateboarding, 11, 184 Ski trail gradients, 41 Snowboarding, 11

General Interest Area, maximizing, 262, 273, 275, 276 Birthday, shared, 508 Container volumes, 544 Dewpoint, 555 Fuel economy, 257 Gas mileage, 130 Government, employment in, 128 Grade, of a road, 41 Grade average, 615 Grade predictions, 577 Grading scale, curving, 518 Infinite paint can (Gabriel’s horn), 526 Juice-box dimensions, 2 Light, maximizing, 275, 276 Pitch, of a roof, 41 Probability, 498, 499, 506 Reservoir filling, 117 Rope around the equator, 287 SAT scores, 516 Shoe size conversions, 169 Stair grade requirements, 47 Surface area, minimizing, 265, 273, 274, 275 View to the horizon, 156 Volume, maximizing, 264, 273, 275 Wedding cost, 47 Wind chill, 555

I-3

Index Absolute maximum or minimum, 202, 250, 251, 253–254, 584 strategies for finding, 255–257 Absolute-value function(s), 60 graph of, 68 Acceleration, 179–180, 418–420 Accumulated future value, 481–483 Accumulated present value, 484–487, 495 Accuracy, and calculator use, 10 Addition Principle, 609 Additive property of definite integrals, 425–427 Algebra, review of, 605–615 Algebraic–graphical connection, 52–53 Anticlastic curve, 566 Antiderivative(s) applications of, 394–395 and area under a graph, 395–396, 411–414 as areas, 399–406 initial conditions and, 393–395 as solutions for differential equations, 526 Antidifferentiation, 390–396. See also Integration properties of, 393 rules of, 391 Applied problems, solving, 613 Approximately equal to, 10 Approximation of area, 402–405 of definite integral, 417–418 Area approximation of, by rectangles, 401–402, 404–405 definite integral as, 405–406 functions, 410 and geometry, 400–401 under a graph, 395–396, 410, 411–414, 415–417 percentiles as, 516–519 of region bounded by two graphs, 427–430 total cost as, 399 total profit as, 401 Arithmetic mean (average), 508, 575 Asymptotes, 235–240 Average cost, 161, 279 Average growth rate, 131 Average profit, 161

I-4

Average rate of change, 42, 121–127, 132. See also Difference quotient(s) Average revenue, 161 Average value of continuous function, 430–432 Average velocity, 131, 179 Axes, 3, 551 Base, 309, 605 logarithmic, 322 one as, 311 Beamon, Bob, 528 Bearley, M. N., 528 Break-even value, 44 Capital, marginal productivity of, 559, 560 Chain Rule, 171, 437, 439 Change of variable, 439 ClearDraw option, 516 Closed and bounded region, 583, 584 Closed interval, 25 Cobb-Douglas production function, 559–560 Coefficient, 53 of correlation, 575 Common logarithm, 324–325 Complex numbers, 52 Composed function, 170 Composition of functions, 169–171, 314 derivative of, 171 Compound interest, 8–10, 23, 339, 345–347 Concavity, 182, 216–217, 566 CONNECTED mode, 59, 237 Constant, limit of, 347 Constant function(s), 34, 37, 53 derivative of, 148 Constant of integration, 390, 393, 527 Constant of proportionality, 36 Constant plus a function, derivative of, 151 Constant Rule, for antidifferentiation, 391 Constant times a function, derivative of, 148–149, 151 Constrained optimization, 579–585 Constraint, 579, 583 Consumer surplus, 474–479 Continuity and differentiability, 140 over (or on) an interval, 113–117

Continuous function, 113, 114, 200–201 area under graph of, 405 average value of, 430–432 Continuous income stream, 481, 482 Continuous random variable, 499–503 expected value of, 508–510 exponentially distributed, 504–505 mean of, 510–511 normally distributed, 513 standard deviation of, 511–512 standard normal distribution of, 513 uniformly distributed, 504, 508 variance of, 511 Convergent improper integral, 494 Cooling, Newton’s Law of, 357–358 Coordinate system, three-dimensional, 551 Coordinates, of point, 3 Correlation coefficient of, 575 squared coefficient of, 617 Cost(s) average, 161–162, 279 fixed, 43 inventory, 269–272 marginal, 277–279 total, 43, 268, 399 variable, 43 Critical point, 566 Critical values, 200–201, 205, 566 Cubic function(s), 53, 68, 74–75 CubicReg feature, 75 Curve fitting, 68–75, 544–545 Curve sketching, 222–229 D-test, 566, 573, 575 Decay exponential, 353, 357 integration of models of, 480–489 radioactive, 353–356 Decay rate, 354 Decreasing function, 198–200, 216 concavity and, 217 derivative and, 200, 203 Definite integral(s), 405–406, 414 additive property of, 425–427 applications involving, 418–420 approximating, with calculator, 417–418 and area between two curves, 427–430 and area under curve, 405–406, 425 of continuous function, 417 properties of, 425–432

INDEX

Delta notation, 279–284 Demand, elasticity of, 371–375, 530–531 Demand curve, 474, 477 Demand function(s), 63–65, 290, 371, 474–475, 477 Density function. See Probability density function Dependent variable, 20 Derivative(s), 93 calculating, 135–139 Chain Rule and, 171 of composition of functions, 171 and concavity, 182, 216–217 of constant function, 148 of constant plus a function, 151 of constant times a function, 148–149, 151 and decreasing function, 200, 203 determining sign of, 208–209 of difference, 150–151 of exponential functions, 314–315, 366–367 first, 198, 206, 219 of function involving e, 315 graphical check of, 161 higher-order, 177–181 higher-order partial, 560–561 and increasing function, 200 as instantaneous rate of change, 134, 149 Leibniz notation for, 144–145, 177, 281, 406, 437 of linear function, formula for, 137 of logarithmic function, 367–368 of natural logarithmic function, 328–331 Newton’s notation for, 144 partial, 556–561, 573–574 of product, 158–159 of quotient, 159–162 second, 198, 219 of sum, 150–151 Derived function, 135 Detection threshold, 531, 532 Difference antiderivative of, 393 derivative of, 150–151 Difference quotient(s), 124–127, 132, 279–280 Difference Rule, 150 Differentiability, and continuity, 140 Differentiable function, 135, 411 Differential(s), 277–284 Differential equation(s), 338, 526–530 separation of variables and, 528–530 solving, 526–527 verifying solutions of, 527–528 Differentiation, 93. See also Derivative(s) checking, 168 implicit, 288–290 order of, 561 steps for, 167

and tangent lines, 147 using limits, 134–140 Diminishing returns, law of, 560 Direct variation, 36–37 Directly proportional variables, 36 Discontinuity, point of, 114, 139 Discounting, 356 Distance, 418–420 as antiderivative of velocity, 394 as area, 399, 400 Distributive law, 608 Divergent improper integral, 494 Domain of function, 13, 14, 17, 26–30 of function of two variables, 548, 552 graphing calculator and, 28–29 of radical function, 62 DOT mode, 59 Double iterated (or double) integral, 588–589 Doubling time, 339 and Rule of 70, 340–341 DrawInv option, 323 e, 311–313 and compound interest formula, 345–347 derivatives of functions involving, 314–315 as logarithmic base, 325, 366, 367 Economic ordering quantity, 272 Elasticity of demand, 371–375, 530–531 Elliptic paraboloid, 552 Endpoints, of interval, 25 Equality, line of, 469, 470, 471 Equation(s) differential, 338, 526–530 exponential, 326–327 logarithmic, 322 logistic, 344, 345 point–slope, 38–39 polynomial, 54–55 rational, 610–612 slope–intercept, 38 solution of, 3, 609–612 Equilibrium point, 64–65, 474, 477 Equivalent inequalities, 612 Euler, Leonhard, 313 Excel (Microsoft), for regression, 616–617 Expected value of continuous random variable, 508–510 Experimental probability, 498–499 Exponent(s), 308, 309, 605 negative integer as, 606 one as, 605 properties of, 606–607 rational, 61 zero as, 605 Exponential decay, 353, 357 Exponential distribution, 504–505 Exponential equation(s), 326–327

Exponential function(s), 309 applications of, 337–347 derivatives of, 313, 314–315, 366–367 graphs of, 308–311, 315–318, 337, 338, 344, 346 properties of, 311, 392 Exponential growth, 337–347, 480–488 Exponential notation, 605–606 Exponential regression, 341–342 Exponential Rule, for antidifferentiation, 391 ExpReg feature, 341 Extended Power Rule, 166–168, 171, 289 Extended Technology Applications Average Price of a Movie Ticket, 90–91 Business: Distribution of Wealth, 469–472 Business of Motion Picture Revenue and DVD Release, 385–388 Curve Fitting and Container Volumes, 545–546 Maximum Sustainable Harvest, 305–306 Minimizing Employees’ Travel Time in a Building, 600–601 Path of a Baseball: The Tale of the Tape, 194–196 Extrema, 201–202 absolute, 250, 251 and critical value, 202 First-Derivative Test for, 203–209, 218 graphing calculator and, 252–253, 254, 256 relative, 201–212 Second-Derivative Test for, 218 Extreme-Value Theorem, 251 for two-variable functions, 583–585 Factoring, 608–609 by grouping, 609 Factors, 605 Family of curves, 527 Fifth derivative, 177 First coordinate, 3 First derivative, 198, 206, 219 First-Derivative Test, 203–209, 218 Fixed costs, 43 fMax or fMin option, 210 fnInt feature, 417, 429 Focus, of elliptic paraboloid, 552 Fourth derivative, 177 Function(s), 13, 14, 20. See also Exponential function(s); Linear function(s); Logarithmic function(s); Probability density function absolute maximum or minimum of, 202, 250, 251, 253–254, 255–257 absolute-value, 60, 68 applications of, 43–44 average value of, 430–432 composed, 170 composition of, 169–171, 314

I-5

I-6

INDEX

constant, 34, 37, 53, 148 continuous, 113, 114, 201, 430–432 critical values of, 200–201, 205, 566 cubic, 53, 68, 74–75 decreasing, 153–154, 198–200, 203, 216, 217 demand, 63–65, 290, 371, 474–475, 477 derived, 135 differentiable, 135, 411 domain of, 13, 14, 17, 26–30 fitting to data, 68–75 graph(s) of, 16–18 increasing, 153–154, 198–200, 203, 216, 217 inverse, 59, 323, 328 iterated, 176 joint probability density, 591 Lagrange, 580 logistic, 344–345, 360 Lorenz, 469, 470 natural logarithmic, 328–331, 392 nonlinear, 50–65 piecewise-defined, 18–20, 97–99, 116–117, 426–427 polynomial, 53–54, 74–75, 235 power, 53, 61–63, 145 production, 559–560 quadratic, 50, 52–53, 68, 74–75 quartic, 68 range of, 13, 14, 17, 26–30 rational, 57–59, 111–112, 234–235, 237 of several variables, 548–552 square-root, 61, 281 supply, 63–65, 474–475, 476, 477 table of values of, 28 third-degree, 53 of two variables, 548–552 values of, 14–15, 28, 200–201 vertical-line test for, 18 zero of, 54 Fundamental Theorem of Integral Calculus, 415 Future value, 481–483 Gabriel’s horn, 526 General solution, of differential equation, 527 Geometric interpretation of function of two variables, 551–553 of multiple integrals, 590–591 of partial derivative, 558–559 Geometry, and area under a graph, 400–401 Gibson, Bob, 249 Gini, Corrado, 470 Gini coefficient, 470, 471 Gini index, 470, 471 Grafly, 553 Graph(s), 3–7, 17 of absolute value function, 60, 68 area below, 399–406, 411–414, 427–430 asymptotes of, 235–240

concavity of, 182, 216–217 of continuous function, 113–114, 200–201, 405 of cubic function, 68 of demand function, 63, 64, 374 of elliptic paraboloid, 552 of exponential function, 308–311, 315–318, 337, 338, 344, 346, 358 of function of several variables, 551–553 of horizontal line, 33–34 of intervals, 26 of inverse functions, 59, 323, 328 of logarithmic function, 322–323, 328, 338, 358 of logistic equation, 344–345, 360 of normal distribution, 512–513 of parabola, 5–6, 50–51 of piecewise-defined function, 18–20 of power function, 54, 63 of product life cycle, 198 of quadratic function, 50–52, 68, 74–75 of quartic function, 68 of rational function, 59 sketching of, 205, 222–229, 241–246 of square-root function, 61 of straight line, 4–5, 35–36, 37, 68 of supply function, 63–65 of vertical line, 34 Graphicus, 55, 56, 140–141, 182, 229–231, 289, 313, 314, 323 Graphing calculator, 6–7 and absolute extrema, 252–253, 254, 256 and approximating definite integrals, 417–418 CLASSIC mode of, 619–620 ClearDraw option of, 516 CONNECTED mode of, 59, 237 CubicReg feature of, 75 and derivatives, 161 and differentiation, 147 and domain and range, 28–29 DOT mode of, 59 DrawInv option of, 323 and exponential equations, 326–327 ExpReg feature of, 341 fMax or fMin option of, 210 fnInt feature of, 417, 429 INTERSECT feature of, 54, 327, 429 and limits, 95–96 and linear regression, 72–73 LinReg feature of, 75 LN key, 326, 327, 328 LOG key, 325 MathPrint operating system for, 618–620 MAXIMUM or MINIMUM option of, 210, 256 and natural logarithms, 326 and polynomial equations, 54–55 QuadReg feature of, 75 REGRESSION feature of, 74 and relative extrema, 209–210

ROOT feature of, 54, 55 ShadeNorm command of, 516

and slope, 34–35 and statistics, 516 TABLE feature of, 16, 95, 96, 210, 254, 279 and tangent line, 147 TRACE feature of, 17, 58, 96, 209, 254 and vertical asymptotes, 237 viewing window of, 6–7, 34–35 WINDOW feature of, 6 Y = editor, 620 Y-VARS feature of, 17, 163, 323 ZERO feature of, 54–55, 327 ZSquare option of, 34 Growth exponential, 310, 337–347, 353, 480–488 integration of models of, 480–489 limited, 344–345 Growth rate, 156, 339, 340, 341–342 Half-life, 353, 354 Half-open interval, 25 Hemisphere, 552 Higher-order derivatives, 177–178 Histogram, 499 Horizontal asymptotes, 235–239 Horizontal line graph of, 33–34 slope of, 40 Hugo, Victor, 135 Implicit differentiation, 288–290 Improper integral(s), 492–495, 512 Income stream, continuous accumulated future value of, 481–483 accumulated present value of, 484–487 Increasing function, 198–200, 216 concavity and, 217 derivative and, 200, 203 Indefinite integral, 390 as solution for differential equation, 526 Independent variable, 20 Indeterminate form, 112 Inelastic demand, 374, 375 Inequalities equivalent, 612 solving, 612–613 Inequality-solving principles, 612 Inequality symbols, 20 Infinity limits involving, 102–105 symbols for, 25, 102 Inflection points, 220–222, 255 Initial condition, 393–395 Inputs, 14, 17, 20 change in, 122 Instantaneous rate of change, 121, 134, 149, 482 Integral(s) application of to probability, 591–592

INDEX

of constant times function, 393 convergent, 494 definite, 405–406 divergent, 494 double iterated, 588–589 evaluating, 414 improper, 492–495, 512 indefinite, 390 multiple, geometric interpretation of, 590–591 recurring, 450 of sum or difference, 393 triple iterated, 593 Integral sign, 390 Integrand, 390 Integration, 390, 414. See also Antidifferentiation and area, 410–417 and change of variable, 439 constant of, 390 formulas for, 436, 454–457 limits of, 414 of a power of x, 391 using substitution, 436–442 using tables of formulas, 454–457 Integration by parts, 445–451 reoccurrence of integral and, 450 repeated, 449–450 tabular, 450–451 Integration-by-Parts Formula, 446 Interest, compound, 8–10, 23, 339, 345–347 Interest rate, 8–10 on annuity, 12 and Rule of 70, 340–341 INTERSECT feature, 54, 327, 429 Interval notation, 24–26, 206 Interval(s), 24–26 Inventory costs, minimizing, 269–272 Inverse functions, 59, 323, 328 Inverse variation, 59 Inversely proportional variables, 59 iPlot, 55–56, 140, 182, 210–212, 313, 314, 412–413 Irrational number, 308 Iterated function, 176 Iterated integration, 588 Jackson, Reggie, 196 Joint probability density function, 591 Labor, marginal productivity of, 559, 560 Lagrange function, 580 Lagrange multipliers, 579–583 Law of diminishing returns, 560 Least common multiple (LCM), 610 Least-squares assumption, 573 Least-squares technique, 572–575 Leibniz, Gottfried Wilhelm von, 144 Leibniz notation, 144–145, 177, 281, 406, 437 Libby, Willard F., 354

Limit(s), 94–99 algebraic, 109–113 of constant, 347 differentiation using, 134–140 existence of, 99 and improper integrals, 494 at infinity, 102–105 of integration, 414 left-handed and right-handed, 95 properties of, 110–111 of rational functions, 111–112 TABLE and TRACE features and, 95–96 “wall” method for determining, 99–101 Limited population growth, 344–345 Limiting value, 351 Line of equality, 469, 470, 471 horizontal, 33–34 point–slope equation of, 38–39 regression, 573, 575, 576 secant, 123, 125, 132, 134, 199 slope of, 37, 39–41 slope–intercept equation of, 38 of symmetry, 50 tangent, 132–134 vertical, 18, 33–34 Linear equations graphs of, 4–5 principles used in solving, 609 Linear function(s), 37, 53, 68, 400–401 applications of, 43–44 average rate of change of, 124 formula for derivative of, 137 Linear regression, 72–73, 575 Linear systems, solving with matrices, 575 LinReg feature, 75 LN key, 326, 327, 328 Local maximum or minimum, 201 LOG key, 325 Logarithm(s), 322 common, 324–325 natural, 325–326 properties of, 323–324 Logarithmic base, 322 e as, 325, 366, 367 Logarithmic equation, 322 Logarithmic function(s) applications involving, 331–333 derivatives of, 367–368 graphs of, 322–323, 325, 328, 329, 332 Logistic equation (logistic function), 344–345, 360 Lorenz, Max Otto, 469 Lorenz function, 469, 470 Mantle, Mickey, 196 Marginal cost, 277–279 Marginal productivity, of labor or capital, 559, 560 Marginal profit, 277–279 Marginal revenue, 277–279 Mathematical model(s), 7–10, 68–75

MathPrint operating system, for TI graphing calculators, 618–620 Matrices, 575 Maxima, 202 absolute, 250–257 and constrained optimization, 579–585 relative, 201–209, 218, 565–570 Maximum–Minimum Principle 1, 251 Maximum–Minimum Principle 2, 253 Maximum–minimum problems, 262–272 strategy for solving, 2, 263 MAXIMUM option, 210, 256 Maximum profit, 268 Mean, of continuous random variable, 510–511 Microsoft Excel, for regression, 616–617 Minima, 202 absolute, 250–257 relative, 201–209, 218, 565–570 MINIMUM option, 210, 256 Multiplication, using distributive law, 608–610 Multiplication Principle, 609 Natural base. See e Natural logarithm(s), 325 calculator and, 326 Natural Logarithm Rule, for antidifferentiation, 391 Natural logarithmic functions, 329–331 derivatives of, 328–331, 392 graphs of, 328 Negative exponent, 606 Negative infinity, 25, 102 Newton, Isaac, 144 Newton’s Law of Cooling, 357–358 Nonlinear functions, 50–65 average rate of change of, 124 Normal distribution, 512–515 Notation delta, 279–284 exponential, 605–606 interval, 24–26, 206 Leibniz, 144–145, 177, 281, 406, 437 limit, 95 set, 24, 26 set-builder, 24, 26 sigma, 403 summation, 402–403 Oblate spheroid, 524 Oblique asymptote, 239–240 Open interval, 25, 200 Ordered pair, 3, 25 Ordered triple, 552 Outputs, 14, 17, 20 change in, 122 Parabola, 5–6, 50–51 Paraboloid, elliptic, 552 Partial derivative(s) finding, 556–558, 573–574

I-7

I-8

INDEX

geometric interpretation of, 558–559 higher-order, 560–561 second-order, 560–561 Particular solution, of differential equation, 527 Percentiles, 516–519 Piecewise-defined function(s), 18–20 and continuity, 116–117 and definite integrals, 426–427 and limits, 97–99 Pixels, 6 Point critical, 566 of discontinuity, 114 equilibrium, 64–65, 474, 477 of inflection, 220–222, 255 of tangency, 133 Point notation, 206 Point–slope equation, 38–39 Polynomial equations, solving, 54–55 Polynomial function(s), 53–54, 235 fitting to data, 74–75 Population growth, 353 limited, 344–345 uninhibited, 338–341 Power function(s), 53, 61–63 derivative of, 145 Power Rule, 145–148 for antidifferentiation, 391–392 Extended, 166–168, 171, 289 Present value, 356–357, 483–487 Prime notation, 144 Principal, 8 Principle of Zero Products, 52, 610 Probability, 498–505, 591–592 expected value and, 508–510 Probability density function, 501–503, 509, 510, 511, 512, 513, 516 constructing, 503 and exponential distributions, 504–505 joint, 591 and uniform distributions, 504 Producer surplus, 474–479 Product Rule, 158–159 for differentiation, 445 Profit, 279 average, 161–162 total, 43, 268, 399 Projection, of curve, 27 Proportionality constant of, 36 direct, 36 inverse, 59 Quadratic formula, 52 Quadratic function(s), 50, 52–53, 68, 74–75 QuadReg feature, 75 Quartic function, 68 Quotient Rule, 159–162

Radioactive decay, 353–356 Range, of function, 13, 14, 17, 26–30 graphing calculator and, 28–29 Rate(s) of change average, 121–124 difference quotients as, 124–127 instantaneous, 121, 134, 149, 482 related, 290–291 Rate of exponential growth, 339 Rational equations, 610–612 Rational exponents, 61 Rational expression, 611 Rational function(s), 57–59, 234–235 graphs of, 59, 237 limits of, 111–112 Real numbers, 24, 25 Recurring integrals, 450 Reduced row echelon form (rref), 574 Reflections, of functions, 416 Region of feasibility, 583, 584 Regression exponential, 341–342 and function in two variables, 572–576 linear, 72–73, 575 and Microsoft Excel, 616–617 REGRESSION feature, 74 Regression line, 573, 575–576 Relative extrema (maxima or minima), 201–209, 219, 221, 565 classifying using second derivatives, 217–220 D-test for, 566, 573, 575 First-Derivative Test for, 203–209 Second-Derivative Test for, 218 Relative frequency graph, 499 Relative maximum or minimum. See Relative extrema Reproduction curve, 305 Revenue, 279 average, 161–162 and elasticity, 374 total, 43, 268, 373 Riemann, G. F. Bernhard, 401 Riemann sum(s), 404, 410, 412, 482, 589 Riemann summmation, 401–405 Right circular cylinder, volume of, 522 ROOT feature, 54, 55 Roster method, 24 Rounding, 10, 327 Rule of 70, 340–341 Saddle point, 566, 568 Scatterplot, 69 Secant line, 123, 125, 132, 134, 199 Second coordinate, 3 Second derivative(s), 177 use of, to classify relative extrema, 217–220 Second-Derivative Test, 218, 566 Second-order partial derivative(s), 560–561 Separation of variables, 528–530 Set, 24, 26

Set-builder notation, 24, 26 Set notation, 24 Sexson, Richie, 196 ShadeNorm command, 516 Sigma notation, 403 Sketching graphs, 205, 222–229, 241–246 Slant asymptote, 239–240 Slope, 35–36, 39–41 applications of, 41–42 computing of, 39–41 of horizontal line, 40 of secant, 132, 134, 199 of tangent, 134, 151–153 undefined, 40 visualizing graphically, 34–35 Slope–intercept equation, 38 Smoothness of graph, 140 Solid of revolution, 522 Solution of equation, 3 Speed, 131, 178, 420. See also Velocity Square-root functions, 61, 281 Squared correlation coefficient, 617 Squaring, of viewing window, 34 Standard deviation, of continuous random variable, 511–512 Standard normal distribution, 513 table of areas for, 621 Standard viewing window, 6 Statistics, calculator and, 516 Stimulus–response model, 531 Straight line, graph of, 4–5, 35–36, 37, 68 Substitution, as integration technique, 436–442 Sum antiderivative of, 393 derivative of, 150–151 Sum–Difference Rule, 150 Summation notation, 402–403 Supply curve, 474, 476, 477 Supply functions, 63–65, 474–475, 476, 477 Surface of revolution, 522 Surplus, consumer and producer, 474–479 Symmetry, line of, 50 TABLE feature, 16, 95, 96, 210, 254, 279

Tabular integration by parts, 450–451 Tangent feature, 147, 182

Tangent line(s), 132–134 finding, 138 graphing calculator and, 147 and partial derivative, 559 slope of, 134, 151–153, 199–200 Technology Connections. See also Extended Technology Applications absolute value, 60 accumulated present value (exploratory), 495 amount of a continuous income stream (exploratory), 483 antiderivatives and area, 395–396

INDEX

approximating definite integrals, 417–418 approximating expressions having irrational exponents, 308 apps for iPhone and iPod Touch, 55–56, 140–141 asymptotes, 237 average value of a function over an interval, 432 calculating a derivative (exploratory), 136 checking derivatives graphically, 161 checking differentiation, 168 concavity (exploratory), 217 consumer surplus (exploratory), 476 creating a table of function values, 28 curves and tangent lines, 138 derivative of a constant function (exploratory), 148 derivatives of exponential functions, 314 derivatives of natural logarithmic functions, 331 derivative of a sum (exploratory), 150 determining domain and range, 28–29 determining a horizontal asymptote, 239 exploring b, 37 exponential growth and decay (exploratory), 353 exponential models using regression: projecting world population growth, 341–342 finding absolute extrema, 252–253, 254, 256 finding the area bounded by two graphs, 429 finding the area under a graph with iPlot, 412–413 finding the derivative of f 1x2 = ex (exploratory), 313 finding function values, 58 finding limits using TABLE and TRACE, 95–96 finding partial derivatives (exploratory), 558 finding relative extrema, 209–210 finding relative extrema with iPlot, 210–212 finding roots, extrema, and inflection points with Graphicus, 229–231 graphing functions defined piecewise, 20 graphing implicit relationships (exploratory), 289 graphing logarithmic functions, 322, 323, 325, 328, 329, 332 graphs and function values, 17 graphs of rational functions, 59

growth (exploratory), 310 increasing and decreasing functions and their derivatives (exploratory), 199 introduction to the use of a graphing calculator: windows and graphs, 6–7 limit of a constant (exploratory), 347 linear regression (exploratory), 575 linear regression: fitting a linear function to data, 72–73 mathematical modeling using regression, 74–75 numerical differentiation and tangent lines, 147 percentiles on a calculator, 517 probability density function (exploratory), 501 relative extrema and critical values (exploratory), 205 relative extrema and first and second derivatives (exploratory), 219 revenue and profit, 279 solving exponential equations, 327 solving linear systems using matrices, 574 solving polynomial equations, 54–55 standard normal distribution (exploratory), 513 statistics on a calculator, 516 TABLE feature, 16, 95–96 tangent-drawing feature (exploratory), 182 tangent lines to curves (exploratory), 133 three-dimensional graphs with Grafly (exploratory), 553 using Y-VARS, 163 visualizing slope, 34–35 Theorem on Limits of Rational Functions, 111 Theoretical probability, 498–499 Third-degree function, 53 Third derivative, 177 Three-dimensional coordinate system, 551 Threshold value, 531, 532 Total cost, 43, 268 as area, 399 Total profit, 43, 268, 418 as area, 401 Total revenue, 43, 268, 373 Total utility, 475 TRACE feature, 17, 58, 96, 209, 254 Trapezoidal Rule, 409, 514 Triple iterated integral, 593 Turning point, 50 Uniform distribution, 504, 508 Uninhibited population growth, 338–341

Union, of sets, 28 Unit elasticity, 374 Utility, 474 total, 475 Values of a function, 200–201 Variable(s), 20 change of, 439 continuous random, 499–503 functions of two, 548–552 separation of, 528–530 Variable costs, 43 Variance, of continuous random variable, 511 Variation direct, 36–37 inverse, 59 Variation constant, 36 Velocity, 178–180, 394, 418–420 average, 131 Vertex, 50 Vertical asymptotes, 235–239 Vertical line graph of, 34 slope of, 40 Vertical-line test, 18 Vertical tangent, 140 Viewing window, 6 squaring, 34 standard, 6 Volume, of solid of revolution, 522–524 “Wall” method for determining limits, 99–101 Weber–Fechner Law, 531 Williams, Ted, 196 WINDOW feature, 6 x-axis, 3 x-intercept, 53, 240–241 y-axis, 3 y-intercept, 37, 240–241 Y = editor, 620 Y-VARS feature, 17, 163, 323 z-axis, 551 z-score (z-value), 515, 621 Zero as exponent, 605 of function, 54 ZERO feature, 54–55, 327 Zero Products, Principle of, 610 Zero slope, 40 ZSquare option, 34

I-9

Additional Instructor’s Answers Chapter R

13.

Exercise Set R.1, p. 10 y 5 4 3 2 1

1.

2. y=x+4

−5 −4 −3 −2 −1−1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

4. y = –3x

– 5 – 4 –3 –2 –1–1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

y 5 4 3 2 1

6.

– 5 – 4 –3 –2 –1–1

8. x+y=5

−5 −4 −3 −2 −1−1

10. 8y – 2x = 4

– 5 – 4 –3 –2 –1–1

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

17.

y 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2 −3 −4 −5

12. 5x − 6y = 12

1 2 3 4 5 x

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

16. x = 2 – y2

−2 −3 −4 −5

y 5 4 3 2 1

18.

y = −5 x + 3 3

1 2 3 4 5 x

19.

x−y=4

−5 −4 −3 −2 −1−1

20. y = 7 − x2

21.

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

41.

22. y + 1 = x3

1 2 3 4 5 x

y 9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1

y = x − 150

42.

200

−200

x = y2 + 2

y = |4 − x |

1 2 3 4 5 x

y = 5 − x2

1 2 3 4 5 x

−2 −3 −4 −5

−2 −3 −4 −5

6x + 3 y = −9

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3

1 2 3 4 5 x

1 2 3 4 5 x

−2 −3 −4 −5

y= x

y 7 6 5 4 3 2 1

y = x2 − 3

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

–2 –3 –4 –5

1 2 3 4 5 x

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

– 5 – 4 –3 –2 –1–1

−2 −3 −4 −5

–2 –3 –4 –5

y 5 4 3 2 1 –2 –3 –4 –5

1 2 3 4 5 x

−2 −3 −4 −5

y 5 4 3 2 1

11.

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

9.

y=

−2 −3 −4 −5

y = x2 − 5

– 5 – 4 –3 –2 –1–1 −1 x 4

−2 −3 −4 −5

y=2 –x – 4 3

y 5 4 3 2 1

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

–2 –3 –4 –5

7.

15.

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

1 2 3 4 5 x

−2 −3 −4 −5

–2 –3 –4 –5

5.

y 5 4 3 2 1

14.

−5 −4 −3 −2 −1−1

y=x−1

−2 −3 −4 −5

y 5 4 3 2 1

3.

y 5 4 3 2 1

y 5 4 3 2 1

y − 7 = x3

1 2 3 4 5 x

y = 25 − |x | 30

200

−50

50

2x + 5y = 10

1 2 3 4 5 x

−200

−30

−2 −3 −4 −5

IA-1

IA-2 43.

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S y = x 3 + 2x 2 − 4x − 13 15

–10

44.

10

10

−10

–15

45.

46.

30

−6

−5 −4 −3 −2 −1−1

y = −2.3x 2 + 4.8x − 9 1

6

y 5 4 3 2 1

26.

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

y 5 4 3 2 1

28.

1 2 x

–2 –3 –4

−5 −4 −3 −2 −1−1

g(x) = – 4x

x = 8 − y2

y 5 4 3 2 1

29.

y1 = 8 − x y2 = − 8 − x

y1 = x − 4 and y2 = − x − 4 3

10

−5 −4 −3 −2 −1−1 8

–3

−10

−10

y 7 6 5 4 3 2 1

31.

Exercise Set R.2, p. 21

19. (b) f 142 = 13, f 132 = 9, f 1- 22 = - 11, f 1k2 = 4k - 3, f 11 + t2 = 4t + 1, f 1x + h2 = 4x + 4h - 3 20. (b) f 152 = 17, f 1- 12 = - 1, f 1k2 = 3k + 2, f 11 + t2 = 3t + 5, f 1x + h2 = 3x + 3h + 2 21. g1- 12 = - 2, g102 = - 3, g112 = - 2, g152 = 22, g1u2 = u2 - 3, g1a + h2 = a 2 + 2ah + h2 - 3, and g1a + h2 - g1a2 = 2a + h, h Z 0 22. g1- 32 = 13, g102 = 4, h g1- 12 = 5, g172 = 53, g1v2 = v 2 + 4, g1a + h2 = a 2 + 2ah + g1a + h2 - g1a2 h2 + 4, and = 2a + h, h Z 0 h 1 1 23. (a) f 142 = , f 1- 32 is undefined, f 102 = , 49 9 1 1 , f 1t + 42 = , f 1a2 = 1a + 322 1t + 722 f 1x + h2 - f 1x2 1 f 1x + h2 = , and = 2 h 1x + h + 32 - 2x - h - 6 , h Z 0 (b) Take an input, square it, add 1x + h + 3221x + 322 six times the input, add 9, and then take the reciprocal of the result. 1 1 24. (a) f 132 = , f 1- 12 = , f 152 is undefined, 4 36 1 1 1 f 1k2 = , f 1t - 12 = , f 1t - 42 = , 1k - 522 1t - 622 1t - 922 1 f 1x + h2 = (b) Take an input, square it, subtract 1x + h - 522 10 times the input, add 25, and then take the reciprocal of the result.

−5 −4 −3 −2 −1−1

y 9 8 7 6 5 4 3 2 1

30.

1 2 3 4 5 x

−2 −3 −4 −5

10

f(x) = x2 − 2

−5 −4 −3 −2 −1−1 y 5 4 3 2 1

32. f(x) = 6 − x2

−5 −4 −3 −2 −1−1 1 2 3 4 5 x

y 8

1 2 3 4 5 x

f(x) = x2 + 4

1 2 3 4 5 x

g(x) = −x2 + 1 1 2 3 4 5 x

y 5 4 3 2 1

g(x) = 1 x3

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

34. g(x) = x 3

4

2

−2 −3 −4 −5

1 –2 –1–1

g(x) = −2x

−2 −3 −4 −5

−2 −3

33.

1 2 3 4 5 x

−2 −3 −4 −5

−20

48.

x = 4 + y2

f(x) = 3x − 1

−2 −3 −4 −5

y 4 3 2 1

27.

25

–4

f(x) = 2x − 5

−2 −3 −4 −5

–2 –1–1

–30

47.

10

−10

y = –9.6x – 100 4.2

–25

y 5 4 3 2 1

25.

y = 23 − 7x

1 2 x

–4

–8 y 5 4 3 2 1

47. (a)

−5 −4 −3 −2 −1−1

x = y2 − 2

1 2 3 4 5 x

−2 −3 −4 −5

49. 50.

f 1x + h2 - f 1x2 h f 1x + h2 - f 1x2 h

y 5 4 3 2 1

48. (a)

−5 −4 −3 −2 −1−1 −2 −3 −4 −5

= 2x + h - 3, h Z 0 = 2x + h + 4, h Z 0

x = y2 − 3

1 2 3 4 5 x

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S y 5 4 3 2 1

55.

– 5 – 4 –3 –2 –1–1

y 5 4 3 2 1

56. y = f(x)

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

y 9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1 y 7 6 5 4 3 2 1

59.

−5 −4 −3 −2 −1−1

y 5 4 3 2 1

58. y = f(x)

−5 −4 −3 −2 −1−1

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

y = f(x)

−5 −4 −3 −2 −1−1

−2 −3 −4 −5 −6 −7

X −3 −1 1 3 5 7 9 X = −3

y = g(x)

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

X

−3 −2.5 −2 −1.5 −1 −.5 0 X = −3

1 2 3 4 5 x y = f(x)

−5 −4 −3 −2 −1−1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

5. y = f(x)

77.

Y1

Exercise Set R.3, p. 31 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 0 1 2 3 4 5

y 5 4 3 2 1

7.

X −3 −2 −1 0 1 2 3 X = −3

Y1 .6 ERR: −1 −.75 −1 ERR: .6

−5 −4 −3 −2 −1−1

y = −2x

y 5 4 3 2 1

−5 −4 −3 −2 −1−1 −2 −3 −4 −5

y 5 4 3 2 1

4.

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

1 2 3 4 5 x

−2 −3 −4 −5

y = −2

6. x = −1.5

y = −4

y 5 4 3 2 1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

1 2 3 4 5 x

−2 −3 −4 −5 y 5 4 3 2 1

8. y = 3.75

1 2 3 4 5 x

−2 −3 −4 −5

9.

1 2 3 4 5 x

−2 −3 −4 −5

−2 −3 −4 −5

3

x=5

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

y 5 4 x = −4.5 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

73. y = 2 # 2x; this is a function.

x=3

y 5 4 3 2 1

y = g(x)

y 5 4 3 2 1

2.

−2 −3 −4 −5

3.

−2 −3 −4 −5 −6

.2 .4444 ERR: −.5714 −.3333 −.2667 −.25

12. 1- 5, 52

y = g(x)

y 5 4 3 2 1

64.

Y1 −10 −8 −14 20 142 400 842

11. 3- 2, 24

y 5 4 3 2 1

1.

−2 −3 −4 −5

Technology Connection, p. 28 1.

Exercise Set R.4, p. 45

y 5 4 3 2 1

62.

72. y = ; 2x; this is not a function 76.

21. (a) 3; (b) 5- 3, - 1, 1, 3, 56; (c) 3; (d) 5- 2, 0, 2, 3, 46 22. (a) - 1; (b) 5- 4, - 3, - 2, - 1, 0, 1, 26 (c) - 2; (d) 5 - 2, - 1, 0, 1, 2, 3, 46 23. (a) 4; (b) 5 - 5, - 3, 1, 2, 3, 4, 56; (c) 5- 5, - 3, 46; (d) 5- 3, 2, 4, 56 24. (a) 2; (b) 5- 6, - 4, - 2, 0,1, 3, 46; (c) 51, 36; (d) 5- 5, - 2, 0, 2, 56 25. (a) - 1; (b) 3- 2, 44; (c) 3; (d) 3- 3, 34 26. (a) About 2.5; (b) 3- 3, 54; (c) about 2.25; (d) 31, 44 27. (a) - 2; (b) 3- 4, 24; (c) - 2; (d) 3- 3, 34 28. (a) About 2.25; (b) 3- 4, 34; (c) about 0; (d) 3- 5, 44 29. (a) 3; (b) 3- 3, 34; (c) about - 1.4 and 1.4; (d) 3- 5, 44 30. (a) 2; (b) 3- 5, 44; (c) 31, 44; (d) 3- 3, 24 31. (a) 1; (b) 3- 5, 52; (c) 33, 52; (d) 5- 2, - 1, 0, 1, 26 32. (a) 2; (b) 3- 4, 44; (c) 10, 24; 0.05 365t (d) 51, 2, 3, 46 58. (a) A1t2 = 3000a1 + b 365

−2 −3 −4 −5

1 2 3 4 5 x

y 3 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

y 5 4 3 2 1

60.

−2 −3 −4 −5

63.

y = f(x)

−2 −3 −4 −5

−2 −3

61.

1 2 3 4 5 x

−2 −3 −4 −5

–2 –3 –4 –5

57.

y = f(x)

IA-3

y = 2.25

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

10. m = −2 y-intercept: (0, 0) 1 2 3 4 5 x

y = −3x

y 5 4 3 2 1

−5 −4 −3 −2 −1−1 −2 −3 −4 −5

m = −3 y-intercept: (0, 0) 1 2 3 4 5 x

IA-4

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S y 5 4 3 2 1

11.

−5 −4 −3 −2 −1−1 −2 −3 −4 −5

f(x) = −0.5x

f(x) = 0.5x

y = 3x − 4

−5 −4 −3 −2 −1−1 −2 −3 −4 −5

y 5 4 3 g(x) = −x + 3 2 1 −5 −4 −3 −2 −1−1

y 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1

14. y = 2x − 5

−2 −3 −4 −5

m=3 y-intercept: (0, −4)

m = −0.5 y-intercept: (0, 0)

20

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5 y 5 4 3 2 1

18.

y=7

m=0 y-intercept: (0, 7)

−5 −4 −3 −2 −1−1 1 2 3 4 5 x

1 4x

3 2

10

$200,000

$70,000

$100,000

2000 −$100,000

4000

10 20 30 40 50 60

x

100

200

y

1.

2

y 5 4 3 2 1

2.

1 y = – x2 2

1

1 2 3 4 5 x –2

m=1 y-intercept: (0, −2.5)

–1

1

2

x

m=0 y-intercept: (0, −5) 1 2 3 4 5 x

y = −5

−5 −4 −3 −2 −1−1

11 3

y = – 1– x 2 2

y 2 1

5. −2

7.

300 x

−$35,000

4.

−2

x

1

2

x

y = −3x2

y 5 4 3 2 1

y = |x|

y = |x − 3| 2 4 6 8 10 x

y = −3x2 + 2

−1 −2 −4 −6 −8 −10 −12 −14 −16

8.

y = |x − 1|

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

y 10 8 6 4 y = x3 + 2 2 −1 −2 −4 −6 −8 −10

1 2 3 4 5 x

y 4 2

6. y = −2 x2 + 1

y 10 8 6 4 y = |x| 2

−2

4

−2 −3

−4 −6 −8 −10

9.

y = − 1 x2

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−1 −1 1 2 −2 −3 −4 −5 −6 −7 y = −2 x2 −8

−10−8 −6 −4 −2−2

1 2 3 4 5 x

y y = x2 7 6 5 4 3 2 y = x2 − 3 1

−2 −3

1 2

4

−2 −3 −4 −5

y y = x2 7 6 5 4 3 2 y = x2 − 1 1

3.

y = 1 x2

−5 −4 −3 −2 −1−1

–1 –2

68. (b) V102 = $5200, V112 = $4687.50, V122 = $4175, V132 = $3662.50, V142 = $3150, V172 = $1612.50, V182 = $1100 69. (b) R1x2 = C1x2 + P1x2 = 13x; he charges $13 per lawn. 72. (a) The number - 700 indicates that the value of the photocopier decreases by $700 per year, and the number 3500 indicates that the original value of the copier was $3500. 80. (b) 0.4 = 40%, so we have M = 40%W. The weight of the muscles is 40% of the body weight. 81. (b) B1W2 = 2.5%W. The weight of the brain is 2.5% of the body weight.

t

Exercise Set R.5, p. 65

P

P

R

A(t) = 0.08t + 19.7

m=2 y-intercept: (0, −5)

C

$35,000

r

26 24 23 22 21 20

R

C

30

1 2 3 4 5 x

1 3x

y

20

88. (a) A102 = 19.7 yr, A112 = 19.78 yr, A1102 = 20.5 yr, A1302 = 22.1 yr, A1502 = 23.7 yr; (c) A(t)

+ 51. y = 52. y = 53. y = 54. y = 55. x = 3 56. x = - 4 4 57. y = 58. y = - 13 59. y = 3 60. y = x - 29 52 66. (a) C1x2 = 20x + 100,000; 67. (a) C1x2 = 80x + 45,000; (b) R1x2 = 45x; (b) R1x2 = 255x; (c) P1x2 = R1x2 - C1x2 = (c) P1x2 = 175x - 45,000 25x - 100,000 y

11r + 5 10

85. (a) Approximately y = 20 9 x - 4372.44; (c) 2012; 86. (a) Approximately y = - 39.25x + 81,839.25

10

−2 −3 −4 −5

−2

D(r) =

1 2 3 4 5 x

y 5 4 g(x) = x − 2.5 3 2 1

16. m = −1 y-intercept: (0, 3)

D(r) 30

y 5 4 3 2 1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

- 47 x - 17 - 14 x - 23 4 91 - 34 3 x + 15

83. (b)

−2 −3 −4 −5

m = 0.5 y-intercept: (0, 0)

−2 −3 −4 −5

17.

y 5 4 3 2 1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

y 5 4 3 2 1

13.

15.

12.

y 8

10. y = x3 1

2

4 x

y = x3 + 1 1 –2

–1

–1 –4

–8

1 y = x3

2

x

IA-5

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

11.

12.

y 3

3 y= x

2 1 2

3

1 4

5 x

y

–1–1 –2

1 2 3 4 x y = x 2 – 4x + 3

–4 –3 –2 –1 –1

y= x−2 1

18.

3 2 1

y 14 12 10 8 6 4 2 −2 −1−2

2

3

4

y 1

5

6

–1 –1 –2 –3 –4

−5 −4 −3 −2 −1−1

y 5 4 3 2 1

22.

−2 −1−1

1 2 3 4 5 6 7 8 x

−2 −3 −4 −5

f(x) = 2x2 − 6x + 1

−2 −1−1

23. g(x) = −3x2 − 4x + 5

y 7 6 5 4 3 2 1

−8 −7 −6 −5 −4 −3 −2 −1−1

y = −x2 − x + 6

35.

1 2 3 4 5 x

–4–3 –2 –1 –1 –2 –3 –4

4 3 2 1 – 4 – 3 – 2––1

1 –2 –3 –4

1 y=— x2 –2 –1

1 2 3 4 5 6 7 8 x

26. 3 y= x

y 9 8 7 6 5 4 3 2 1

2 y = – –x x 1 2 3 4

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

1 2 x

4 3 2 1 1 2

x

y 10 8 6 4 2

36.

−10−8 −6 −4 −2−2

1 2 3 4 5 x

g(x) =

x2 − 25 x−5

2 4 6 8 10 x

−4 −6 −8 −10

−2 −3 −4 −5

140 120 100 80 60 40 20

150

40,000

50

80,000 x

2000 4000 6000 8000 w 1 2 3 4 5 x

99. (a) 99,130 particles>cm3, 108,347 particles>cm3, 127,322 particles>cm3 P (b) 150,000

100. (c)

y=

−3 x

P = 1000 t5/4 + 14,000

f(h) = 0.144h1/2

1

90,000 60,000

1 2 3 4 5 x

f(h)

2

120,000

100

30,000 10

20

30

40

t

Technology Connection, p. 72

y 5 4 1 y= 3 x−1 2 1 −5 −4 −3 −2 −1−1

H(w) = 0.059 w

100

−2 −3 −4 −5

30.

1 2 x

−2 −3

200

y 5 4 3 2 1

−5 −4 −3 −2 −1−1

28.

−8 −7 −6 −5 −4 −3 −2 −1−1

1 ; 22 - 1 ; 22 , or - 0.207, 1.207 42. , or - 1.207, 2 2 - 4 ; 210 0.207 43. , or - 2.387, - 0.279 3 5 ; 233 1 1 44. , or - 0.186, 2.686 67. 6 68. 6 17 4 2e 2m 19 1 1 69. 70. 4 97. (b) R(x) R(x) = 11.74x0.25 200 2x 2 - 3 2y 2 + 7 180 160 98. H(w) 0.92 250

−2 −3 −4 −5

y

29.

f(x) = 3x2 − 6x + 4

−8 −7 −6 −5 −4 −3 −2 −1−1

y

27.

y 7 6 5 4 3 2 1

41.

1 2 x

1 2 3 4 x

x2 + 7x + 10 x+2

1 2 3 x

−2 −3 −4 −5

g(x) = −2x2 − 3x + 7

y 4 3 2 y = –x 2 1

g(x) =

y 5 4 x2 − 1 3 f(x) = 2 x−1 1 −5 −4 −3 −2 −1−1

24.

−2 −3

25.

y 9 8 7 6 5 4 3 2 1

34.

−2 −3 −4 −5

−2 −3

21.

y 5 4 2 x + 5x + 6 3 f(x) = 2 x+3 1 −7 −6 −5 −4 −3 −2 −1−1

1 2 3 4 x

1 2 3 4 5 x

−2 −3 −4 −5

1 2 3 4 5 6 7 8 x

y 7 6 5 4 3 2 1

y 5 4 3 2 1

−5 −4 −3 −2 −1−1

x

33.

y = x2 − 6x + 5

20.

y = – x 2 + 2x – 1

y= 1 |x |

1 2 3 4 x

–2

−4 −6

19.

32.

3

y = 

x

1 y= x

2

y= x−1 1

17.

y 2

31.

y

y = 2.7x + 63.8

1. (a) 96

1 2 3 4 5 x

6

11 80

200 h

IA-6

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S y 5 4 3 2 1

29. (b)

Technology Connection, p. 74 y = −0.00005368295x 4 + 0.037566680x 3 − 3.4791715x 2 + 105.81080x − 916.68952

1. (a)

130

−5 −4 −3 −2 −1−1

10

31. (a) 3- 4, 52;

45

(b) 12, q 2;

0

2. (a) y = −62.8327x 2 + 5417.8404x − 57264.7856

y = f(x) 1 2 3 4 5 x

−2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5

−5 −4 −3 −2 −1 0 1 2 3 4 5

40. (c) P1x2 = 9.5x - 4000

62500

y $8000

R

6000

C

4000 P

2000 10

200 400 600

75

(b)

−4000

y = −1.6519x 3 + 145.6606x 2 − 2658.3088x + 36491.7730

−1−1 −2 −3 −4 −5

75 0

(c)

y=

−0.0771x 4

+

11.3952x 3

−

639.2276x 2

+ 17037.1915x − 135483.9938

10

75 0

(c) y=−

y 5 4 3 2 1

−5 −4 −3 −2 −1−1

21.

1 2 3 4 5 x

−2 −3 −4 −5

22.

y 10 8 6 4 2 −10−8 −6 −4 −2−2

−4 −6 −8 −10

y 9 8 f(x) = (x − 2)2 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1

f(x) =

− 16 x+4

x2

2 4 6 8 10 x

y= x−3 y = x3

1 x

23.

y 5 4 3 2 1 −1−1 −2 −3 −4 −5

49. (b)

1 2 3 4 5 x

g(x) = x + 1

M 177 176 175 174 173 172 171 170

y 5 4 3 2 1

(d)

−5 −4 −3 −2 −1−1

2 4 6 8 10 x

y=

x2 + x − 6 x−2

1 2 3 4 5 x

−2 −3 −4 −5

50. (a) M = 0.2 r + 160

P 1000 800 600 400

50

52.

3

g(x) = x + 2

−2 −3 −4 −5

1 2 3 4 5 6 7 8 x

1 2 3 4 5 x

1 2 3 4 5 x

y 5 4 3 2 1 −10−8 −6 −4 −2−1

y 5 4 3 2 1

y = (x − 1)3

−2 −3 −4 −5

(b)

−2 −3 −4 −5

Chapter Review Exercises, p. 85 y = |x + 1|

−5 −4 −3 −2 −1−1

1 2 3 4 5 6 7 8 9 x

−5 −4 −3 −2 −1−1

9

20.

x

Vertex at 13, - 12

1400

5 200

y=

−2 −1−1

y = 93.2857x 2 − 1336x + 5460.8286

y 5 4 3 2 1

(b)

y 9 f(x) = x2 − 6x + 8 8 7 6 5 4 3 2 1

42. (a)

62500

3.

y 5 4 3 2 1

41. (a)

62500

10

x

−2000

0

60

70

80

r

10

−200

53.

f(x ) = x 3 − 9x 2 + 27x + 50 400

y=

3

20

30

40

|4 − x 2| + 1 6

1 2 3 4 5 6 7 8 9 x −10

10 −10 −400

Zero: x = - 1.25; domain: ; range: 

10 −1

Zero: none; domain: ; range: 31, q 2

x

IA-7

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

57. (a) y = 37.58x + 294.48; y = - 0.59x 2 + 74.61x - 117.72; y = 0.02x 3 - 2.60x 2 + 125.71x - 439.65; y = 0.003x 4 - 0.324x 3 + 11.46x 2 - 88.51x + 507.84 (b) y1 = 37.58x +2 294.48

−5 −4 −3 −2 −1−1

y2 = −0.59x + 74.61x − 117.72 y3 = 0.02x 3 − 2.60x 2 + 125.71x − 439.65 y4 = 0.0028x 4 − 0.324x 3 + 11.46x 2 − 88.51x + 507.84

2400

67. 60 0

Chapter R Test, p. 88 y 12

–12

–8

y

17.

x : -1 y 5 4 3 2 1 −2 −1−1

2–1 f(x) = x——– x +1

4

–1

–4

4

8

1

x

x:3

–2

x:4

 xx

2 2

69.

+ 2, for x  0 – 2, for x  0 5 4 3 2 1

–3 –2 –1 –1 –2 –3

1 2 3

F(x) =

1 x−3

−7 −6 −5 −4 −3 −2 −1−1

lim G1x2 = 1;

x : -1

lim G1x2 does not exist

x : -2

30

40

x

1 2 3 4 5 x f(x) =

1 −2 x

−5 −4 −3 −2 −1−1

27. Zeros: ; 28 L ;2.828,

y=

; 210 L ;3.162; domain: ; range: 3- 1, q 2

3

|9 − x 2| − 1

x:0

x:0

g(x) =

12

−10

10

1 +4 x+2

−7 −6 −5 −4 −3 −2 −1−1

−2 −1−1

Exercise Set 1.1, p. 106 y 5 4 3 2 1

63.

−5 −4 −3 −2 −1−1

y 9 8 7 6 5 4 3 2 1

64. f(x) = |x | 1 2 3 4 5 x

−2 −3 −4 −5

lim f 1x2 = 0;

−5 −4 −3 −2 −1−1

x : -1

x : -2

x:0

lim g1x2 does not exist

1 2 3 4 5 x

lim f 1x2 = 0

x:3

74. F

1 2 3 4 5 x

G 1 2 3 4 5 6 7 x

−2

x:1

lim F1x2 does not exist

x:1

y 8 7 6 5 4 3 2 1

−3 −2 −1−1

lim F1x2 = 3; lim+ F1x2 = 1;

x : 1-

1 +2 x−3

lim g1x2 = 2;

−2 −3 −4 −5

f(x) = x2

lim f 1x2 = 1;

x:0

lim f 1x2 = 2

−5 −4 −3 −2 −1−1

g(x) =

x: q

lim g1x2 does not exist

73.

1 2 3 4 5 6 7 8 x

−2 −3

1 2 3 x

x : -2

Chapter 1

y 7 6 5 4 3 2 1

72.

lim g1x2 = 4; y 5 4 3 2 1

1 2 3 4 5 x

lim f 1x2 does not exist

x: q

−12

1 +3 x

lim f 1x2 = 3;

x: q

y 9 8 7 6 5 4 3 2 1

f(x) =

−2

x: q

71.

(c) y = - 1.94x 2 + 102.74x + 1253.49

y 8 7 6 5 4 3 2 1

70.

lim f 1x2 does not exist

20

1 2 3 x

−2 −3 −4 −5

lim f 1x2 = - 2;

10

1 x+2

G(x) =

1 2 3 4 5 6 7 8 x

−2 −3 −4 −5 −6 −7

2600 2500 2400 2300 2200 2100 2000 1900 1800 1700

y 5 4 3 2 1

68.

−5 −4 −3 −2 −1−1

y

22. (a)

lim g1x2 = 1

x:0

y 3 2 1

–12 f(x) =

lim g1x2 = 4;

lim F1x2 = 1

–8

1 2 3 4 5 x

x : -3

lim F1x2 does not exist;

–1 12 x

–4

21.

g(x) = x2 − 5

−2 −3 −4 −5

8

g(x) = | x | + 1

−2 −3 −4

lim g1x2 = - 4

0

f(x) = 8 x

−5 −4 −3 −2 −1−1

x:0

y1

14.

1 2 3 4 5 x

lim g1x2 = - 5;

y3

y 6 5 4 3 2 1

66.

−2 −3 −4 −5

y2 y4

y 5 4 3 2 1

65.

lim G1x2 = 1; lim+ G1x2 = 3;

x : 2-

x:2

lim G1x2 does not exist

x:2

IA-8

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S y 7 6 5 4 3 2 1

75.

−2 −1−1

y 5 4 3 2 1

76.

g

−5 −4 −3 −2 −1−1

−2 −3

x : 1-

lim g1x2 does not exist

x:1

x:3

x:3

y 4

77.

f

lim f 1x2 = - 1; lim+ f 1x2 = - 1;

lim g1x2 = 1; lim+ g1x2 = 0;

x : 3-

1 2 3 4 5 x

−2 −3 −4 −5

1 2 3 4 5 6 7 8 x

x:1

lim f 1x2 = - 1

y

78.

Exercise Set 1.4, p. 141

3

1. (a) and (b)

3 2 2

F G

1

1 −2

−3

−1

1

2

−2

−1

x

lim G1x2 = 1 y 5 4 3 2 1

−5 −4 −3 −2 −1−1

2 x

lim F1x2 = - 1

x : -1

79.

1 −1

−1

x : -1

80.

8 6

1 2 3 4 5 x

−2 −3 −4 −5

lim H1x2 does not exist;

x:0

lim H1x2 = 2

G

4 2 −4

−2

2

4

x

lim G1x2 = 1; lim G1x2 = 9

x : -1

x:3

x:1

Exercise Set 1.3, p. 128 1. (a) 8x + 4h; (b) 48, 44, 40.4, 40.04 2. (a) 10x + 5h; (b) 60, 55, 50.5, 50.05 3. (a) - 8x - 4h; (b) - 48, - 44, - 40.4, - 40.04 4. (a) - 10x - 5h; (b) - 60, - 55, - 50.5, - 50.05 5. (a) 2x + h + 1; (b) 13, 12, 11.1, 11.01 6. (a) 2x + h - 1; -2 (b) 11, 10, 9.1, 9.01 7. (a) # ; x 1x + h2 -9 2 1 4 40 (b) - 35 8. (a) # ; , - 15 , - 51 , - 501 x 1x + h2 9 3 6 60 2 (b) - 35, - 10, - 17, - 167 11. (a) - 3x - 3xh - h2; (b) - 109, - 91, - 76.51, 75.1501 12. (a) 36x 2 + 36xh + 12h2; (b) 1308, 1092, 918.12, 901.8012 14. (a) 2x + h - 4; (b) 8, 7, 6.1, 6.01; 17. About 0.3% per yr; about - 0.5% per yr; about - 0.07% per yr 18. About 1.6% per yr; about - 4.8% per yr; about - 1.22% per yr 19. About 0.35% per yr; about - 0.56% per yr; about - 0.05% per yr 20. About 2.7% per yr; about 2.8% per yr; about 2.7% per yr 21. About 3.7% per yr; about 2.6% per yr; about 3.25% per yr 22. About 0.98% per yr; about 0.9% per yr; about 0.94% per yr 23. About 0.97% per yr; about 3% per yr; about 1.9% per yr 24. About - 3.5% per yr; about - 3.4% per yr; about - 3.5% per yr 27. (a) 70 pleasure units> unit of product, 39 pleasure units> unit of product, 29 pleasure units> unit of product, 23 pleasure units> unit of product 28. (a) 300 units> thousands of dollars, 180 units> thousands of dollars, 120 units> thousands of dollars, 100 units> thousands of dollars 37. (b) 1.09 represents the average growth rate, in hectares> g, of home range with respect to body weight when the mammal grows from 200 to 300 g 39. (a) 1.25 words> min,

2. (a) and (b) y

y 9 8 (−2, 6) 7 6 5 4 3 2 1 −2

y

H

1.25 words> min, 0.625 words> min, 0 words> min, 0 words> min 43. (a) 125 million people> yr for both countries (c) A: 290 million people> yr, - 40 million people> yr, - 50 million people> yr, 300 million people> yr; B: 125 million people> yr in all intervals 51. 5ax 4 + 10ax 3h + 10ax 2h2 + 5axh3 + ah4 + 4bx 3 + 6bx 2h + 4bxh2 + bh3 2 55. 22x + 2h + 1 + 22x + 1 -1 56. 2x2x + h A 2x + 2x + h B

−1 −1

3 (−2, 2)

f

f 2 1

−2

(1, 1.5) 1

2

(1, 12)

−1

1

x-axis is tangent to curve at (0, 0).

x-axis is tangent to curve at (0, 0).

(c) f ¿1x2 = 3x; (d) - 6, 0, 3 3. (a) and (b)

(c) f ¿1x2 = x; (d) - 2, 0, 1 4. (a) and (b)

x-axis is tangent to curve at (0, 0). y 1 −2

−1 −1 −2 −3 −4 −5 −6 −7 −8 (−2, −8)−9

−5 −4 −3 −2 −1−2 −4 −6 (−2, −8) −8 −10

1

2

x

−5 −4 −3 −2 −1−1

−2

−1 −2 −4 −6 −8 −10 −12 −14 −16 −18

(1, −2)

f

(−2, −12)

1 2 3 4 5 x x-axis is tangent to curve at (0, 0).

f

1 2 3 4 5 x

−2

(c) f ¿1x2 = 2; (d) 2, 2, 2

2 x (1, −3)

f

6. (a) and (b) f (−2, 8)

(1, 1)

1

(c) f ¿1x2 = - 6x; (d) 12, 0, - 6

f

(c) f ¿1x2 = 3x 2; (d) 12, 0, 3 7. (a) and (b) All tangent lines are identical to the graph of the original function. y 8 7 6 5 4 3 2 1

x

x-axis is tangent to curve at (0, 0). y 2

(c) f ¿1x2 = - 4x; (d) 8, 0, - 4 5. (a) and (b) y 10 8 6 4 2

2

−1

x

y 10 8 6 4 2

−5 −4 −3 −2 −1−2 −4 −6 −8 −10

x-axis is tangent to curve at (0, 0). 2 3 4 5 x (1, −1)

(c) f ¿1x2 = - 3x 2; (d) - 12, 0, - 3 8. (a) and (b) All tangent lines are identical to the graph of the original function. y 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1

f

1 2 3 4 5 x

−2

(c) f ¿1x2 = - 2; (d) - 2, - 2, - 2

IA-9

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

9. (a) and (b) All tangent lines are identical to the graph of the original function.

10. (a) and (b) All tangent lines are identical to the graph of the original function. y 5 4 3 2 1

y 4 3 2 1 −2 −1−1

1 2 3 4 5 6 7 8 x

−2 −3 −4 −5 −6

1 1 1 1 ; (d) , , 2 2 2 2 11. (a) and (b)

(c) f ¿1x2 = y 8 7 6 5 4 3 2 1

1 2 (0, 0)

y 10 8 6 4 (0, −2) 2

−3 −2 (−2, 0)

−1 −2 −4 −6 −8 −10

3 x

−5 −4 −3

−1−1 −2 1 −2, − ( 2 ) −3 −4 −5

(0, 0) 1

2

3 x

(1, 0)

y

40 (−2, 31)

1

x

f

−2

2

1 ; x2 1 (d) - 4, does not exist, - 1

4

x

120

0

5 0

A - 23, 2 + 223 B , or approximately 1- 1.73, 5.462 78. A 22, 1 - 422 B , or approximately 11.41, - 4.662; A - 22, 1 + 422 B , or approximately 1 - 1.41, 6.662 85. A - 2 + 23, 43 - 23 B , or approximately 1- 0.27, - 0.402; A 2 - 23, 43 + 23 B , or approximately 1- 3.73, 3.072 86. A 1 + 26, - 11 3 - 326 B , or approximately 13.45, - 11.022; A 1 - 26, - 11 + 326 B , or approximately 3 1- 1.45, 3.682 106. f ¿1x2 = 3x 2 + a; if a 7 0, f ¿1x2 is always positive, but if a 6 0, f ¿1x2 can be negative. d d 0 120. 1 = x = 0x -1 = 0 dx dx 123. y = x 4 − 3x 2 + 1 124. y = 1.6x 3 − 2.3x − 3.7 2 −4 −2

f (1, 2)

−6

(c) f ¿1x2 = -

1 (d) - , does not exist, - 2 2 54. g¿1x2 is not defined at x = 0 because the graph has vertical slope there; g1x2 is differentiable for all x Z 0. 55. (a) lim F1x2 = 5, F122 = 5; therefore, lim F1x2 = F122;

126.

y=

1

−1

2 ; x2

1- 0.692, - 2.6392, 10.692, - 4.7612

10, 12, 1- 1.225, - 1.252, 11.225, - 1.252 125. y = 10.2x 4 − 6.9x 3

1 2 3 4 5 x

−3 −4 −5

4

2

−2

y 5 4 3 2 1

x:2

5

2

(c) f ¿1x2 = 10x - 2; (d) - 22, - 2, 8 16. (a) and (b) There is no tangent line at x = 0.

−5 −4 −3 −2 −1−1 (−2, −1) −2

0 0

(d) y = 1.67t + 23.3, $1.67 million>yr; y = - 8.17t + 33.17, - $8.17 million>yr; y = - 4.003t + 29.003, - $4 million>yr

−20

(1, 1) 1 2 3 4 5 x

(0, 7)

20 (1, 10)

−4

5 0

Exercise Set 1.5, p. 154

60 (1, 3)

120

77. A 23, 2 - 223 B , or approximately 11.73, - 1.462;

−3 −2 −1 −1 −2

f

(c) f ¿1x2 = -

f

(c) f ¿1x2 = 2x - 1; (d) - 5, - 1, 1 14. (a) and (b)

(c) f ¿1x2 = 4x + 3; (d) - 5, 3, 7 15. (a) and (b) There is no tangent line for x = 0. y 5 4 3 2 1

y 8 7 6 5 4 3 2 1

(1, 2)

(c) f ¿1x2 = 2x + 1; (d) - 3, 1, 3 13. (a) and (b) f

3 3 3 3 ; (d) , , 4 4 4 4 12. (a) and (b)

(−2, 6)

−3 −2 −1 −1 −2

f

(c) f ¿1x2 =

f

120

0 1 2 3 4 5 x

−2 −3 −4 −5

f

(−2, 2)

−5 −4 −3 −2 −1−1

(b) y = 21.545t + 3.455

64. (a)

1

−4

4

−1

10, 02, 10.507, - 0.2252

−3

1- 0.346, - 2.1912, 11.929, 2.3582 128. 10

100

127. f

5x 2 + 8x − 3 3x 2 + 2 3

f´

f

x:2

(b) no, the graph has a corner there 56. (a) lim G1x2 = 1, G112 = 1; therefore,

−6

6

x:1

−4

4 f´

lim G1x2 = G112 (b) yes, G¿112 = 3

x:1

−100

f ¿112 = 45

−10

f ¿112 = - 2

IA-10

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S 2

129.

114. 13x 3 + 6x - 2213x 4 + 72 + x19x 2 + 6213x 4 + 72 + x13x 3 + 6x - 22112x 32, or 72x 7 + 108x 5 - 30x 4 + 84x 3 + - 4x x2 + 1 b + 2 84x - 14 115. 3x 2 a 2 1x 3 - 82, or x - 1 1x - 122 3x 6 - 4x 4 - 3x 2 + 32x 1x 2 - 122 6t 2 t3 - 1 116. 5t 4 a 3 b + a 3 b 1t 5 + 32 t + 1 1t + 122 - 2x 1 118. (b) g¿1x2 = 119. (b) g¿1x2 = 2 1x + 122 1x - 122 126. 127. 5 20

1.5

130.

f´

f

f

−2

2 −1

2 f´

−1.5

−4

f ¿112 = 1

f ¿112 = 1 5

131.

6

132.

f´ f´

f

f −5

5

−4

4

f

f´

f´

f´ f

−3

f ¿ 112 = 0

−3

−3

f ¿112 = 1.2

4 f' f −2

2

−5

+ 11

22t 3x - 16x 3 - 2 24. 1x 3 - 122 6

39.

1x

1>2

+ 32 A

6x 2>3 A 2x + 3 B 2

3

−2

2

1- 0.596, - 0.8942, 10, - 12 0.15 130.

1>3

- 8 + 2x - 15x 1>6

- 72 A

1 -1>2 2x

B

47.

2>3

2

−1.5

, or

3 x - 5B2 A2

x1x + 22 2x -1 + 1 , or , for x Z 0 -1 2 1x + 12 1x - 122

1t 2 - 2t + 421- 2t + 32 - 1- t 2 + 3t + 5212t - 22 1t 2 - 2t + 422

,

- t - 18t + 22 1t 2 - 2t + 422 1- t 2 + 4t + 1216t + 22 - 13t 2 + 2t - 121- 2t + 42 48. , 1- t 2 + 4t + 122 14t 2 + 4t + 6 49–96. Left to the student or 2 1t - 4t - 122 97. (a) y = 2; (b) y = 12 x + 2 98. (a) y = 12; 12 7 21 (b) y = 25 x + 25 99. (a) y = x + 5; (b) y = 21 4 x - 4 2 - 41t 2 - 12 - 50012t - 92 110. (a) P¿1t2 = 111. (a) T¿1t2 = 2 2 12t + 92 1t 2 + 122 30x 2 + 224x - 24 15x 2 + 112x - 12 112. , or , for x Z 0 2 2 18 + 10x 2 215x 2 + 422 2

or

f´

f

1x 1>3 - 52 A 12 x -1>2 B - 1x 1>2 + 42 A 13 x -2>3 B

41.

–1

1- 0.2, - 0.752, 10.2, 0.752 5 131.

f´

6x A 2 3 x - 5B - 2x -1 - 2x 42. , or 2 , for x Z 0 1x + x -122 1x + 122 2 - 21x - 3x - 12 5x 2 - 6x + 5 45. 46. 2 2 1x + 12 1x 2 - 122 or

1 f´

12x + 32 x + 3x 2 + 2x 38. + 12x 2 1x 2 + 122

40.

–1

4

- 2x15x 3 - 3x - 152

A 2x + 3 B 2

6 - 2x + 21x 1>6

f

f´

f

−3

23.

B - 1x

129. 1

f´

4

1 -2>3 3x

There are no points at which the tangent line is horizontal.

2

Exercise Set 1.6, p. 163 17

4

−20

10, 02, 1- 1.414, - 42, 11.414, - 42 128.

−2

12. 32t +

−4 f

(b) f ¿1x2 = 3x 2 - x; x = 0, x = 23; (c) 0 6 x 6 23; (d) no, all x such that f ¿1x2 = 0 were found in part (b)

133. (a)

3

,

1.5

−0.15

10.4, 0.031252, 10, 02, 1- 0.4, 0.031252

Exercise Set 1.7, p. 173

f

−3

3

−4

1- 1, - 22, 11, 22

17. 9x 211 + x 322 - 32x 712 + x 823 18. 15x 213 + x 324 28x 611 + x 723 26. 15x + 22312x - 3271120x - 282, or 415x + 22312x - 327130x - 72 5x 4 + 6 5x 4 + 6 29. , or 2>3 4 5 2 3x 1x + 622>3 32 3 1x + 6x2 2 - 24x 21x 2 - 12 2x - 2x 2 + 2 b # 2 32. 3a 2 , or x + 1 1x + 122 1x 2 + 124 -7 13 33. 34. 3>2 1>2 21x + 32 14 - x2 212x + 321>215 - x23>2 3 2 99 35. 20012x - 3x + 4x + 12 13x 2 - 3x + 22 36. 20417x 4 + 6x 3 - x2203128x 3 + 18x 2 - 12 -1 1 4 - x 3 -2>3 x 4 - 2x 3 + 8x - 4 39. 40. a b 3>2 1>2 3 x - x2 1x - x 222 1x - 12 1x + 12 3 6 - 12x + 32 16x + 612 15x - 42 1120x + 1072 41. 42. 6 13x - 22 16x + 124 2 2x - 3x x + 1 61. (b) 65. f 1x2 = , g1x2 = x 3 x - 1 11 + x26 71. f ¿1x2 = 632x 3 + 14x - 52245 36x 2 + 814x - 524

IA-11

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

72. f ¿1x2 =

3 A - x + 4x + 22x + 1 B

# a - 5x

b 22x + 1 1 3 73. f ¿1x2 = a2x b 2 221 - 3x 23x + 21 - 3x 50012x - 0.12 4000x 77. P¿1x2 = 31x 2 + 222>3 2x 2 - 0.1x 95w 85. (a) D1c2 = 4.25c + 106.25, c1w2 = L 2.199w; 43.2 (b) 4.25 mg>unit of creatine clearance; (c) 2.199 units of creatine clearance> kg 1 1 # a1 + 1 b 87. 1 + + 22x 22x 23x + 2x 4 2 12t + 27t + 48 92. 94. 3x 21x 2 + 123>2 + 3x 41x 2 + 121>2, 21t 4 + 3t 2 + 823>4 1 or 13x 2 + 6x 4221 + x 2 95. 11 - x221 - x 2 2 31x - x - 1221x 2 + 4x - 12 1 96. 97. 1x 2 + 124 11 + u223>2 2 x + x - 2 98. 12x + 123>21x 2 - 4x21>2 5

2

4

1

+ 4 +

100. 2436x13 - x25 + 24313 - x241- 6x + 32, or - 7236x13 - x25 + 24313 - x2412x - 12 8 10 103. 104.

58. (b) - 3.52, - 1.25, 0.031, 0.00018 65. k1k - 121k - 221k - 321k - 42x k - 5 3 -6 18 67. f ¿1x2 = , f –1x2 = , f ‡1x2 = , 1x + 222 1x + 223 1x + 224 - 72 -5 10 68. f ¿1x2 = , f –1x2 = f 41x2 = , 5 2 1x + 22 1x - 22 1x - 223 - 30 120 f ‡1x2 = , f 41x2 = 4 1x - 22 1x - 225 3 3 74. 75. a

−5

5

1 - 2.14476, - 7.7282, 12.14476, 7.7282 107.

a

22x - 1 5 A 22x - 1 + x 3 B 4312x - 12-1>2 + 3x 24

Exercise Set 1.8, p. 182

v1t2 switches at t = 0. 77.

3 a

s

3

v

s

−2

4

−9

−3

v1t2 switches at t = 1.

v1t2 switches at t = 0.604 and t = - 1.104.

x 2 + 4x - 21 = x : -7 x + 7 1x + 721x - 32 lim = x : -7 x + 7 lim 1x - 32 = - 10

y 10 8 6 4 2

5

−10 −8 −6 −4 −2−2

17. lim

2 4 6 8 10 x

x : -7

−4 f −6 −8 −10

−10

22. Not continuous, since lim g1x2 does not exist , or

141x 2 + 3x25113x 2 + 39x + 272 61x 3 + 2x24151x 4 + 72x 2 + 202 1012x 2 - 3x + 1281152x 2 - 228x + 852 1013x 2 + 2x + 123181x 2 + 54x + 112 31x 2 + 22 41x 2 - 32 45x 4 - 54x 2 - 3 23. 24. 27. 41x 2 + 125>4 91x 2 - 124>3 161x 3 - x25>4 6 3 40x + 56x - 2 28. 29. 85 x -3>4 - 14 x -3>2 30. 43 x -2>3 + 14 x -3>2 91x 4 + x24>3 53. (a) The velocity at t = 20 sec is greater, since the slope of a tangent line is greater there. (b) The acceleration is positive, since the velocity (slope of a tangent line) is increasing over time. 54. (a) The velocity is greatest at time 0. The slope of a tangent line is greatest there. (b) The acceleration is negative, since the slopes of tangent lines are decreasing. 55. (a) $146,000>month, $84,000>month, - $4000>month; (b) - $68,000>month2, - $56,000>month2, - $32,000>month2 56. (a) 2 items>day, 14 items>day, 74 items>day; (b) 6 items>day2, 18 items>day2, 42 items>day2 19. 20. 21. 22.

v

−3

1- 1.47481, 9.48782

5 A 22x - 1 + x 3 B 4 A 3x 2 22x - 1 + 1 B

3

−3

v1t2 switches at t = - 1.29 and t = 1.29. 9 76.

f´ −8

−3

a

f −5

f

v

−3

16. 3

s

s

Chapter Review Exercises, p. 190

f´ −3

v

x : -2

29. Not continuous, since lim g1x2 does not exist x : -2

- x 2 + 16x + 8 50. (a) AC1x2 = 5x -1>2 + 100x -1, 18 - x22 AR1x2 = 40, AP1x2 = 40 - 5x -1>2 - 100x -1 50 56. 1- 1.7137, 37.4452, 10, 02, f f´ 11.7137, - 37.4452 43.

−3

3

−50

Chapter 1 Test, p. 192

1. (b) lim- f 1x2 = 12; lim+ f 1x2 = 12 ; lim f 1x2 = 12 2.

x:6 y 18 16 14 12 10 x 2 − 36 8 f(x) = x−6 6 4 2

−8 −6 −4 −2−2

2 4 6 8 10 12 x

x:6

x:6

38. (a) AR = 50, AC = x -1>3 + 750x -1, AP = 50 - x -1>3 - 750x -1; (b) average cost is dropping at approximately $11.74 per item.

IA-12 43.

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

50

Exercise Set 2.1, p. 212 f

f´

1. Relative minimum at 1- 2, 12

0

5

f(x) = x 2 + 4x + 5

−30

11.0836, 25.10292 and 12.9503, 8.62472

Extended Technology Application, p. 195 1.

160

160

y 10 9 8 7 6 5 4 3 2 1

−6 −5 −4 −3 −2 −1−1

f(x) = x 2 + 6x − 3

1 2 x

3. Relative maximum at A - 12, 214 B 0

500 0

0

500 0

Xscl = 50, Yscl = 20 3

Xscl = 50, Yscl = 20

2. (a) y = - 0.0000045x + 0.000204x + 0.7806x + 4.6048; (b) 160 (c) acceptable fit;

0

2

500 0

y 7 6 f(x) = 5 − x − x 2 5 4 3 2 1 −5 −4 −3−2 −1−1 −2 −3 −4 −5

Xscl = 50, Yscl = 20 2

(e) dy>dx = - 0.0000135x + 0.000408x + 0.7806; (f ) approximately 1256, 1422; at about 256 ft from home plate, the ball reached its maximum height of approximately 142 ft. 3. (a) y = - 0.0000000024x 4 - 0.0000026x 3 - 0.00026x 2 + 0.8150x + 4.3026; (b) 160

1 2 3 4 5 x

5. Relative minimum at 1- 1, - 22 y 5 4 3 g(x) = 1 + 6x + 3x 2 2 1 −5 −4 −3 −2 −1−1

2. Relative minimum at 1- 3, - 122

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10 −12 −14

Xscl = 50, Yscl = 20

(e) dy>dx = - 0.0000000096x 3 - 0.0000078x 2 0.00053x + 0.815; (f) approximately (257, 142); at about 257 ft from home plate, the ball reached its maximum height of approximately 142 ft. dy 303.75 - 0.003x 2 4. (a) (c) = ; 160 dx 2202,500 - x 2 (d) approximately (318, 152); at about 318 ft from home plate, the ball reached its maximum height of approximately 152 ft. 0

y 5 4 3 f(x) = 2 − 3x − 2x 2 2 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

6. Relative minimum at 1- 2, - 132 y 6 4 2 −10 −8 −6 −4 −2 −2 −4 −6 −8 −10 −12 −14

1 2 3 4 5 x

7. Relative minimum at 11, 12; relative maximum at A - 13, 59 27 B

2 4 6 8 10 x

F(x) = 0.5x 2 + 2x − 11

−5 −4 −3 −2 −1−1

9. Relative minimum at 11, 42; relative maximum at 1- 1, 82

Technology Connection, p. 208 f(x) = 2 − (x − 1)2/3

f´(x) = − 2 (x − 1)−1/3 3 4

2.

4

−4

6

−2

−4

6

−2

The derivative is not defined at 11, 22.

y 9 8 7 6 5 4 3 2 1

1 2 3 4 5 x

−2 −3 −4 −5

G(x) = x 3 − x 2 − x + 2

Chapter 2

8. Relative minimum at A 23, 113 27 B ; relative maximum at A - 1, 13 2 B

y 5 4 3 2 1

500 Xscl = 50, Yscl = 20

1.

x

500 0

0

2

4. Relative maximum at A - 34, 258 B

−2 −3 −4 −5

0

y 10 8 6 4 2

y 9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2 −3

f(x) = x 3 − 3x + 6

1 2 3 4 5 x

−5 −4 −3 −2 −1−1 −2 −3 g(x) = x 3 +

1 2 3 4 5 x

1 2 x − 2x + 5 2

10. Relative minimum at 12, - 42; relative maximum at 10, 02 y 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2 −3 −4 −5

1 2 3 4 5 x

f(x) = x 3 − 3x2

IA-13

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

11. Relative minimum at 10, 02; relative maximum at 1- 1, 12 y 5 4 3 2 f(x) = 3x2 + 2x3 1 −5 −4 −3 −2 −1−1

1 2 3 4 5 x

10 −3 −2 −1 −10

g(x) = 2x 3 − 16

−5 −4 −3 −2 −1−1

y

10 1 2 3 4 5 6 7 x

−10 −20 G(x) = x 3 − 6x2 + 10

y 14 12 10 8 6 4 2

3 x

−2 −3 −4 −5

17. Relative maximum at 27 A 34, 256 B

g(x) = x 3 − x4 0.5

1.0

x

19. No relative extrema exist. y 10 8 6 4 2 −4 −6 −8 −10

1 3 x − 2x 2 + 4x − 1 3

1

2

x

−10 −8 −6 −4 −2

2 F(x) = x − 1 2 4 6 8 10 x

−2

−2

f(x) = 1 – x 2/3

0.5

−1

1

2 x

−0.5 −1

27. Relative minimum at 10, - 82

20. No relative extrema exist.

−5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

1 2 3 4 5 x

1 F(x) = − x3 + 3x 2 − 9x + 2 3

y 3 2 1 −2 −3

−16 −12 −8 −4 −1 −2 −3 −4 −5 −6 −7

4

8 x

28. Relative maximum at 10, 52

4

2

5 F(x) = 2 x +1

1 −10 −8 −6 −4 −2 −1

1 2 3 4 5 x

−4 −3−2 −1−1

y 3 f(x) = (x + 3)2/3 − 5 2 1

3 −8 G(x) = 2 x +1

29. Relative minimum at 1- 1, - 22; relative maximum at 11, 22

−4 −6 −8 −10

26. Relative minimum at 1- 3, - 52

y 5

y 1

−2.5 −1.5 −0.5 0.5 1.5 2.5 x −0.5 −1.0 −1.5 −2.0 f(x) = x 4 − 2x 3 −2.5

3

G(x) = x + 2 2 4 6 8 10 x

−1

1

18. Relative minimum at A 32, - 27 16 B

−10 −8 −6 −4 −2

−1

1 2 3 4 5 x

−2

1

3

y

y 2.5 2.0 1.5 1.0 0.5

24. No relative extrema exist.

2

25. Relative maximum at 10, 12

−10 −15 −20 −25

−5 −4 −3 −2 −1−2

−1 −2 −4 −6

y

1

y 10 8 6 4 2 1 2 3 4 5 x

y

F(x) = 1 − x3

16. Relative minimum at 1- 3, - 152; relative maximum at 11, 172

−5 −4 −3 −2 −1−5

−2

23. No relative extrema exist. 1 2 3 4 5 x

f(x) = 12 + 9x − 3x2 − x3

f(x) =

2

f(x) = 3x 4 − 15x 2 + 12

y 25 20 15 10 5

−5 −4 −3 −2 −1−2

1

g(x) = 2x 4 − 20x2 + 18

14. No relative extrema exist.

1 2 3 4 5 x

20

−1.0 −0.5−0.1 −0.2 −0.3 −0.4 −0.5

20

1 2 3 4 5 x

−2 −3 −4 −5

y 5 4 3 2 1

15. Relative minimum at 14, - 222; relative maximum at 10, 102

y 0.5 0.4 0.3 0.2 0.1

y 30

f(x) = x 3 + 3x

22. Relative minima at 210 27 a, - b and 2 4 210 27 a , - b; 2 4 relative maximum at 10, 122

−30

y 15 10 5

−3 −2 −1

21. Relative minima at A - 25, - 32 B and A 25, - 32 B ; relative maximum at 10, 182

−20

13. No relative extrema exist.

−10 −15 −20 −25 −30 −35

y 5 4 3 2 1 −5 −4 −3 −2 −1−1

−2 −3 −4 −5

−5 −4 −3 −2 −1−5

12. No relative extrema exist.

2 4 6 8 10 x

30. Relative minimum at 10, 02 y 1.5 1

x2 g(x) = 2 x +1

0.5 1 2 3 4 x 4x g(x) = ——– x2 + 1

−5 −4 −3 −2 −1 −0.5 −1 −1.5

1 2 3 4 5 x

IA-14

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

31. No relative extrema exist.

32. No relative extrema exist.

y 2 1

f(x) = (x +

−5 −4 −3 −2 −1 −1

1 2 3 4 5 x

1)1/3

1 2 3 4 5 x −10

−3

−2

34. Relative maximum at 10, 12

33. Relative minimum at 1- 1, 22

Relative minima at 1- 2, 12 and 12, 12; relative maximum at 10, 2.5872 f(x) = x 9 − x 2 100. 5

y 1.5

y 10 9 8 7 6 5 4 3 2 1

1 0.5 −8

−4 −0.5

−4

8

12 x

4

8

x

−1.5

−5

Relative minimum at 1- 2.12, - 4.52; relative maximum at 12.12, 4.52 f(x) = | 2x − 5 | 102. f(x) = | x − 2 | 101.

87. Relative minimum at 11.94, 15,8822; relative maximum at 17.05, 17,7732

88. Relative maximum at 1150, 22,5062 –4 25,000 N(a) = −a2 + 300a + 6 20,000

17,000 16,000

10,000

15,000

5,000 6

8

–3

10 t

100

200

300 a

|

103.

7 –2

–2

Relative minimum at A 52, 0 B ; increasing on A 52, q B ; decreasing on A - q , 52 B ; f ¿ does not exist at x = 52

Relative minimum at 12, 02; increasing on 12, q 2; decreasing on 1- q , 22; f ¿ does not exist at x = 2

15,000

4

6

N(a)

E(t) = −28.31t3 + 381.86t2 − 1162.07t + 16,905.87

2

5

4

g(x) = √x 2 + 2x + 5

E(t)

4

1 F(x) = √x 2 + 1

−1 4

10 0

−2

−1

18,000

|4 − x 2| + 1

1

−5 −4 −3−2 −1

−12 −8 −4 −1

3

6

y 3 2

3

f (x) = x

f(x) =

99.

f(x) = x 2 − 1

|

3

89. Relative maximum at 16, 102.22 T(t) 105 104 103 102 101 100 99 98 97 96 95

2

–2 T(t) = −0.1t2 + 1.2t + 98.6 –2

Relative maximum at 10, 12; relative minima at 1- 1, 02 and 11, 02; increasing on 1- 1, 02 and 11, q 2; decreasing on 1 - q , - 12 and 10, 12; f ¿ does not exist at x = - 1 and x = 1 104. f(x) = | x 2 − 3x + 2 |

1 2 3 4 5 6 7 8 9 10 11 12 t

3 f (x) = −x 6 − 4x 5 + 54x 4 + 160x 3 − 641x 2 − 828x + 1200

97.

7000

−8

8

Relative minima at 1- 3.683, - 2288.032 and 12.116, - 1083.082; relative maxima at 1- 6.262, 3213.82 and 1- 0.559, 1440.062 and 15.054, 6674.122

f (x) = x 4 + 4x 3 − 36x 2 − 160x + 400 800

−10

10

−400

Relative minima at 1- 5, 4252 and 14, - 3042; relative maximum at 1- 2, 5602

3

A 32, 14 B ; relative minima at 11, 02 and 12, 02; increasing on A 1, 32 B and 12, q 2; decreasing on 1- q , 12 and A 32, 2 B ; f ¿ does not exist at x = 1 and x = 2 –1

Relative maximum at

− 3000

98.

0

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

|

105.

f(x) = 9 − x 2

IA-15

Technology Connection, p. 231

|

10

f´(x) = 3x 5 − 5x 3

f´(x) = 15x 4 − 15x 2

10

10

−3 –4

3

−3

3

4 –1

Relative maximum at 10, 9); relative minima at 1- 3, 02 and 13, 02; increasing on 1- 3, 02 and 13, q 2; decreasing on 1- q , - 32 and 10, 32; f ¿ does not exist at x = - 3 and x = 3 106. f(x) = | − x 2 + 4x − 4 |

−10

−10

f ˝(x) = 60x 3 − 30x 10

5 −3

0

−10

4

Exercise Set 2.2, p. 231

–1

Relative minimum at 12, 02; increasing on 12, q 2; decreasing on 1- q , 22 107. 108. f(x) = | x 3 − 1 | f(x) = | x 4 − 2 x 2 | 10

9

–2

–2

3 –1

3

2

7. Relative minimum is f A 12 B = - 1; relative maximum is f A - 12 B = 3. 8. Relative minimum is f 122 = - 17; relative maximum is f 1- 22 = 15. y f(x) = x 3 − 12x 9. Relative minimum at 12, - 162, 16 relative maximum at 1- 2, 162; 12 8 inflection point at 10, 02; increasing 4 on 1- q , - 22 and 12, q 2, decreas−5 −4 −3 −2 −1 1 2 3 4 5 x ing on 1- 2, 22; concave down on −4 1- q , 02, concave up on 10, q 2 −8 −12 −16

–1

Relative minimum at 11, 02; increasing on 11, q 2; decreasing on 1 - q , 12; f ¿ does not exist at x = 1

Relative maxima at 1- 1, 12 and 11, 12; relative minima at 1- 1.41, 02, 10, 02, and 11.41, 02; increasing on 1- 1.41, - 12, 10, 12, and 11.41, q 2; decreasing on 1- q , - 1.412, 1- 1, 02, and 11, 1.412; f ¿ does not exist at x = - 2 and x = 2

10.

y 50 f(x) = x 3 − 27x 40 30 20 10 −5 −4 −3 −2 −1 −10 −20 −30 −40 −50

Technology Connection, p. 226

11.

1. Relative minimum at 11, - 12; inflection points at 10, 02, 10.553, - 0.5122, 11.447, - 0.5122, and 12, 02

y 50 40 30 20 10

6

6

−1

−5 −4 −3 −2 −1 −10 −20 −30 −40 −50

f´

f

3

−1

3

1 2 3 4 5 x

1 2 3 4 5 x

Relative minimum at 13, - 542, relative maximum at 1 - 3, 542; inflection point at 10, 02; increasing on 1- q , - 32 and 13, q 2, decreasing on 1- 3, 32; concave down on 1- q , 02, concave up on 10, q 2 Relative minimum at 12, - 512, relative maximum at 1- 2, 452; inflection point at 10, - 32; increasing on 1- q , - 22 and 12, q 2, decreasing on 1- 2, 22; concave down on 1- q , 02, concave up on 10, q 2

f(x) = 3x 3 − 36x − 3 −2

−2 6

3 −2

y

f(x) = 2x3 – 3x 2 – 36x + 28

100 80 60 40 20

f˝

−1

12.

−5 −4 −3 −2 −1 −20 −60 −80 −100

2 3 4 5 6 x

Relative minimum at 13, - 532, relative maximum at 1- 2, 722; inflection point at A 12, 19 2 B ; increasing on 1- q , - 22 and 13, q 2, decreasing on 1- 2, 32; concave down on A - q , 12 B , concave up on A 12, q B

IA-16 13.

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

y f(x) = 8– x 3 – 2x + 1– 2 3

3

1 −1

1 x −1

Relative minimum at 1- 6, - 282, relative maximum at 10, 802; inflection point at 1- 3, 262; increasing on 1- 6, 02, decreasing on 1- q , - 62 and 10, q 2; concave up on 1- q , - 32, concave down on 1- 3, q 2

y 80

14. f(x) = 80 − 9x 2 − x 3

60 40 20

−8 −7 −6 −5 −4 −3 −2−1

1 2 x

−20

15.

y 10 5 −2−1−5

Relative minimum at 10, - 42, relative maximum at 12, 02; inflection point at 11, - 22; increasing on 10, 22, decreasing on 1- q , 02 and 12, q 2; concave up on 1- q , 12, concave down on 11, q 2

f(x) = – x 3 + 3x 2 – 4 1 2 3 4 x

−10 −15 −20

16.

y 2 1 −3 −2 −1

17.

f(x) = 1

−1 −2 −3 −4 −5 −6

−x 3

+ 3x − 2

2

3 x

y 6 −1

1

2

3

4 x

−6 −12 −18 −24 −30

f(x) = 3x 4 − 16x 3 + 18x 2

18. −2

−1

1

2 x

−6 −12 −18 −24 −30

f(x) = 3x 4 + 4x 3 − 12x 2 + 5

19.

f(x) = x4 – 6x 2 y 1 −3 −2 −1

−1 −2 −3 −4 −5 −6 −7 −8 −9 −10

1

2

3

x

20.

−2

2x 2

−x 4

1

−1

1

2

x

−1 −2

and 10, 12, decreasing on 1- 1, 02 and 11, q 2; concave down on 1 1 b and a , q b, a - q, 23 23 1 1 concave up on a , b 23 23 f(x) = x 3 – 2x 2 – 4x + 3 Relative minimum at 12, - 52, relative maximum at A - 23, 121 27 B ; 7 inflection point at A 23, - 27 B; increasing on A - q , - 23 B and 12, q 2, decreasing on A - 23, 2 B ; 1 2 3 4 x concave down on A - q , 23 B , concave up on A 23, q B

−3

21.

y 10 8 6 4 2

−3 −2 −1 −2 −4 −6 −8 −10

22.

Relative minima at 10, 02, and 13, - 272, relative maximum at 11, 52; inflection points at 10.451, 2.321) and 12.215, - 13.3582; increasing on 10, 12 and 13, q 2, decreasing on 1- q , 02 and 11, 32; concave up on 1- q , 0.4512 and 12.215, q 2, concave down on 10.451, 2.2152

23.

Relative minima at A - 23, - 9 B and A 23, - 9 B , relative maximum at 10, 02; inflection points at 1- 1, - 52 and 11, - 52; increasing on A - 23, 0 B and A 23, q B , decreasing on A - q , - 23 B and A 0, 23 B ; concave up on 1- q , - 12 and 11, q 2, concave down on 1- 1, 12

Relative minimum at 10, 02, relative maxima at 1- 1, 12 and 11, 12; 1 5 inflection points at a , b and 23 9 1 5 a , b ; increasing on 1- q , - 12 23 9

y 2

f(x) =

Relative minimum at 1- 1, - 42, relative maximum at 11, 02; inflection point at 10, - 22; increasing on 1- 1, 12, decreasing on 1- q , - 12 and 11, q 2; concave up on 1- q , 02, concave down on 10, q 2

Relative minima at 1- 2, - 272 and 11, 02, relative maximum at 10, 52; inflection points at 1- 1.215, - 13.3582 and 10.549, 2.3212; increasing on 1- 2, 02 and 11, q 2, decreasing on 1- q , - 22 and 10, 12; concave up on 1- q , - 1.2152 and 10.549, q 2, concave down on 1- 1.215, 0.5492

y 6 −3

Relative minimum at A 12, - 13 B , relative maximum at A - 12, 1 B ; inflection point at A 0, 13 B ; increasing on A - q , - 12 B and A 12, q B , decreasing on A - 12, 12 B ; concave down on 1- q , 02, concave up on 10, q 2

y 7 6 5 4 3 2 1 −1 −1 −2 −3

1

2

3

4

5 x

f(x) = x 3 − 6x 2 + 9x + 1

y 3 2 1 −2 −1.5 −1 −0.5 −1 f(x) = 3x 4 + 4x 3 −2

0.5

1 x

Relative minimum at 13, 12, relative maximum at 11, 52; inflection point at 12, 32; increasing on 1- q , 12 and 13, q 2, decreasing on 11, 32; concave down on 1- q , 22, concave up on 12, q 2 Relative minimum at 1- 1, - 12; inflection points at A - 23, - 16 27 B and 10, 02; increasing on 1- 1, q 2, decreasing on 1- q , - 12; concave up on A - q , - 23 B and 10, q 2, concave down on A - 23, 0 B

−3

24.

−3 −2 −1

25.

Relative minimum at A 32, - 27 16 B ; inflection points at 10, 02 and 11, - 12; increasing on A 32, q B ,

y 5 4 3 2 1 −1 −2 −3

1

2

3 x

f(x) =

x4

− 2x 3

y 600 450 300 150 −15 −10 −5 −150 −300 −450 −600 −750 −900

5

10 15 x

f(x) = x 3 − 6x 2 − 135x

decreasing on A - q , 32 B ; concave down on 10, 12, concave up on 1- q , 02 and 11, q 2

Relative minimum at 19, - 9722, relative maximum at 1- 5, 4002; inflection point at 12, - 2862; increasing on 1- q , - 52 and 19, q 2, decreasing on 1- 5, 92; concave down on 1- q , 22, concave up on 12, q 2

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

26.

Relative minimum at 18, - 9722, relative maximum at 1- 6, 4002; inflection point at 11, - 2862; increasing on 1- q , - 62 and 18, q 2, decreasing on 1- 6, 82; concave down on 1- q , 12, concave up on 11, q 2

y 400 200 −12 −8 −4 −200

4

8 12 16 x

−400 −600 −800

33.

−1

−1000 f(x) = x 3 − 3x 2 − 144x − 140

27.

Relative minimum at 13, - 172; inflection points at 10, 102 and 12, - 62; increasing on 13, q 2, decreasing on 1- q , 32; concave down on 10, 22, concave up on 1- q , 02 and 12, q 2

y 40 30 20 10 −2 −1 −10

1

2

3

4 x

f(x) = x 4 − 4x 3 + 10

1

2

f(x) =

1

f(x) = (x + 1) 2/3 2

−2

2

y 6 5 4 3 2 1 −1.5 −1 −0.5−1 −2 −3 −4

No relative extrema; inflection point at 12, 22; increasing on 1- q , q 2; concave down on 1- q , 22, concave up on 12, q 2

3

4 x

y 3

f(x)

= x3

0.5 1

+ 3x + 1

2 x

1 x

−0.5

1 −4 −3 −2 −1

1

2

−1 −2 −3

y 100 80 60 40 20 −3 −2 −1 −20 −40 −60 −80 −100

No relative extrema; inflection point at 10, 12; increasing on 1- q , q 2; concave down on 1- q , 02, concave up on 10, q 2

f(x) = 20x 3 – 3x 5

1

2

3

x

Relative minimum at 11, 02; no inflection points; increasing on 11, q 2, decreasing on 1- q , 12; concave down on 1- q , 12 and 11, q 2

1 2 3 4 5 6 x

y 2

f(x) = (x −

−2

2

4

3)1/3

6

8

−1

10 x

−1

No relative extrema; inflection point at 13, - 12; increasing on 1- q , q 2; concave up on 1- q , 32, concave down on 13, q 2

−2 −3 −4

Relative minimum at 1- 1, - 22, relative maximum at 11, 22; inflection points at 1- 0.707, - 1.2372, 10, 02, and 10.707, 1.2372; increasing on 1- 1, 12, decreasing on 1- q , - 12 and 11, q 2; concave down on 1 - 0.707, 02 and 10.707, q 2, concave up on 1- q , - 0.7072 and 10, 0.7072

x

Relative minima at 10, 02 and 11, 1 02, relative maximum at A 12, 16 B; inflection points at 10.211, 0.0282 and 10.789, 0.0282; increasing on A 0, 12 B and 11, q 2, decreasing on 1- q , 02 and A 12, 1 B ; concave down on 10.211, 0.7892, concave up on 1- q , 0.2112 and 10.789, q 2

−1

y 5

38.

f(x) = 5x 3 − 3x 5

Relative minima at 10, 02 and 13, 02, relative maximum at A 32, 81 16 B ; inflection points at 10.634, 2.252 and 12.366, 2.252; increasing on A 0, 32 B and 13, q 2, decreasing on 1- q , 02 and A 32, 3 B ; concave down on 10.634, 2.3662, concave up on 1- q , 0.6342 and 12.366, q 2

Relative minimum at 1- 1, 02; no inflection points; increasing on 1 - 1, q 2, decreasing on 1- q , - 12; concave down on 1- q , - 12 and 1- 1, q 2

y

2

1.5 x

1 −1

1.5

y 4

37.

2

32.

−1

1

31.

1

1

−8

−2

0.5

f(x) = x 2 (1 − x)2

f(x) = (x − 1)2/3

f(x) = x 3 − 6x 2 + 12x − 6

30.

4 x

3

−2

−6

−1 −0.5

36.

4 3 x − 2x 2 + x 3

2

−4

3

y 0.5

3 x

y 4

−1

2

0.25

No relative extrema; inflection point at A 12, 16 B ; increasing on 1- q , q 2; concave down on A - q , 12 B , concave up on A 12, q B

−1 −1 −2 −3 −4 −5

29.

1

−1

34.

35.

y 5 4 3 2 1 −2

f(x) = x 2 (3 − x)2

−0.25

−20

28.

y 7 6 5 4 3 2 1

IA-17

Relative minimum at 1 - 2, - 642, relative maximum at 12, 642; inflection points at A - 22, - 2822 B , A 22, 2822 B , and 10, 02; increasing on 1- 2, 22, decreasing on 1- q , - 22 and 12, q 2; concave down on A - 22, 0 B and A 22, q B , concave up on A - q , - 22 B and A 0, 22 B

f(x) = (x − 2)1/3 + 3

4 3 2 1 −2

39.

2

4

6

8

No relative extrema; inflection point at 12, 32; increasing on 1- q , q 2; concave up on 1- q , 22, concave down on 12, q 2

10 x

−1 y 6 5 4 3 2 1

−2 −1

−2 −3 −4

f(x) = −2(x − 4)2/3 + 5

2

4

6

8 10 12 x

Relative maximum at 14, 52; no inflection points; increasing on 1- q , 42, decreasing on 14, q 2; concave up on 1- q , 42 and 14, q 2

IA-18

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S y f(x) = −3(x − 2)2/3 + 3 3 2 1

40.

−3 −2 −1−1

1 2 3 4 5 6 7 x

−2 −3 −4 −5 −6 −7

46.

Relative minimum at 10, - 42; inflection points at A - 213, - 3 B and A 213, - 3 B ; decreasing on 1- q , 02, increasing on 10, q 2; concave down on A - q , - 213 B

y 1 −6 −4 −2

2

4

6 x

−1 −2

f(x) =

−4 x2 + 1

A 213, q B , concave up on A - 213, 213 B

−3

y 3

41.

2

f(x) = x√4 − x2

1 −3 −2 −1

1

2

3 x

−1 −2 −3

42.

y 1 0.5 −1.5 −1 −0.5

f(x) = −x√1 − x2 0.5

1

1.5 x

−0.5 −1

43.

Relative maximum at 12, 32; no inflection points; increasing on 1- q , 22, decreasing on 12, q 2; concave up on 1- q , 22 and 12, q 2

y x f(x) = ——– x2 + 1

0.5

−3−2 −1

1 2 3 x

−0.5

44.

y 5 4 3 2 1 −10 −8 −6 −4 −2 −1

f(x) =

8x x2 + 1

2 4 6 8 10 x

−2 −3 −4 −5

45.

f(x) =

x2

3 +1

1 2 −1

decreasing on A - 212 , 212 B ; concave up on 10, 12, concave down on 1 - 1, 02

1 Relative minimum at a - 1, - b, 2 1 relative maximum at a1, b; inflection 2 23 b and 10, 02 points at a - 23, 4 23 b ; increasing on and a 23, 4 1- 1, 12, decreasing on 1- q , - 12 and 11, q 2; concave up on A - 23, 0 B and A 23, q B , concave down on A - q , - 23 B and A 0, 23 B Relative minimum at 1- 1, - 42, relative maximum at 11, 42; inflection points at A - 23, - 223 B and 10, 02 and A 23, 223 B ; increasing on 1- 1, 12, decreasing on 1- q , - 12 and 11, q 2; concave up on A - 23, 0 B and A 23, q B , concave down on

inflection points at A - 213, 94 B and

3

−6 −4 −2

Relative minimum at A 212 , - 12 B , relative maximum at A - 212 , 12 B ; inflection point at 10, 02; increasing on A - 1, - 212 B and A 212 , 1 B ,

A - q , - 23 B and A 0, 23 B Relative maximum at 10, 32;

y 4

2

Relative minimum at A - 22, - 2 B , relative maximum at A 22, 2 B ; inflection point at 10, 02; increasing on A - 22, 22 B , decreasing on A - 2, - 22 B and A 22, 2 B ; concave up on 1- 2, 02, concave down on 10, 22

4

6 x

and

−4

103.

104.

y 1400

y 1300 1200 1100 1000 900 800 700 600 500 400 300 200 100

R(x)

C(x)

1200

R(x)

1000 800

P(x)

600

P(x)

400 200 0 20 40 60 80 100 120 x

C(x)

10 20 30 40 50 60 70 80 90100 x

105.

106.

p(x)

y 25

13 12 11 10 9 8 7

20 f 15 10

p

5 10

20

30

40

x 2

f (x) = 3x 2/3 − 2x

122.

6 10 14 18 22 26 30 x f (x) = 4x − 6x 2/3

123. 4

10 −1 −4

10 −8

−4

Relative maximum at 10, 02; relative minimum at 11, - 22 f (x) = x 2 (1 − x )3 125.

Relative minimum at 10, 02; relative maximum at 11, 12 124. f (x) = x 2(x − 2) 3 2

−1

0.2

3

−0.5

−2

1.5

−0.2

Relative minimum at 10, 02; relative maximum at 10.4, 0.0352 127. f (x) = (x − 1) 2/3 − (x + 1) 2/3

Relative minimum at A 45, - 1.106 B ; relative maximum at 10, 02 f (x) = x − x 126. 3

A 213, 94 B ; increasing on 1- q , 02,

6

2

decreasing on 10, q 2; concave up

A 213, q B , concave down on A - 213, 213 B on A - q , - 213 B and

10

−10 0 −1

Relative minimum at 10.25, - 0.252

4 −2

Relative maximum at 1- 1, 1.5872; relative minimum at 11, - 1.5872

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

Technology Connection, p. 240

1. x-intercepts: 10, 02, 13, 02, and 1 - 5, 02; y-intercept: 10, 02 2. x-intercepts: 10, 02, 11, 02, and 1- 3, 02; y-intercept: 10, 02

Exercise Set 2.3, p. 247 23.

y 10 8 6 5 f(x) = − 4 x 2 −10 −8 −6 −4 −2 −2

2 4 6 8 10 x

Increasing on 1- q , 02 and 10, q 2 No relative extrema Asymptotes: x = 0 and y = 0 Concave up on 1- q , 02; concave down on 10, q 2 No intercepts y 24.

−10 −8 −6 −4 −2 −2 −4 −6 −8

f(x) = 4 – x

2 4 6 8 10 x

Decreasing on 1- q , 02 and 10, q 2 No relative extrema Asymptotes: x = 0 and y = 0 Concave down on 1- q , 02; concave up on 10, q 2 No intercepts y 25. 5 4 3 2 1

−4 −2−1 −2 −3 −4 −5

f(x) =

1 x−5

2 4 6 8 10 12 x

Decreasing on 1- q , 52 and 15, q 2 No relative extrema Asymptotes: x = 5 and y = 0 Concave down on 1- q , 52; concave up on 15, q 2 y-intercept: A 0, - 15 B y 26. 5 4 3 2 1

−2−1 −1 −2 −3 −4 −5

y 5 4 3 2 1 −6 −5 −4 −3 −2 −1−1 f(x) =

−4 −6 −8 −10

8 6 4 2

27.

–2 f(x) = ——– x–5 x 1 2 3 4 5 6 7 8

Increasing on 1- q , 52 and 15, q 2 No relative extrema Asymptotes: x = 5 and y = 0 Concave up on 1- q , 52; concave down on 15, q 2 y-intercept: A 0, 25 B

1 2 3 4 x

−2 −3 −4 −5

1 x+2

Decreasing on 1- q , - 22 and 1- 2, q 2 No relative extrema Asymptotes: x = - 2 and y = 0 Concave down on 1- q , - 22; concave up on 1- 2, q 2 y-intercept: A 0, 12 B y 28. 5 4 3 2 1

−2 −1 −1 −2 −3 −4 −5

1 f(x) = ——– x–3

1 2 3 4 5 x

Decreasing on 1- q , 32 and 13, q 2 No relative extrema Asymptotes: x = 3 and y = 0 Concave down on 1- q , 32; concave up on 13, q 2 y-intercept: A 0, - 13 B y 29. 5 4 3 2 1

−6 −4 −2−1 −2 −3 −4 −5

f(x) =

−3 x−3

2 4 6 8 10 x

Increasing on 1- q , 32 and 13, q 2 No relative extrema Asymptotes: x = 3 and y = 0 Concave up on 1- q , 32; concave down on 13, q 2 y-intercept: 10, 12 y 30. –2 f(x) = ——– x+5

8 7 6 5 4 3 2 1

−10−9−8−7−6−5−4−3−2−1 −1

1 2 x

−2 −3 −4 −5 −6 −7 −8

Increasing on 1- q , - 52 and 1- 5, q 2 No relative extrema Asymptotes: x = - 5 and y = 0 Concave up on 1- q , - 52; concave down on 1- 5, q 2 y-intercept: A 0, - 25 B

IA-19

IA-20

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

31.

35.

y 8 7 6 5 4 3 2 1

f(x) =

−5 −4 −3 −2 −1 −2

3x − 1 x

−4 −6

−5 −4 −3 −2 −1−1

Increasing on 1- q , 02 and 10, q 2 No relative extrema Asymptotes: x = 0 and y = 3 Concave up on 1- q , 02; concave down on 10, q 2 x-intercept: A 13, 0 B y 32.

−2 −1

4 3

2

10 8 6 4 2

x

−10 −8 −6 −4 −2 −2

1

x

2

10 8 6 4 2

f(x) =

x x+2

2 4 6 8 10 x

−4 −6 −8 −10

Increasing on A - q , - 22 B and A 22, q B ; decreasing on A - 22, 0 B and A 0, 22 B Relative minimum at A 22, 222 B ;

relative maximum at A - 22, - 222 B ; Asymptotes: x = 0 and y = x Concave down on 1- q , 02; concave up on 10, q 2 No intercepts y 34. 9 9 6

f(x) = x + – x

−3

(3, 6)

3

6

Increasing on 1- q , - 22 and 1- 2, q 2 No relative extrema Asymptotes: x = - 2 and y = 1 Concave up on 1- q , - 22; concave down on 1- 2, q 2 x- and y-intercept: 10, 02 y 38. Decreasing on54 1- q , 32 and 13, q 2 x 3 f(x) = ——– x–3 2 1 −5 −4−3−2−1 1 2 3 4 5 x −1 −2 −3 −4 −5

3 −6

−2 −1

Increasing on 1- q , 02; decreasing on 10, q 2 No relative extrema Asymptotes: x = 0 and y = 0 Concave up on 1- q , 02 and 10, q 2 No intercepts y 37.

−10 −8 −6 −4 −2 −2

2 4 6 8 10 x

−4 −6 −8 −10

−9

2 f(x) = — x2

1

x

Decreasing on 1- q , 02 and 10, q 2 No relative extrema Asymptotes: x = 0 and y = 2 Concave down on 1- q , 02; concave up on 10, q 2 x-intercept: A - 12, 0 B y 33. 2 f(x) = x +

Decreasing on 1- q , 02; increasing on 10, q 2 No relative extrema Asymptotes: x = 0 and y = 0 Concave down on 1- q , 02 and 10, q 2 No intercepts y 36.

2 1

f(x) = −12 x

−10

2x + 1 f(x) = ——– x

−2 −4

1 2 3 4 5 x

−8

1 2 3 4 5 x

−2

10 8 6 4 2

y 2

9

x

−3 −6 (–3, – 6) −9

Increasing on 1- q , - 32 and 13, q 2; decreasing on 1- 3, 02 and 10, 32 Relative minimum at 13, 6); relative maximum at 1- 3, - 62 Asymptotes: x = 0 and y = x Concave down on 1- q , 02; concave up on 10, q 2 No intercepts

No relative extrema Asymptotes: x = 3 and y = 1 Concave down on 1- q , 32; concave up on 13, q 2 Intercept: 10, 02

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S y 0.25

39.

43.

IA-21

y 8

−5 −4 −3 −2 −1

1 2 3 4 5 x

−0.25

−1 x2 + 2

f(x) =

−0.5

4 2 1

−0.75

Decreasing on 1- q , 02; increasing on 10, q 2 Relative minimum at A 0, - 12 B Asymptote: y = 0 Concave up on A - 223, 223 B ; concave down on A - q , - 223 B and A 223, q B Inflection points: A - 223, - 38 B and y-intercept: A 0, - 12 B y 40.

A 223, - 38 B

0.5

( ) 1 0, – 3

0.4

−8 −6 −4 −2

−3 −2 −1 −0.1

1

2

6

Inflection points: A - 1, 14 B and A 1, 14 B y-intercept: A 0, 13 B 41. y

2 4 6 8 10 x

−2 −3

Decreasing on 1 - q , - 32, 1- 3, 32, and 13, q 2 No relative extrema Asymptotes: x = 3 and y = 0 Concave down on 1- q , - 32 and 1- 3, 32; concave up on 13, q 2 y-intercept: A 0, - 13 B y 42. 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2 −3 −4 −5

2 4 6 8 10 x

−10

Increasing on 1- q , - 12 and 1- 1, q 2 No relative extrema Asymptotes: x = - 1 and y = 1 Concave up on 1- q , - 12; concave down on 1- 1, q 2 x-intercept: 12, 02; y-intercept: 10, - 22 y 45. 10

1 −6 −4 −2 −1

x−2 x+1

20

x+3 f(x) = x2 − 9

2

f(x) =

−6

3 x

Increasing on 1- q , 02; decreasing on 10, q 2 Relative maximum at A 0, 13 B Asymptote: y = 0 Concave down on 1- 1, 12; concave up on 1- q , - 12 and 11, q 2

8 x

10

−10 −8 −6 −4 −2 −2

0.1

6

Increasing on 1- q , - 22 and 1- 2, q 2 No relative extrema Asymptotes: x = - 2 and y = 1 Concave up on 1- q , - 22; concave down on 1- 2, q 2 x-intercept: 11, 02; y-intercept: A 0, - 12 B y 44.

2

0.2

4

−4

1 x2 + 3

f(x) =

2 −2

0.3

3

x–1 f(x) = ——– x+2

6

f(x) =

x−1 x2 − 1

1 2 3 4 5 x

Decreasing on 1- q , - 12, 1- 1, 12 and 11, q 2 No relative extrema Asymptotes: x = - 1 and y = 0 Concave down on 1- q , - 12; concave up on 1- 1, 12 and 11, q 2 y-intercept: 10, 12

−6 −5 −4 −3 −2 −1

1 2

x

−10 x2 – 4 f(x) = ——– x+3

−20

Increasing on A - q , - 3 - 25 B and A - 3 + 25, q B , or approximately 1- q , - 5.2362 and 1- 0.764, q 2; decreasing on A - 3 - 25, 3 B and A - 3, - 3 + 25 B , or approximately 1- 5.236, - 32 and 1- 3, - 0.7642 Relative maximum at A - 3 - 25, - 6 - 225 B or approximately 1- 5.236, - 10.4722; relative minimum at A - 3 + 25, - 6 + 225 B , or approximately 1- 0.764, - 1.5282 Asymptotes: x = - 3 and y = x - 3 Concave down on 1- q , - 32; concave up on 1- 3, q 2 x-intercepts: 1- 2, 02, 12, 02; y-intercept: A 0, - 43 B y 46. 10 8 x2 − 9 6 f(x) = 4 x+1 2 −10 −8 −6 −4−2 −2 −4 −6 −8 −10

2 4 6 8 10 x

Increasing on 1- q , - 12 and 1- 1, q 2 No relative extrema Asymptotes: x = - 1 and y = x - 1 Concave up on 1- q , - 12; concave down on 1- 1, q 2 x-intercepts: 1- 3, 02 and 13, 02; y-intercept: 10, - 92

IA-22 47.

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

Concave down on 1- q , - 12; concave up on 1- 1, 12 and 11, q 2 x-intercept: 1- 2, 02; y-intercept: 10, 12 y 51.

y 5 4 3 2 1 −4 −3 −2 −1 −2 −3 −4 −5

3

1 2 3 4 5 6 x

2 1 −3 −2 −1

x+1 f(x) = 2 x − 2x − 3

Decreasing on 1 - q , - 12, 1- 1, 32, and 13, q 2 No relative extrema Asymptotes: x = 3 and y = 0 Concave down on 1- q , - 12 and 1- 1, 32; concave up on 13, q 2 y-intercept: A 0, - 13 B y 48. 5 4 3 2 1 −8 −6 −4 −2 −1 −2 −3 −4 −5

2

4

6 x

2

3 x

1 f(x) = ——– x2 – 1

−2 −3

Increasing on 1- q , - 12 and 1- 1, 02; decreasing on 10, 12 and 11, q 2 Relative maximum at 10, - 12 Asymptotes: x = - 1, x = 1, and y = 0 Concave up on 1- q , - 12 and 11, q 2; concave down on 1- 1, 12 y-intercept: 10, - 12 y 52. 10 f(x) = 2 x +4 1

Decreasing on 1 - q , - 52, 1- 5, 32, and 13, q 2 No relative extrema Asymptotes: x = - 5 and y = 0 Concave down on 1- q , - 52; concave up on 1- 5, 32 and 13, q 2 y-intercept: A 0, 15 B y 49. 7 6 5 4 3 2 1 −7−6 −5 −4 −3−2 −1 −1 −2 −3 −4 −5 −6 −7

2x 2 f(x) = ——–— x 2 – 16

1 2 3 4 5 6 7 x

Increasing on 1- q , - 42 and 1- 4, 02; decreasing on 10, 42 and 14, q 2 Relative maximum at 10, 02 Asymptotes: x = - 4, x = 4, and y = 2 Concave up on 1- q , - 42 and 14, q 2; concave down on 1- 4, 42 x- and y-intercept: 10, 02 y 50. 5 4 2 3 f(x) = x + x − 2 2x 2 − 2 2 1

−2 −3 −4 −5

1

3

x−3 f(x) = 2 x + 2x − 15

−5 −4 −3 −2 −1−1

−1

1 2 3 4 5 x

Decreasing on 1 - q , - 12, 1- 1, 12, and 11, q 2 No relative extrema Asymptotes: x = - 1 and y = 12

−10−8 −6 −4 −2

2 4 6 8 10 x

−1

Increasing on 1- q , 02; decreasing on 10, q 2 Relative maximum at 10, 2.52 Asymptote: y = 0 2 2 Concave up on a - q , b and a , q b, or 23 23 approximately 1- q , - 1.15472 and 11.1547, q 2; concave down 2 2 on a , b, or approximately 1- 1.1547, 1.15472 23 23 2 15 2 15 Inflection points: a , b and a , b, or 23 8 23 8 approximately 1- 1.1547, 1.8752 and 11.1547, 1.8752 y-intercept: 10, 2.52 y 53. x2 + 1 f(x) = ——–– x

4 2

−6 −4 −2

2

4

6 x

−2 −4

Increasing on 1- q , - 12 and 11, q 2; decreasing on 1- 1, 02 and 10, 12 Relative maximum at 1- 1, - 22; relative minimum at 11, 22 Asymptotes: x = 0 and y = x Concave down on 1- q , 02; concave up on 10, q 2 No intercepts

IA-23

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

54.

y 7 6 5 x3 f(x) = 2 4 x −1 3 2 1 −6 −5−4 −3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6 −7

Increasing on A - q , - 23 B and A 23, q B ;

decreasing on A - 23, - 1 B , 1- 1, 12, and A 1, 23 B - 323 b , or approximately Relative maximum at a - 23, 2 323 1- 1.732, - 2.5982; relative minimum at a 23, b, 2 or approximately 11.732, 2.598) Asymptotes: x = - 1, x = 1, and y = x Concave down on 1- q , - 12 and 10, 12; concave up on 1- 1, 02 and 11, q 2 Inflection point: 10, 02 x- and y-intercepts: 10, 02 y 55. 10 8 6 4 2 −10 −8 −6 −4 −2−2

A(x) = 3x +

4

65. (c)

48,000 C(p) = ——–– 100 – p

20

66. (c)

300

80 x

8 12 16 20 24 28 x

C(p) $20,000 18,000 16,000 14,000 12,000 10,000 8,000 6,000 4,000 2,000

A(x) 350

(c) Slant asymptote: y = 3x. As x, the number of units produced, increases, the average cost approaches 3x.

40

60

80

100

p

68. (c) 1 5000 A(x) = − x + 400 − 2 x

250 200 150 50

f(x) =

x2 − 9 x−3

2 4 6 8 10 x

Increasing on 1- q , 32 and 13, q 2 No relative extrema No asymptotes No concavity x-intercept: 1 - 3, 02; y-intercept: 10, 32 y 56. 2

−10 −8 −6 −4 −2−2

A(x) 100 90 80 70 60 50 40 30 20 10

100

−4 −6 −8 −10

10 8 6 4 2

64. (b)

f(x) =

x − 16 x+4

200

400

600

800 x

A(t) 100 90 80 70 60 50 40 30 20 10

A(t) =

0

2

4

100 t2 + 1

6

8

10 12 t

69. (b) lim E1n2 = q . The pitcher gives up one or more runs but n:0 gets no one out (0 innings pitched). (c) E = 2.00; pitcher gave up an average of 2 earned runs per game (9 innings). 80.

f (x) = x 2 + 1 x2 20

81.

f (x) = 2

x x2 + 1

−8 −5

8

5 0

82. 2 4 6 8 10 x

−2

83.

3 2 f (x) = x + 4x + x − 6 x2 − x − 2 20

3 2 f (x) = x + 2x − 15x x 2 − 5x − 14 35

−4 −6 −8 −10

Increasing on 1- q , - 42 and 1- 4, q 2 No relative extrema No asymptotes No concavity x-intercept: 14, 02; y-intercept: 10, - 42 - 3x 2 + 15 x2 - 2 59. g1x2 = 2 60. g1x2 = x - 1 x 2 + 2x x - 9 61. h1x2 = 2 x + x - 6 63. (c) V(t) 2 50 40 30 20 10

25t V(t) = 50 – ——— (t + 2)2

5 10 15 20 25 30 35 40 45 50 55 60 65 70 t

−10

14

−10

−8

84.

−10

85.

3 2 f (x) = x + 2x − 3x x 2 − 25 30

−20

20

f (x) =

|

1 −2 x

5

|

20 4

−4 −30

−1

IA-24 86.

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

88. Asymptote: y = x 2 - 6

2 f (x) = x − 3 2x − 4 10

−10

5 f(x) = x + x − 9 and y = x 2 − 6 x 3 + 6x 8

f

10 −5

5 y

−10 −8

x 2 - 10x + 1 89. (b) f ¿1x2 = ; critical values: ' 0.101 and 1x 2 + x - 622 ' 9.899

Technology Connection, p. 253

1. On 3- 2, 14, absolute minimum is - 8 at x = - 2, and absolute maximum is 2.185 at x = - 0.333; on 3- 1, 24, absolute minimum is 1 at x = - 1 and x = 1, and absolute maximum is 4 at x = 2

Exercise Set 2.4, p. 257 9. Absolute maximum: 8 at x = 3; absolute minimum: - 17 at x = - 2 10. Absolute maximum: 6 at x = 1; absolute minimum: 2 at x = - 1 11. Absolute maximum: 15 at x = - 2; absolute minimum: - 13 at x = 5 14. Absolute maximum: 24 for 4 … x … 13; absolute minimum: 24 for 4 … x … 13 15. Absolute maximum: 4 at x = - 1; absolute minimum: - 12 at x = 3 16. Absolute maximum: 10 at x = - 1; absolute minimum: 1 at x = 2 18. Absolute maximum: 4 at x = 1; absolute minimum: - 23 at x = 4 19. Absolute maximum: 50 at x = 5; absolute minimum: - 4 at x = 2 20. Absolute maximum: 24 at x = 3; absolute minimum: 4 at x = 1 21. Absolute maximum: 2 at x = - 1; absolute minimum: - 110 at x = - 5 22. Absolute maximum: 325 at x = - 5; absolute minimum: 0 at x = 0 23. Absolute maximum: 513 at x = - 8; absolute minimum: - 511 at x = 8 24. Absolute maximum: 2000 at x = 10; absolute minimum: - 2000 at x = - 10 25. Absolute maximum: 17 at x = 1; absolute minimum: - 15 at x = - 3 26. Absolute maximum: 10 at x = 0; absolute minimum: - 22 at x = 4 27. Absolute 3 maximum: 32 at x = - 2; absolute minimum: - 27 16 at x = 2 27 3 28. Absolute maximum: 256 at x = 4; absolute minimum: - 2 at x = - 1 29. Absolute maximum: 13 at x = - 2 and x = 2; absolute minimum: 4 at x = - 1 and x = 1 30. Absolute maximum: 12 at x = - 3 and x = 3; absolute minimum; - 13 at x = - 2 and x = 2 31. Absolute maximum: - 1 at x = 5; absolute minimum: - 5 at x = - 3 32. Absolute maximum: 1 at x = 0; absolute minimum: - 3 at x = - 8 and x = 8 1 33. Absolute maximum: 20 20 at x = 20; absolute minimum: 2 at x = 1 34. Absolute maximum: - 4 at x = - 2; absolute minimum: - 17 35. Absolute 2 at x = - 8 maximum: 45 at x = - 2 and x = 2; absolute minimum: 0 at x = 0 36. Absolute maximum: 2 at x = 1; absolute minimum: - 2 at x = - 1 37. Absolute maximum: 3 at x = 26; absolute minimum: - 1 at x = - 2 38. Absolute maximum: 4 at x = 64; absolute minimum: 2 at x = 8 57. Absolute maximum: 223 at x = - 23; absolute minimum: 1025 58. Absolute maximum: at - 223 at x = 23 3 1025 at x = 25 x = - 25; absolute minimum: 3

63. Absolute maximum: 2000 at x = 20; absolute minimum: 16 0 at x = 0 and x = 30 64. Absolute maximum: 3725 27 at x = 3 ; absolute minimum: 0 at x = 0 and x = 8 69. Absolute maximum: 3 at x = - 1; absolute minimum: - 38 at x = 12 70. Absolute maximum: 3 at x = 1; absolute minimum: - 38 at 71. Absolute maximum: 2 at x = 8; absolute x = - 12 minimum: 0 at x = 0 72. Absolute maximum: 2 at x = 4; absolute minimum: 0 at x = 0 75. Absolute maximum: - 1 at x = 1; absolute minimum: - 5 at x = - 1 76. Absolute maximum: 59 at x = - 10; absolute minimum: - 41 at x = 10 77. No absolute maximum; absolute minimum: - 5 at x = - 1 78. Absolute maximum: 19 at x = - 2: no absolute minimum 79. Absolute maximum: 1 at x = - 1 and x = 1: absolute minimum: 0 at x = 0 80. No absolute maximum; absolute minimum: 0 at x = 0 83. Absolute maximum: - 10 3 + 223 at x = 2 - 23; absolute minimum: - 10 223 3 at x = 2 + 23 84. Absolute maximum: 10 3 + 223 at x = - 2 - 23; absolute minimum: 10 223 3 at x = - 2 + 23 85. No absolute maximum; absolute minimum: - 1 at x = - 1 and x = 1 86. No absolute maximum; absolute minimum: 0 at x = - 1 and x = 1 1 5000 104. (a) A1x2 = - x + 400 x 2 y 107. Absolute maximum: 3 at x = 1; absolute minimum: - 5 at x = - 3 4 2 −4

−2

2

4

x

1

2

x

−2 −4

108. Absolute maximum: 10 at x = 2; absolute minimum: 0 at x = 0

y 10 8 6 4 2 −2

−1

109. Absolute maxima: 1 at x = 0 and 1 at x = 2; absolute minimum: - 15 at x = - 4 y −4

−2

2

x

−5

−10

−15

110. Absolute maximum: 265 L 8.062 at x = 67; absolute minimum: 1 at x = 3 y

12 8 4 −2

2 −4

67 x

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

111. (a)

y

y4 = y2 − x2

62.

x 3 = y 2 (2 − x )

63.

8

1.5

6

IA-25

4 2

−1

−2

2

4

6

−8

−1.5 y2 = x3

64.

113. Absolute maximum: 326 at x = 3; absolute minimum: - 2 223 2 at x = - 2 114. Absolute maximum: at x = ; 9 3 absolute minimum: 0 at x = 0 and x = 1 123. (a) P1t2 = t + 8.857; P172 = 15.857 mm Hg; (b) P1t2 = 0.117t 4 - 1.520t 3 + 6.193t 2 - 7.018t + 10.009; P172 = 24.86 mm Hg; P10.7652 = 7.62 mm Hg is the smallest contraction

6

−2

5

−6

Chapter Review Exercises, p. 301 3 14. Relative maximum: 25 4 at x = - 2

Exercise Set 2. 5, p. 273 56. Area is minimized when the circumference of the circle is 10.56 in. and the perimeter of the square is 13.44 in.; area is maximized when the circumference of the circle is 24 in. and the perimeter of the square is 0 in. 3x 2 20 x 60. (b) C¿1x2 = + 8 and A¿1x2 = - 2; 100 50 x (c) minimum = $11>unit at x 0 = 10 units; C¿1102 = $11>unit. 62. Minimum: 6 - 422, or approximately 0.343, at x = 2 - 22 L 0.586 and y = - 1 + 22 L 0.414 63. x = - 22, y = 0, Q = - 322 L - 4.24 aQ 2bQ 64. Order times; lot size: A 2b A a

y 6 5 4 3 2 1 −7 −6 −5 −4 −3−2 −1−1

f(x) = 4 − 3x − x2

15. Relative minima: 2 at x = - 1 and 2 at x = 1; relative maximum: 3 at x = 0 f(x) = x4 – 2x 2 + 3 y 8 7 6 5 4 3 2 1

1. (b) R11002 = $500, C11002 = $190, P11002 = $310; (c) R¿1x2 = 5; C¿1x2 = 0.002x + 1.2; P¿1x2 = - 0.002x + 3.8; (d) R¿11002 = $5, C¿11002 = $1.40, P¿11002 = $3.60 2. (b) R1202 = $800, C1202 = $90, P1202 = $710; (c) R¿1x2 = 50 - x; C¿1x2 = 4; P¿1x2 = 46 - x; (d) R¿1202 = $30, C¿1202 = $4, P¿1202 = $26

−2 −1

42. 50. 53. 55.

x4 = y2 + x6

61.

0.5

3

–8x f(x) = ——– x2 + 1

−5−4−3−2−1 −1 −2 −3 −4

1 2 3 4 5 x

17. No relative extrema y 8 7 6 5 4 3 2 1 −1 −1 −2

−4

5

−3

x

4 3 2 1

- 5x 2y 2 + 3y

- 3xy 2 - 2y

x 2 + y2 = 4

2

y

21. 22. 3xy 2 4x 2y + 3x 3x 3y - 2x dV dV dR dR (a) 43. (a) = 1041.67R = 952.38R dt dt dt dt x11 - y 22 1 - y3 2 2x 51. , or , or 3y 21x + 122 3y 21x + 12 y1x 2 + 122 y11 + x 22 5x 4 - 31x - y22 - 31x + y22 6y 2 - 5x 4 + 6x 2 , or 31x + y22 - 31x - y22 - 5y 4 y15y 3 - 12x2 2 2 2 2 2x1y 3 - x 32 - 61y - xy + x 2 y - x 5 56. 57. , or 12y - x23 y3 y3 y5

60.

1

16. Relative minimum: - 4 at x = 1; relative maximum: 4 at x = - 1

Exercise Set 2.7, p. 292 10.

1 2 3 x

−2 −3 −4

Exercise Set 2.6, p. 285

7 ;3

4

8 x

−2

4x 2 - 2y 3

−2

1

−1

1

−0.5

f(x) = 4 + (x – 1)3

1

2

3 x

IA-26

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

1 18. Relative minimum: 76 27 at x = 3 ; relative maximum: 4 at x = - 1 y 10 9 8 f(x) = x 3 + x 2 – x + 3 7 6 5 4 3 2 1 −2

1

f(x) = 3x 2/3

8 7 6 5 4 3 2 1

−3 −2 −1 −10 −20 −30 −40

f(x) =

3

2x 3

–

3x 2

y 5 4 3 2 1

−1

1

−1

2 x

f(x) = x4 − 2x2

−2

Concave up on A - q , - 213 B and

concave down on A - 213, y 26.

23 B

A 213, q B ;

1

5

4 3 2 1 1 2 3 x

−2

−1

1

−1

2 x

−2 f(x) =

1 2 3 x

−4 −6 −8 −10 −12

1 3 x + 3x 2 + 9x + 2 3

No relative extrema Inflection point at 1- 3, - 72 Increasing on 1- q , q 2 Concave down on 1- q , - 32; concave up on 1- 3, q 2 y 23. 20 16 12 8 4 −8 −12 −16 −20

1 −2

Relative minima: - 1 at x = - 1 and - 1 at x = 1; relative maximum: 0 at x = 0 Inflection points at A - 213, - 59 B and A 213, - 59 B Increasing on 1- 1, 02 and 11, q 2; decreasing on 1- q , - 12 and 10, 12

y 8 6 4 2

−4 −2−4

2

– 12x + 10

1 2 3 4 x

−3−2 −1 −1

f(x) =

4 x

Relative minimum: - 35 at x = 2; relative maximum: 19 at x = - 1 Inflection point at A 12, - 8 B Increasing on 1- q , - 12 and 12, q 2; decreasing on 1- 1, 22 Concave down on A - q , 12 B ; concave up on A 12, q B y 25.

1 2 3 4 x

f(x) = x 3 – 3x + 2

−7 −6 −5 −4 −3−2 −1−2

3

5

21. Relative maximum: 4 at x = - 1; relative minimum: 0 at x = 1

22.

2

4

20. Relative maximum: 17 at x = - 1; relative minimum: - 10 at x = 2 y 40 30 20 10

1

f(x) = 4x3 − 6x2 − 24x + 5

2 x

19. Relative minimum: 0 at x = 0

−4 −3−2 −1

y 24 18 12 6 −3 −2 −1 −6 −12 −18 −24 −30 −36

−1 −1

y

24.

3x4 +

2x3 − 3x2 + 1

1 Relative minima: - 1 at x = - 1 and 11 16 at x = 2 ; relative maximum: 1 at x = 0 Inflection points at 1- 0.608, - 0.1472 and 10.274, 0.8332 Increasing on 1- 1, 02 and A 12, q B ; decreasing on 1- q , - 12 and A 0, 12 B Concave down on 1- 0.608, 0.2742; concave up on 1- q , - 0.6082 and 10.274, q 2 y 27. 80 70 60 50 40 30 20 10

−6 −5 −4 −3 −2 −1 −10 −20 2 4 6 8 10 12 14 16 x

f(x) = x2 − 10x + 8

Relative minimum: - 17 at x = 5 Decreasing on 1- q , 52; increasing on 15, q 2 Concave up on 1- q , q 2

f(x) =

1 2 3 4 x

1 5 3 4 4 3 x + x − x +8 4 3 5

Relative minimum: 457 60 at x = 1; 1208 relative maximum: 15 at x = - 4 Inflection points at 1- 2.932, 53.7012, 10, 82, and 10.682, 7.7692 Increasing on 1- q , - 42 and 11, q 2; decreasing on 1- 4, 12 Concave down on 1- q , - 2.9322 and 10, 0.6822; concave up on 1- 2.932, 02 and 10.682, q 2

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

28.

32.

y 10 8 6 4 2 −5 −4 −3 −2 −1 −2 −4 −6

1 2 3 4 5 x f(x) =

2x + 5 x+1

No relative extrema Decreasing on 1- q , - 12 and 1- 1, q 2 Concave down on 1- q , - 12; concave up on 1- 1, q 2 Asymptotes: x = - 1 and y = 2 x-intercept: A - 52, 0 B ; y-intercept: 10, 52 y 29. 5 4 3 2 1

−4 −3 −2 −1−1

1 2 3 4 5 6 x

−2 −3 −4 −5

f(x) =

x x−2

No relative extrema Decreasing on 1- q , 22 and 12, q 2 Concave down on 1- q , 22; concave up on 12, q 2 Asymptotes: x = 2 and y = 1 x-intercept: 10, 02; y-intercept: 10, 02 y 30. 5 4 3 2 1

−10 −8 −6 −4 −2−1

2 4 6 8 10 x

−2 −3 −4 −5

f(x) =

5 x2 − 16

5 Relative maximum at A 0, - 16 B Decreasing on 10, 42 and 14, q 2; increasing on 1- q , - 42 and 1- 4, 02 Concave down on1- 4, 42; concave up on 1- q ,- 42 and 14, q 2 Asymptotes: x = - 4, x = 4, and y = 0 5 y-intercept: A 0, - 16 B y 31. 5 4 3 2 1

−4 −3 −2 −1−1

1 2 3 4 5 6 x

−2 −3 −4 −5

f(x) = −

x+1 x2 − x − 2

No relative extrema Increasing on 1- q , - 12, 1- 1, 22, and 12, q 2 Concave up on 1- q , - 12 and 1- 1, 22; concave down on 12, q 2 Asymptotes; x = 2 and y = 0 y-intercept: A 0, 12 B

y 6 5 4 3 2 1 −4 −3−2 −1 −1 −2 −3 −4 −5 −6

IA-27

x 2 – 2x + 2 f(x) = ——–——– x–1

1 2 3 4 x

Relative minimum at 12, 22; relative maximum at 10, - 22 Decreasing on 10, 12 and 11, 22; increasing on 1- q , 02 and 12, q 2 Concave down on 1- q , 12; concave up on 11, q 2 Asymptotes: x = 1 and y = x - 1 y-intercept: 10, - 22 33. y 10 8 6 4 2 −10 −8 −6 −4 −2−2 −4 −6 −8 −10

2 4 6 8 10 x f(x) =

x2 + 3 x

Relative minimum at A 23, 223 B ; relative maximum at A - 23, - 223 B Decreasing on A - 23, 0 B and A 0, 23 B ; increasing on A - q , - 23 B and A 23, q B Concave down on 1- q , 02; concave up on 10, q 2 Asymptotes: x = 0 and y = x No intercepts 34. Absolute maximum: 66 at x = 3; absolute minimum: 2 at 16 x = 1 35. Absolute maximum: 75 23 27 at x = 3 ; absolute minima: 0 at x = 0 and x = 8 36. No absolute maxima; absolute minimum: 1022 at x = 522 37. No absolute maxima; absolute minima: 0 at x = - 1 and x = 1 39. Q = - 1 when x = - 1 and y = - 1 40. Maximum profit is $451 when - 3y - 2x 2 4 30 units are produced and sold. 48. ; 2y 2 + 3x 5 50. $600>day, $450>day, $150>day 52. Absolute maxima: 4 at x = 2 and x = 6; absolute minimum: - 2 at x = - 2 3x 5 - 21x - y23 - 21x + y23 53. 54. Relative maximum 21x + y23 - 21x - y23 - 3y 5 at 10, 02; relative minima at 1- 9, - 94772 and 115, - 37,1252 3x + 3 55. f 1x2 = (answers may vary) 56. Relative maxima at x + 2 1- 1.714, 37.4452; relative minimum at 11.714, - 37.4452 57. Relative maximum at 10, 1.082; relative minima at 1- 3, - 12 and 13, - 12 58. (a) Linear: y = 6.998187602x - 124.6183581 Quadratic: y = 0.0439274846x 2 + 2.881202838x - 53.51475166 Cubic: y = - 0.0033441547x 3 + 0.4795643605x 2 - 11.35931622x + 5.276985809 Quartic: y = - 0.00005539834x 4 + 0.0067192294x 3 - 0.0996735857x 2 - 0.8409991942x - 0.246072967 (b) The quartic function best fits the data. (c) The domain is 326, 1024. Very few women outside of the age range from 26 to 102 years old develop breast cancer. (d) Maximum: 466 per 100,000 women at x = 79.0 years old

IA-28

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

6.

Chapter 2 Test, p. 303

f(x) = 2x4 – 4x 2 + 1 y 5 4 3 2 1

1. Relative minimum: - 9 at x = 2 Decreasing on 1- q , 22; increasing on 12, q 2 f(x) = x 2 – 4x – 5 y 1 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

−2

1

−1 −2

2 x

Relative maximum: 1 at x = 0; relative minima: - 1 at x = - 1 and x = 1 Inflection points: A - 213, - 19 B and A 213, - 19 B y 7. f(x) = (x – 2)3 + 3

2. Relative minimum: 2 at x = - 1; relative maximum: 6 at x = 1 Decreasing on 1- q , - 12 and 11, q 2; increasing on 1- 1, 12 y 8 7 f(x) = 4 + 3x − x3 6 5 4 3 2 1 −3 −2 −1 −1 −2 −3

1

2

3 x

−4−3−2 −1

1

2

−3 −2 −1 −1 −2 −3 −4 −5

4 x

3

1 2 3 x

Relative maximum: 92 at x = at x = - 292 Inflection point: 10, 02 y 9.

2 4 6 8 10 12 x

22; relative minimum: - 2 9

20 2 f(x) = ——– 15 x – 1 10 5 −1

4. Relative maximum: 4 at x = 0 Increasing on 1- q , 02; decreasing on 10, q 2 y 4 3 2 1

–1 –2.5 –5 –7.5 –10

5 4 f(x) = x 9 – x2 3 2 1

y 3 f(x) = (x − 2)2/3 − 4 2 1 −8 −6 −4 −2 −1 −2 −3 −4 −5

10 7.5 5 2.5

No relative extrema Inflection point: 12, 32 y 8.

3. Relative minimum: - 4 at x = 2 Decreasing on 1- q , 22; increasing on 12, q 2

5.

−1

1 2 3 4 5 x

1

2

3 x

No relative extrema Asymptotes: x = 1 and y = 0 10. y

16 f(x) = ——– x2 + 4

5 4 3 2 1

1 2 3 4 x

y 8 7 6 5 4 f(x) = x 3 + x 2 – x + 1 3 2 1

−5 −4 −3 −2 −1−1

1 2 3 4 5 x

−2 −3 −4 −5

f(x) =

1 2 3 x −3 −2−1 −1 −2 −3

Relative maximum: 2 at x = - 1; relative minimum: 22 27 at x = Inflection point: A - 13, 38 B 27

−5 −10 −15 −20

1 3

−8 x2 − 4

Relative minimum: 2 at x = 0 Asymptotes: x = - 2, x = 2, and y = 0

9

IA-29

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

11.

3. (a)

y 7 6 5 4 3 2 1

x2 – 1 f(x) = ——–– x −3

−2

−1

1

−1 −2 −3 −4 −5 −6 −7

2

3

x

–2

250 0

15 10 5 –4

–5 –10 –15

2000 Xscl = 200, Yscl = 200

(b) 50,000; (c) 25,000

(b) 400,000; (c) 400,000

5. (a)

6. (a) y = - 0.0011P 3 + 0.0715P 2 - 0.0338P + 4 (b) 50

f(x) = x – 3 x+2 0 2

50 0

4 x

1. (a)

12

200 0

Xscl = 25, Yscl = 25

(b) 60,000; (c) 90,000

Xscl = 5, Yscl = 5

Exercise Set 3.1, p. 319 y

1.

2.

16 12 y = 4x

8 4 –1

1

2

x

y 10 9 8 7 6 5 4 3 2 1 −5−4−3−2−1

3. y = (0.25)

x

4.

y 10 9 8 7 6 5 4 3 2 1

−5−4−3−2−1

5.

7.

0

70 0

Chapter 3

y 10 9 8 7 6 5 4 3 2 1 −5−4−3−2−1

200

25

Xscl = 1, Yscl = 5

0

(c) 33,841

2. (a)

0

Xscl = 10, Yscl = 50

(b) 30,513; (c) 205,923

Extended Technology Application, p. 306

(b) 4500; (c) 20,250

0 0

Xscl = 50, Yscl = 50

250

No relative extrema Asymptotes: x = - 2 and y = 1 13. Absolute maximum: 9 at x = 3; no absolute minimum 14. Absolute maximum: 2 at x = - 1; absolute minimum: - 1 at x = - 2 15. Absolute maximum: 28.49 at x = 4.3; no absolute minimum 16. Absolute maximum: 7 at x = - 1; absolute minimum: 3 at x = 1 17. There are no absolute 1 extrema. 18. Absolute minimum: - 13 12 at x = 6 19. Absolute minimum: 48 at x = 4 23. Dimensions: 40 in. by 40 in. by 10 in.; maximum volume: 16,000 in3 24. Order 35 times per year; lot size, 35 33. Absolute minimum: 0 at x = 0; relative maximum: 25.103 at x = 1.084; relative minimum: 8.625 at x = 2.95 34. Relative minimum: - 0.186 at x = 0.775; relative maximum: 0.186 at x = - 0.775 35. (a) Linear: y = - 0.7707142857x + 12691.60714 Quadratic: y = - 0.9998904762x 2 + 299.1964286x + 192.9761905 Cubic: y = 0.000084x 3 - 1.037690476x 2 + 303.3964286x + 129.9761905 Quartic: y = - 0.000001966061x 4 + 0.0012636364x 3 - 1.256063636x 2 + 315.8247403x + 66.78138528 (b) Since the number of bowling balls sold cannot be negative, the domain is 30, 3004. This is supported by both the quadratic model and the raw data. The cubic and quartic models can also be used but are more complicated. (c) Based on the quadratic function, the maximum value is 22,575 bowling balls. The company should spend $150,000 on advertising.

0

2000

0

No relative extrema Asymptotes: x = 0 and y = x y 12.

–6

4. (a)

150

y 10 9 8 7 6 5 4 3 2 1 −5−4−3−2−1

6. f(x) = 32 x

()

8.

g(x) =

()

1 2 3 4 5 x

1 2 3 4 5 x

y 8 7 6 5 4 3 2 1 −5−4−3−2−1−1 −2

1 2 3 4 5 x

2 x 3

1 2 3 4 5 x

y 10 9 y = (0.2)x 8 7 6 5 4 3 2 1 −5−4−3−2−1

1 2 3 4 5 x

y = 5x

y 8 7 6 5 4 3 2 1 −5−4−3−2−1−1 −2

f(x) = 43 x

()

1 2 3 4 5 x

g(x) = 34 x

()

1 2 3 4 5 x

IA-30 9.

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

10.

y 10 9 8 7 6 5 4 3 2 1

x

f(x) = (2.5)

−5−4−3−2−1

−5 −4 −3 −2 −1−1

2

y 10 9 8 7 6 5 4 3 2 1 −5−4−3−2−1

2

56. f(x) = e2x

2 4 6 8 10 x

- 7x

g(x) = e−2x

,

y 10 9 8 7 6 5 4 3 2 1

−5−4−3−2−1

1 2 3 4 5 x

57.

58.

(1/2)x

−5−4−3−2−1

y 10 9 8 7 6 5 4 3 2 1 −3−2−1

1 2 3 4 5 x

f(x) = e(1/3)x

1 2 3 4 5 6 7 x

No critical values No inflection points Increasing on 1- q , q 2 Concave up on 1- q , q 2

59.

60.

−5−4−3−2−1

f(x) = 12 e−x

y 10 9 8 7 6 5 4 3 2 1 −5−4−3−2−1

1 2 3 4 5 x

g(x) = 13e−x

No critical values No inflection points Decreasing on 1- q , q 2 Concave up on 1- q , q 2

61.

62.

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9

1 2 3 4 5 6 x

F(x) = −e(1/3)x

No critical values No inflection points Decreasing on 1- q , q 2 Concave down on 1- q , q 2

q = 240e−0.003x

q = 75e 0.004x

100 200 300 400 160 240 320

x

x

88. (c) C(t) = 10t 2 e –t

P(t) 100 90 80 70 60 50 40 30

P(t) = 40 + 60e −0.7t

1 2 3 4 5 6 7 8 9 10 t

f(x) = e −x

2

1.1

1

−3

7

0

3 −0.1

Relative minimum at 10, 02; Relative maximum at 10, 12 relative maximum at 12, 0.54132 109. f (x) = f '(x) = f "(x) = e x 4

1 2 3 4 5 x

No critical values No inflection points Decreasing on 1- q , q 2 Concave up on 1- q , q 2 y 1

q 500 450 400 350 300 250 200 150 100 50

(c) C¿1t2 = 10te -t12 - t2; - 1x 3 + x 2 + x - 12e -x 98. 11 + x 222 107. 108. f(x) = x 2 e − x

−2

1 2 3 4 5 x

No critical values No inflection points Increasing on 30, q 2 Concave down on 30, q 2

1 2 3 4 5 6 7 8 9 10 t

No critical values No inflection points Increasing on 1- q , q 2 Concave up on 1- q , q 2 y 10 9 8 7 6 5 4 3 2 1

C 6 5 4 3 2 1

f(x) = 3 − e−x x≥0

−2 −3 −4

86. (b)

80

87. (b)

y 4 3 2 1 −5 −4 −3 −2 −1−1

q 240 220 200 180 160 140 120 100 80 60 40 20

1 2 3 4 5 x

No critical values No inflection points Decreasing on 1- q , q 2 Concave up on 1- q , q 2

g(x) = e

1 2 3 4 5 x

No critical values on 30, q 2 No inflection points Increasing on 30, q 2 Concave down on 30, q 2 85. (b)

No critical values No inflection points Increasing on 1- q , q 2 Concave up on 1- q , q 2 y 10 9 8 7 6 5 4 3 2 1

64.

g(x) = 2(1 − e−x) x≥0

−2 −3

−2

1 2 3 4 5 x

y 3 2 1

f(x) = (1.2)x

−10−8 −6−4−2−1

53. 14x 2 + 3x2e x - 7x12x - 72 + 18x + 32e x 2 or 18x 3 - 22x 2 - 13x + 32e x - 7x 55.

63.

y 8 7 6 5 4 3 2 1

y 1 −5 −4 −3 −2 −1−1

(1/2)x

G(x) = −e

−3 −1

110.

1 2 3 4 5 x

−2 −3 −4 −5 −6 −7 −8 −9

No critical values No inflection points Decreasing on 1- q , q 2 Concave down on 1- q , q 2

3

f(x) = f "(x) = e − x

f '(x) = − e −x

4

1 −3

−3

3

3 −1

−4

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

111.

f(x) = 2e 0.3x

f '(x) = 0.6e 0.3 x

10

10

−10

10

−10

10 −2

−2 f "(x) = 0.18e 0.3x

−10

10 −2

112.

price increases to 60¢ in 2016, the cost for the years 2016–2018 would be 3 * $0.60 * 10,000, or $18,000. If the price increases to 69¢ in 2019, the cost for the years 2019–2020 would be 2 * $0.69 * 10,000, or $13,800. Thus, the total cost of regular first-class stamps for the years 2010–2020 would be $13,200 + $15,300 + $18,000 + $13,800 for a total of $60,300. Thus, by buying Forever Stamps, the firm would save $60,300 - $48,400, or $11,900. 31. (d) 43. (c) P(x) 100 90 80 70 60 50 40 30 20 10

10

f(x) = 1000e − 0.08x −10

P(t) 6000 5500 5000 4500 4000 3500 3000 2500 2000 1500 1000 500

100 P(x) = ——–———– 1 + 49e –0.13x

10 20 30 40 50 60 x

f "(x) = −80e −0.08x

2000

0

f "(x) =

−200 6.4e −0.08x

113.

20

(

1 f (x) = 1 + −− x

)

p(t) 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

2500 P(t) = ——–———–– 1 + 5.25e –0.32t

5 10 15 20 25 30 t

x

5780 1 + 4.78e−0.4t

46. (c)

P(t) 3000 2500 2000 1500 1000 500

40 0

P(t) =

2 4 6 8 10 12 14 16 18 20 t

40

44. (c)

−10

IA-31

3

p(t) = 1 − e−0.28t

2 4 6 8 10 12 14 16 18 20 t

−10

40 0

0

0

40

47. (b) at 7 months, the (c) percentage of doctors who are prescribing the medication is growing by 2.4% per month;

Exercise Set 3.2, p. 334

ln 12x2 + ln 17x2 2e x e 2x + 2e x ln x 66. + 2e 2x ln x 74. x x x 500 79. (b) N¿1a2 = , N¿1102 = 50 units per $1000 spent on a 200 advertising 80. (b) N¿1a2 = , N¿1102 = 20 units per a - 20 $1000 spent on advertising 87. (e) S¿1t2 = t + 1 - 15 0.37 88. (e) S¿1t2 = 89. (c) v¿1p2 = p t + 1 43ln 1x + 5243 1 94. 95. x + 5 x ln 13x2 # ln 1ln 13x22 -1 1 -2 97. , or 1 - t 1 + t 11 - t211 + t2 1 1 105. , or 2x A 1 - 2x B A 1 + 2x B 2x A 1 - x B 65.

Exercise Set 3.3, p. 347 28. k = 0.1802, or 18.02%; R1t2 = 1.265e 0.1802t, where t0 = 1986; $137.04 billion 29. (d) For the years 2010–2020, the total cost of Forever Stamps is 11 * $4400, or $48,400. For the years 2010–2012, the cost of regular first-class stamps is 3 * $4400, or $13,200. If the price of a regular postage stamp increases to 51¢ in 2013, the cost of postage for the years 2013–2015 would be 3 * $0.51 * 10,000, or $15,300. If the

P(t) 100 90 80 70 60 50 40 30 20 10

P(t) = 100(1 – e –0.4t)

2 4 6 8 10 12 14 16 t

48. (c)

N(t) 800 700 600 500 400 300 200 100

20

49. (c)

An unrestricted growth model is inappropriate since the population of the university is bounded.

568.803 N(t) = 1 + 62.200e −0.092 t

40

60

80 100 t

29.47232081 N(t) = 1 + 79.56767122e−0.809743969t 30

0

15 0

1898.885181e -0.809743969t 11 + 79.56767122e -0.809743969t22 69. (a) R102 = $2 million; this represents the initial revenue of the corporation at its inception. (b) lim R1t2 = $4000 million = Rmax; this represents the upper (d) N¿1t2 =

t: q

limit of the revenue of the company over all time. It is never actually attained.

IA-32

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

Exercise Set 3.4, p. 360

24. (a) V1t2 = 34,001.78697 # 0.6702977719x;

V1t2 = 34,001.78697e -0.4000332297t; (b) $2067.20, $622.56; 25. (b) Q(t) 0.10 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01

Q(t) = (Q0 − 0.00055)e 0.163 t + 0.00055 1 2 3 4 5 6 7 8 9 10 t

26. (b)

Q(t) 0.10 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01

Q(t) = (Q0 − 0.00055)e 0.163t + 0.00055

1 2 3 4 5 6 7 8 9 10 t

5. (a) E1x2 = 1; (b) 1, unit elasticity; (c) total revenue is independent of x. 6. (a) 1; (b) 1, unit elasticity; (c) total revenue is independent of x. x 7. (a) E1x2 = ; (b) 0.10, inelastic; (c) $400 21600 - x2 x 8. (a) E1x2 = ; (b) 2.5, elastic; (c) $200 21300 - x2 9. (a) E1x2 = 0.25x; (b) 2.5, elastic; (c) $4 10. (a) E1x2 = 0.05x; (b) 4, elastic; (c) $20 2x 11. (a) E1x2 = ; (b) 0.5, inelastic; (c) $3 x + 3 8 2x 12. (a) E1x2 = ; (b) , elastic; (c) $6 x + 6 7 13. (b) approximately 19¢; (c) prices greater than 19¢; (d) prices less than 19¢ x12 - 2x2 x150 - 50x2 14. (a) E1x2 = , or 2 63,000 + 50x - 25x 2520 + 2x - x 2 9 29 (b) , inelastic; (d) , elastic 122 28

Chapter Review Exercises, p. 382

27. (b) 1,953,564 farms, 1,753,573 farms 28. (c) 230 months

30.

Exercise Set 3.5, p. 368 5 9 6 18. 19. 15x + 12 ln 6 19x - 22 ln 32 16x - 72 ln 10 5 3x 2 + 1 4x 3 - 1 20. 21. 3 22. 4 15x + 42 ln 10 1x + x2 ln 8 1x - x2 ln 9 2 1 23. 24. 31ln 621x + 5x 2>32 A x - 22x B ln 7 x 6 5x 25. # 26. # + 6x # ln 6 # log 7 x + 5x # ln 5 # log 2 x x ln 7 x ln 2 1 1 27. 51log 12 x24 a 28. 71log 9 x26 a b b x ln 12 x ln 9 14x + 127x # ln 7 - 4 # 7x 29. 14x + 122 3 6 # 52x - 1 3 31. + 1ln 5252x - 1 # 6x 2 # log 16x + 52 16x + 521ln 102

12

17.

2x + 8 # ln 4 # 8x 3 # log 17x + 32 + 7 # 4 1ln 10217x + 32 7x # 9 # 1log 4 x28 33. 1ln 727x # 1log 4 x29 + x # ln 4 13x 5 + x25 35. 513x 5 + x24115x 4 + 12 # 1log 3 x2 + # 4

4

32. 4 2x

+8

ln 3 x 44. (d) dI>dR = I010R1ln 102 45. (c) The power mower is 109 times louder than a just audible sound. ex x 54. ae x1ln x2 + bx e x

Exercise Set 3.6, p. 376 x 5 ; (b) , inelastic; (c) $200 400 - x 11 x 19 2. (a) E1x2 = ; (b) , inelastic; (c) $250 500 - x 231 x 3. (a) E1x2 = ; (b) 11.5, elastic; (c) $25 50 - x 57 x 4. (a) E1x2 = ; (b) , inelastic; (c) $125 250 - x 193 1. (a) E1x2 =

31.

y 16 f(x) = 4 x

g(x) = 13 x

()

8 4 −1

1

2

x

y 10 9 8 7 6 5 4 3 2 1

−5−4−3−2−1

46. (d)

51.

p 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

1 2 3 4 5 x f (x) =

p(t) = 1 – e –0.7t

e1/x

(1 + e1/x )2 0.5

−4

4 −0.1

2 4 6 8 10 12 14 16 t

Chapter 3 Test, p. 384 3e x + 3e x # ln x x (e) P(t) 6.

100 90 80 70 60 50 40 30 20 10

19. (c) P¿1t2 =

672e -0.28t ; 11 + 24e -0.28t22

24.

f (x) =

e x − e −x e x + e −x 2

100 P(t) = ——–———– 1 + 24e –0.28t −4

10

20

30

40

t

4

−2

26. (a) y = 686996.582 # 1.086747476x, y = 686996.582 # e 0.0831892684x

Extended Technology Application, p. 387 1. Linear: R1t2 = - 5.2444t + 52.3333 Quadratic: R1t2 = 0.756x 2 - 13.5603x + 68.965 Cubic: R1t2 = - 0.2107x 3 + 4.2327x 2 - 29.5954x + 87.044 Exponential: R1t2 = 68.455210.78682x

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

The linear and cubic functions both reach R = 0 too fast. The value from the quadratic function decreases as we might expect over time, but then it makes a dramatic rise. Most movies do not have this revenue pattern. The exponential function shows a steady decrease and approaches 0 as a limit, but never reaches it. It is a reasonable assumption that G gets smaller and smaller over time. Eventually, box office revenue reaches 0. Gross revenue thereafter comes from DVD rentals, TV rights, and electronic outlets, such as iTunes. 2. G = $4.90, $3.85, $3.03, $2.39, $1.88, $1.48, $1.16, $0.91, all in millions 3. R = $239.79, $243.64, $246.67, $249.06, $250.94, $252.43, $253.58, $254.49, all in millions. There are costs, such as marketing and shipping costs, associated with distributing a movie to theaters. Eventually, movie executives want the jump in revenue that comes with DVD and electronic rentals. 251.1 4. R1t2 = 1 + 3.4687e -0.4183t 364.3354 5. R¿1t2 = , which represents the rate of 11 + 3.4687e -0.4183t22 change of the total revenue; lim R¿1t2 = 0, which means that

Technology Connection, p. 432 1. 16

Chapter 4 Exercise Set 4.1, p. 396 75 L 2.344 sec 72. (a) Alphaville: P1t2 = 45t + 5000, 32 Betaburgh: Q1t2 = 3500e 0.03t; (b) Alphaville: P1102 = 5450, Betaburgh: Q1102 = 4725 ' y Q Populations are the same 1 57002 (c) in 2006 1t L 16.52. P Population

71. (c) t =

5000

3000 1000 0

81.

10 Years

20

x

1 # ln x + C , or log x + C ln 10

Exercise Set 4.2, p. 407 4

40. a kf 1x i2 = kf 1x 12 + kf 1x 22 + kf 1x 32 + kf 1x 42 x=1

= k3 f 1x 12 + f 1x 22 + f 1x 32 + f 1x 424 4

= k a f 1x i2;

n

x=1

Á + kf 1x n2 a kf 1x i2 = kf 1x 12 + kf 1x 22 + kf 1x 32 + = k3 f 1x 12 + f 1x 22 + f 1x 32 + Á + f 1x n24

i=1

n

= k a f 1x i2 i=1

Exercise Set 4.3, p. 421 21. Total amount of the drug, in milligrams, in v cubic centimeters e2 - 3 of blood 56. , or approximately 2.195 2

Over the interval 30, 24, the areas under f 1x2 = x 4 and yav = 3.2 are equal.

f(x) = x 4 and yav = 3.2

0

2

−1

Exercise Set 4.4, p. 433 56. (b) 73 is the score Antonio receives, on average, when he studies for 7–10 hr; (c) 80 is the score Bonnie receives, on average, when she studies for 6–10 hr 80 76. (a) g −3

t: q

eventually the total revenue does not change. From the logistic function, it would be about $251.1 million, but the table shows about $254 million. 6. From the logistic function, it would be about $745 million, but the table shows about $762 million.

IA-33

f 3

−80

Exercise Set 4.5, p. 443 1 + C 411 + x 324 92. x + 1 + 2 ln 1x + 12 + C, or x + 2 ln 1x + 12 + K, where K = 1 + C 93. t - 4 - ln 1t - 42 + C, or 1ln x2-n + 1 t - ln 1t - 42 + K, where K = - 4 + C 94. + C -n + 1 40. -

Exercise Set 4.6, p. 452

8. 1- x - 12e -x + C, or - 1x + 12e -x + C, or - xe -x - e -x + C x 4 ln x x4 + C 4 16 x3 + C 11. 14 x 2 ln x - 18 x 2 + C 12. x 3 ln x 3 13. 1x + 52 ln 1x + 52 - x + C 14. 1x + 42 ln 1x + 42 - x + C x2 x2 15. a + 2xb ln x - 2x + C 2 4 x2 x2 16. a + xb ln x - x + C 2 4 x2 x2 17. a - xb ln x + x + C 2 4 x2 x2 18. a - 2xb ln x + 2x + C 2 4 2 4 3>2 19. 3 x1x + 22 - 15 1x + 225>2 + C 4 20. 23 x1x + 523>2 - 15 1x + 525>2 + C x4 x 4 ln 2 x 4 ln x x4 x4 21. ln 12x2 + C , or + + C 4 16 4 4 16 3 3 x x 22. + C 23. x 2e x - 2xe x + 2e x + C ln 15x2 3 9 24. x1ln x22 - 2x ln x + 2x + C x -4 x -4 25. 12 x 2e 2x - 12xe 2x + 14 e 2x + C 26. 1ln x2 + C 4 16 1 3 -2x 3 2 -2x 3 3 -2x -2x 27. - 2 x e - 4x e - 4 xe - 8e + C 5 4 4x 5 3 4x 15 2 4x 4x 28. 14 x 5e 4x - 16 x e + 16 x e - 15 x e + 64 128 xe 15 4x 1 4 4 8 4 3x 3 3x 2 3x + C 29. 3 1x + 42e - 9 x e + 9 x e - 27 xe 3x + 512 e 8 3x 3 -x 2 -x -x -x + C 30. - x e - 3x e - 5xe - 6e + C 81 e 9.

x 3 ln x x3 + C 3 9

10.

IA-34

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

31. 83 ln 2 - 79 32. 4 ln 2 - 15 16 33. 14 ln 14 - 10 ln 10 - 4 34. 12 ln 12 - 7 ln 7 - 5 2 2 43. 125 15x + 125>2 - 75 15x + 123>2 + C; they are the same. 1>2 44. (a) 2x1x - 32 - 43 1x - 323>2 + C; (b) 231x - 323>2 + 61x - 321>2 + C 5 9>5 50. 16 113t 2 - 48214t + 724>5 - 325 + 288 t14t + 72 1625 2 14>5 3 51. 14t + 72 + C 127x + 83x 2213x + 827>6 16,128 7 4 1296 2 13>6 + 1729 x13x + 8219>6 91 181x + 83213x + 82 2x 3 2x 3 x3 2592 25>6 52. + C 1ln x22 1ln x2 + + C 43,225 13x + 82 3 9 27 2x n + 1 2x n + 1 xn+1 53. ln x + 1ln x22 + C 2 n + 1 1n + 12 1n + 123 xn+1 xn+1 54. ln x + C 55. Let u = x n and n + 1 1n + 122 dv = e x dx. Then du = nx n - 1 dx and v = e x . Next, use integration by parts. 56. Let u = 1ln x2n and dv = dx . Then n du = 1ln x2n - 1 dx and v = x. Next, use integration by parts. x

Exercise Set 4.7, p. 457 4. ln ƒ x + 2x 2 + 9 ƒ + C 7. 3 - x - 3 ln ƒ 3 - x ƒ + C 9.

6. - 12 ln `

2 + 24 + x 2 ` + C x

1 8. + ln ƒ 1 - x ƒ + C 1 - x

1 1 x + ln ` ` + C 818 - x2 64 8 - x

C x2x 2 + 9 + 9 ln ƒ x + 2x 2 + 9 ƒ D + C 11. 1ln 32 x + x ln x - x + C 12. A ln 45 B x + x ln x - x + C 10.

1 2

x5 x5 1ln x2 + C 14. - 12 x 3e -2x - 34 x 2e -2x - 34 xe -2x 5 25 x4 x4 x5 3 -2x + C 15. 1ln x2 + C 16. x 51ln x2 + C 8e 4 16 5 13.

17. ln ƒ x + 2x 2 + 7 ƒ + C

18. - 3 ln `

1 + 21 - x 2 ` + C x

2 x 2 + ln ` ` + C 5 - 7x 5 5 - 7x x - 1>2 5 1 x 20. ln ` ` + C 21. - ln ` ` + C 5 7x + 2 4 x + 1>2 19.

22.

3 2

C t2t 2 -

1 9

-

1 9

ln ƒ t + 2t 2 +

1 9

ƒD + C

23. m 2m + 4 + 4 ln ƒ m + 2m + 4 ƒ + C 3 5 5 + C 24. 1- ln x - 12 + C 25. 2 1ln x2 + x 2x 4x 2 4 3 2 26. x1ln x2 - 4x1ln x2 + 12x1ln x2 - 24x1ln x2 + 24 + C 27. x 3e x - 3x 2e x + 6xe x - 6e x + C 1 28. 32 ln ƒ x + 2x 2 + 25 ƒ + C 29. 15 13x - 1211 + 2x23>2 + C x 2 30. 135 19x - 4212 + 3x23>2 + C 33. - 4 ln ` ` + C 3x - 2 1 -3 34. + ln ƒ 2x - 3 ƒ + C 412x - 32 4 -1 1 x 35. + ln ` ` + C 21x - 22 4 x - 2 2

2

36. 12 C e x 2e 2x + 1 + ln ƒ e x + 2e 2x + 1 ƒ D + C -3 37. -x + ln ƒ e -x - 3 ƒ + C e - 3 38. 14 C ln x21ln x22 + 49 + 49 ln 1ln x + 21ln x22 + 492 D + C

Chapter Review Exercises, p. 466 x3 8 8 x + 81 + C 35. e 3x A 13 x 4 - 49 x 3 + 49 x 2 - 27 B +C 3 7 + x 1 2 2 5x 36. ln ` ` + C 37. 15 x 2e 5x - 25 xe 5x + 125 e + C 14 7 - x x 1 1 38. + ln ƒ 7x + 1 ƒ + C 49 7 49 ln x 1 39. ln ƒ x + 2x 2 - 36 ƒ + C 40. x 7 a b + C 7 49 1 8x 41. 64 e 18x - 12 + C 46. 10x 3e 0.1x - 300x 2e 0.1x + 6000xe 0.1x - 60,000e 0.1x + C 2 1 48. 75 15x - 8224 + 5x + C 54. 10 C ln ƒ t 5 + 3 ƒ D 2 + C 1 -x 55. - 2 ln 11 + 2e 2 + C 56. 1ln 1x22 + C, or 14 1ln x22 + C ln x 1 57. x 92 a b + C 58. 1x - 32 ln 1x - 32 92 8464 1 59. 1x - 42 ln 1x - 42 + C + C 31ln x23 60. 37 1x + 327>3 - 94 1x + 324>3 + C 1 61. 16 12x + 122 - 14 12x + 12 + 18 ln 12x + 12 + C 34. x 3 ln x -

Chapter 4 Test, p. 468 x4 x4 x4 ln x 4 + C , or x 4 ln ƒ x ƒ + C 4 4 4 1 x 19. ln ` ` + C 26. x 5e x - 5x 4e x + 7 7 - x 20x 3e x - 60x 2e x + 120xe x - 120e x + C 6 27. 16 e x + C 28. 23 x 3>21ln x2 - 49 x 3>2 + C 8 + x 1 29. ln a b + C 16 8 - x 30. - 10x 4e -0.1x - 400x 3e -0.1x - 12,000x 2e -0.1x 240,000xe -0.1x - 2,400,000e -0.1x + C 34. 1x + 32 ln 1x + 32 - 1x + 52 ln 1x + 52 + C 108 2 3 35. 18x 3 + 10215x - 422>3 x 15x - 425>3 + 10 125 243 81 x15x - 428>3 15x - 4211>3 + C 625 34,375 2 36. 27 13x - 223>2 + 49 13x - 221>2 + C 16 37. x + 8 ln x + C x 17.

Chapter 5 Technology Connection, p. 474

The point of intersection is 12, 92; this is the equilibrium point.

30

1.

D

0

S

5 0

Exercise Set 5.1, p. 479 19. (a) 16, $22; (b) (c) $7.62; (d) $7.20

8

D

Consumer surplus Producer surplus 0

S

12 0

IA-35

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

20. (a) 127, $62; (b) (c) $121.50; (d) $40.50

16 Consumer surplus Producer surplus

D S

0

50 0

21. (a) Linear

1 2 - x 1 1 ; f 1x2 = 19. ; f 1x2 = 21. 2 2 ln 3 x ln 3 1 1 1 22. 35. (a) k L ; f 1x2 = L 0.000017 ln 2 x ln 2 59,119.3 1 36. (a) k L L 0.000013 42. (b) n = 23 76,770.2

Exercise Set 5.5, p. 520

20

1. m = E1x2 = 5, E1x 22 =

0

8 0

37. (c) 4%: $688,339, 6%: $582,338, 8%: $498,815, 10%: $432,332

Exercise Set 5.3, p. 496 y = xe − 0.1x

5

0

100 0

Exercise Set 5.4, p. 506 3

1 x2 3 9 1 x dx = c d = = 1 8 1 8 8 L1 4 1 x2 1 2. 2x dx = c 2 # d = 3x 2410 = 12 - 02 = 1 2 0 L0 1>3 1 3. 3 dx = 33x41>3 = 3a - 0b = 1 0 3 L0 8 3 5 1 x 8 8 4. = = 1 dx = c d = 5 3 5 5 5 L3 5 4 3 2 x3 4 1 5. x dx = c d = 14 3 - 032 = 1 64 64 64 0 L0 3 3 2 3 x3 3 x3 3 33 13 26 6. x dx = c # d = c d = = = 1 26 3 1 26 1 26 26 26 L1 26 e 1 7. dx = 3ln x4e1 = ln e - ln 1 = 1 - 0 = 1 L1 x 1 1 1 1 1 1 8. e x dx = c ex d = e1 e0 e 1 e 1 e 1 e 1 0 L0 1 e - 1 e = = 1 = e - 1 e - 1 e - 1 1 1- 123 3 2 x3 1 13 1 1 9. x dx = c d = = + = 1 2 -1 2 2 2 2 L-1 2 3 1 1- 22 1 2 1 x3 1 13 1 -8 10. x dx = c # d = = = 1 3 3 -2 9 9 9 9 L-2 3 1.

q

11.

L0

3e -3x dx = lim

b: q L 0

b

3e -3x dx = lim 3- e -3x4b0 b: q

= lim 3- e -3b - 1- e 024 = lim a -

1 + 1b = 1 e 3b b 4 -4x b 12. 4e dx = lim 4e -4x dx = lim c e d q q b: L b: -4 0 0 L0 1 -4b -4 # 0 = lim 3- e - 1- e 24 = lim a - 4b + 1b = 1 b: q b: q e b: q q -4x

8 8 222 , E1x 22 = 8, s2 = , s = 3 9 3 9 2 1 1 2 4. m = E1x2 = 2, E1x 2 = , s = , s = 2 2 A2 13 11 211 2 2 5. m = E1x2 = , E1x 2 = 5, s = ,s = 6 36 6 14 5 13 1 13 6. m = E1x2 = , E1x 22 = , s2 = ,s = 9 2 162 9A 2 3 3 3 7. m = E1x2 = 0, E1x 22 = , s2 = , s = 5 5 A5 5 11 2 51 1 51 8. m = E1x2 = - , E1x 22 = ,s = ,s = 4 5 80 4A 5 6 27 27 ln 5 - 36 9. m = E1x2 = , E1x 22 = , s2 = , ln 5 ln 5 1ln 522 227 ln 5 - 36 4.8 2.4 10. m = E1x2 = s = , E1x 22 = , ln 5 ln 4 ln 4 4.8 ln 4 - 5.76 24.8 ln 4 - 5.76 ,s = s2 = ln 4 1ln 422 2 b + a b + ab + a 2 61. m = E1x2 = , E1x 22 = , 2 3 2 1b - a2 b - a ,s = s2 = 12 223 3. m = E1x2 =

Exercise Set 5.2, p. 489

53.

79 2 4 2 ,s = ,s = 3 3 23 97 25 5 11 2. m = E1x2 = , E1x 22 = , s2 = ,s = 2 3 12 223

b: q

62. m = E1x2 =

3a 3a 2 a 23 , E1x 22 = 3a 2, s2 = ,s = 2 4 2

Exercise Set 5.6, p. 525 y

21. y = 4 − x2

22.

y

(h, r)

y = r2 − x2, r > 0 r 2

−r −2

x 2

r

x

The graphs are semicircles. Their rotation about the x-axis creates spheres of radius 2 and r, respectively.

Let y =

r x. h

Exercise Set 5.7, p. 533 1. y = x 5 + C; y = x 5, y = x 5 - 1, y = x 5 + p (answers may vary) 2. y = x 6 + C; y = x 6, y = x 6 - 1, 6 3. y = 12 e 2x + 12 x 2 + C; y = x + p (answers may vary) 1 2x 1 2 1 2x 1 2 y = 2 e + 2 x , y = 2 e + 2 x - 3, y = 12 e 2x + 12 x 2 + 3 (answers may vary) 4. y = 14 e 4x - 12 x 2 + 2x + C; y = 14 e 4x - 12 x 2 + 2x, y = 14 e 4x - 12 x 2 + 2x - 4, y = 14 e 4x - 12 x 2 + 2x + 57.2 (answers may vary)

IA-36

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

5. y = 8 ln x - 13 x 3 + 16 x 6 + C; y = 8 ln x - 13 x 3 + 16 x 6, y = 8 ln x - 13 x 3 + 16 x 6 + 5, y = 8 ln x - 13 x 3 + 16 x 6 - 17 (answers may vary) 6. y = 3 ln x + 13 x 3 - 15 x 5 + C; y = 3 ln x + 13 x 3 - 15 x 5, y = 3 ln x + 13 x 3 - 15 x 5 - 7, y = 3 ln x + 13 x 3 - 15 x 5 + 53 (answers may vary) 7. y = 13 x 3 + x 2 - 3x + 4 1 2 3 8. y = x - 2 x + 5x + 6 9. f 1x2 = 35 x 5>3 - 12 x 2 - 61 10 10. f 1x2 = 57 x 7>5 + 12 x 2 - 115 14 12. y¿ = ln x - 5; y– = 1>x 11. y¿ = ln x + 3; y– = 1>x 1 1 0 y– 0 y– x x 1 1 1 1 0 0 x x x x 0 0 0 0 13. y¿ = 4e x + 3xe x; y– = 7e x + 3xe x y– 2y¿ + y 0 0 17e x + 3xe x2 - 214e x + 3xe x2 + 1e x + 3xe x2 17e x - 8e x + e x2 + 13xe x - 6xe x + 3xe x2 0 0 14. y¿ = - e x + xe x; y– = xe x y– 2y¿ + y 0 0 1xe x2 - 21 - e x + xe x2 + 1- 2e x + xe x2 1xe x - 2xe x + xe x2 + 12e x - 2e x2 0 0 19. y = 22x 2 + C1, y = - 22x 2 + C1, where C1 = 2C 20. y = 212 x 2 + C and y = - 212 x 2 + C 21. y = 212x + C1 and y = - 212x + C1, where C1 = 2C 3 22. y = 221x + C1, where C1 = 3C 120 33. C1x2 = 2.6x - 0.01x 2 + 120, A1x2 = 2.6 - 0.01x + x 20 51. y1 = 210x + C1 and y2 = - 210x + C1, C1 = 100 where C1 = 2C −20

C1 = 5

40 C1 = −200

−20

Extended Technology Application, p. 545 1. y = 0.00525488582427x 3 - 0.31949926791313x 2 + 5.2617546608767x - 8.994864719578

Chapter 6 Exercise Set 6.1, p. 554 33.

34.

z

z z = x2 + y2

z = y2

y x

y x

35.

36.

z

z

z = (x4  16x2)ey

2

y

y

x x z = 4(x2 + y2 ) − (x2 + y2 )2

37.

38.

z

z

z = x3 − 3xy2 z=

1 x2 + 4y2

x y

y x

Exercise Set 6.2, p. 562 y y x 1 1 16. ; 1 + ln 1xy2 17. + ;- 2 x y 3x 3x 2 y y 1 x 1 18. ;- 2 + y 5x 5x 2 y 35. 4y 2e 2xy; 4xye 2xy + 2e 2xy; 4xye 2xy + 2e 2xy; 4x 2e 2xy 36. y 2e xy; xye xy + e xy; xye xy + e xy; x 2e xy y 3>5 0p 0p x 2>5 41. (b) = 960a b , = 1440a b ; x y 0x 0y 0p 0p (c) ` = 7680, ` = 360 0x 132, 10242 0y 132, 10242 y 0.379 0p 0p x 0.621 42. (b) = 1117.8a b = 682.2a b , ; x y 0x 0y 0p 0p (c) ` L 966, ` L 867 0x 12500, 17002 0y 12500, 17002 0P 43. (b) = - 0.005075w -1.638r 1.038s 0.873t 2.468, 0w 0P = 0.008257w -0.638r 0.038s 0.873t 2.468, 0r 0P = 0.006945w -0.638r 1.038s -0.127t 2.468, 0s 0P = 0.019633w -0.638r 1.038s 0.873t 1.468 0t - 4xt 2 4x 2t 57. fx = 2 ;f = 2 2 2 t 1x - t 2 1x - t 222 4 2 - x + 3x t + 2xt - x3 - x2 58. fx = ; f = t 1x 3 + t22 1x 3 + t22 59. fx =

1 2x11 + 22t2

; ft =

- 1 - 22x 2t11 + 22t22

A D D ITI O N A L I N STR U C TO R ’ S A N SW E R S

3 4 t5 5 4 ; f = 2x 3t 4 Bx t 4 61. fx = 4x -1>3 - 2x -3>4t 1>2 + 6x -3>2t 3>2; ft = - 4x 1>4t -1>2 - 18x -1>2t 1>2 - 20xt 21x 2 + t 224 20x 2t1x 2 + t 224 62. fx = ; f = t 1x 2 - t 226 1x 2 - t 226 - 6y -2 2 -2 2 6x ; fxy = 3 + 3 ; fyx = 3 + 3 ; fyy = 4 63. fxx = x4 y x y x y 2y 2 - 2xy - 2xy 2x 2 64. fxx = ;f = ;f = ;f = 3 xy 3 yx 3 yy 1x - y2 1x - y2 1x - y2 1x - y23 60. fx =

Exercise Set 6.3, p. 570 20. (a) R = 78p1 - 6p 12 - 6p1p2 + 66p2 - 6p 22; (b) p1 = 5, or $50, p2 = 3, or $30; (c) q1 = 39, or 3900 units, q2 = 33, or 3300 units

Exercise Set 6.5, p. 586 3 27 L 1.6 ft, h = 2r L 3.2 ft; about 48.3 ft2 A 2p 3 99 20. r = L 2.5 in., h = 2r L 5.0 in.; about 117.8 in2 A 2p 21. (a) R = 64p1 - 4p 12 - 4p1p2 + 56p2 - 4p 22; (b) p1 = 6, or $60, p2 = 4, or $40; (c) q1 = 32, or 3200 units, q2 = 28, or 2800 units; (d) $304,000 31. (a) P1x, y2 = 45x + 50y, 180 acres of celery and 120 acres of lettuce, $14,100 profit; (b) 270 acres of lettuce and zero acres of celery, $14,100 profit 32. Minimum = - 92 at A 292, - 292 B and A - 292, 292 B

19. r =

34. Maximum = 23 at 35. Maximum =

8 27

1 , 1 , 1 B A 23 23 23

at A ; 223, ; 223, ; 223 B

36. Maximum = 6 at

A 23, 43, - 43 B

Chapter Review Exercises, p. 598 37.

Cumulative Review, p. 602

29. Relative maximum at 1- 1, 32, relative minimum at 11, - 12; point of inflection at 10, 12; increasing on 1 - q , - 12 and 11, q 2, decreasing on 1- 1, 12; concave down on 1- q , 02, y concave up on 10, q 2 5 4 f(x) = x3 – 3x + 1 3 2 1

-5 -4 -3 -2 -1-1

1 2 3 4 5 x

-2 -3 -4 -5

30. Relative maxima at 1 - 1, - 22 and 11, - 22, relative 1 22 minimum at 10, - 32; points of inflection at a ,b 9 23 22 1 and a ,b ; increasing on 1- q , - 12 and 10, 12, 9 23 decreasing on 1- 1, 02 and 11, q 2; concave down on 1 1 a - q, b and a , q b , concave up on 23 23 y 1 1 2 a, b f(x) = 2x 2 – x4 – 3 1 23 23 -2 -1 -1 -2 -3 -4 -5 -6

1

2

x

31. Relative maximum at 11, 42; relative minimum at 1- 1, - 42; points of inflection at A - 23, - 223 B and 10, 02 and A 23, 223 B ; decreasing on 1- q , - 12 and 11, q 2, increasing on 1- 1, 12; concave down on A - q , - 23 B and A 0, 23 B , concave up on A - 23, 0 B and A 23, q B ; horizontal asymptote y at y = 0 4 8x f(x) = ——– 3 x2 + 1 2 1

z z = x2 + 4y2

–5–4–3–2–1 –1 –2 –3 –4

1 2 3 4 5

32. Relative maximum at 10, - 22; no points of inflection; vertical asymptotes at x = - 2 and x = 2; horizontal asymptote at y = 0; increasing on 1- q , - 22 and 1- 2, 02, decreasing on 10, 22 and 12, q 2; concave up on 1- q , - 22 and 12, q 2, cony cave down on 1- 2, 22

x y

10 8 6 4 2

Chapter 6 Test, p. 599 15.

IA-37

z 1 1 z = x − − y2 − − x 3 2 3

−10 −8 −6 −4 −2−2

f(x) =

8 x2 − 4

2 4 6 8 10 x

−4 −6 −8 −10

x

Appendix A, p. 614 y

x 8y 20 1 8 20 -16 x y z , or 81 81z 16 9y 14z 4 57. 9x -16y 14z 4, or x 16

55.

y 21z 15 1 -9 21 15 x y z , or 125 125x 9 625x 16 58. 625x 16y -20z -12, or 20 12 y z 56.

Summary of Important Formulas for Differentiation

d k x = kx k - 1. dx

1. Power Rule. For any real number k,

2. Derivative of a Constant Function. If F1x2 = c, then F¿1x2 = 0. 3. Derivative of a Constant Times a Function. If F1x2 = cf1x2, then F¿1x2 = cf ¿1x2. 4. Derivative of a Sum. If F1x2 = f1x2 + g1x2, then F¿1x2 = f ¿1x2 + g¿1x2. 5. Derivative of a Difference. If F1x2 = f1x2 - g1x2, then F¿1x2 = f ¿1x2 - g¿1x2. 6. Derivative of a Product. If F1x2 = f1x2g1x2, then F¿1x2 = f1x2g¿1x2 + g1x2f ¿1x2. 7. Derivative of a Quotient. If F1x2 = F¿1x2 =

f1x2 , then g1x2 g1x2f ¿1x2 - f1x2g¿1x2 . 3g1x242

8. Extended Power Rule. If F1x2 = 3g1x24k, then

F¿1x2 = k3g1x24k - 1g¿1x2. 9. Chain Rule. If F1x2 = f3g1x24, then F¿1x2 = f ¿3g1x24g¿1x2. Or, if y = f1u2 and u = g1x2, then dy dy = dx du

# du . dx

Summary of Important Formulas for Differentiation (continued)

10.

d x e = ex dx

11.

d f 1x2 e = e f 1x2 # f ¿1x2 dx

12.

d 1 ln x = x , x 7 0 dx

13.

f ¿1x2 d ln f1x2 = , f1x2 7 0 dx f1x2

14.

d 1 ln ƒ x ƒ = x , x Z 0 dx

15.

f ¿1x2 d ln ƒ f1x2 ƒ = , f1x2 Z 0 dx f1x2

16.

d x a = 1ln a2a x dx

17.

d 1 log a x = dx ln a

# 1x ,

18.

1 d log a ƒ x ƒ = dx ln a

#

x 7 0

1 x, x Z 0

Table of Integrals ■ ■ ■

1.

Antiderivative of a constant: 1 k dx  kx  C Antiderivative of a constant times a function: 1 k # f1x2 dx  k 1 f1x2 dx Sum/difference property of antidifferentiation: 1 7f1x2 — g1x28 dx  1 f1x2 dx — 1 g1x2 dx L

xn+1 + C, n Z - 1 n + 1

x n dx =

dx 2. = ln x + C, x 7 0 L x 3. 4. 5.

6. 7. 8. 9. 10.

L L L

L L L L L

u dv = uv -

L

17.

v du

1 a2

x n e ax dx =

# e ax1ax

20. - 12 + C

n x ne ax x n - 1e ax dx + C a aL

1ln x2n dx = x1ln x2n - n

L

1ln x2n - 1 dx + C , n Z - 1

ln x 1 d + C, n Z - 1 n + 1 1n + 122

ax 11. a x dx = + C, a 7 0, a Z 1 ln a L 12. 13. 14. 15.

21. 22.

ln x dx = x ln x - x + C

x n ln x dx = x n + 1 c

1 L 2x 2 + a 2 1 L 2x - a 2 2

1 L x2 - a2 1 La - x 2

2

L x2a + x

2

1 L x2a 2 - x 2

dx = -

1 a + 2a 2 + x 2 ln ` ` +C a x

dx = -

a + 2a 2 - x 2 1 ln ` ` +C x a

a x a x dx = 2 + - 2 ln ƒ a + bx ƒ + C a + bx b b b L x a 1 19. dx = 2 + 2 ln ƒ a + bx ƒ + C 2 b 1a + bx2 b L 1a + bx2

1 # ax e + C a

xe ax dx =

1 2

18.

e x dx = e x + C e ax dx =

16.

dx = ln ƒ x + 2x 2 + a 2 ƒ + C dx = ln ƒ x + 2x 2 - a 2 ƒ + C

dx =

1 x - a ln ` ` +C 2a x + a

dx =

1 a + x ln ` ` +C 2a a - x

23. 24.

1 1 x dx = ln ` ` +C a a + bx x1a + bx2 L 1 L x1a + bx2

2

L

L L

26.

1 x 1 + 2 ln ` ` +C a + bx a1a + bx2 a

2x 2 ; a 2 dx =

1 2

C x 2x 2 ; a 2 ; a 2 ln ƒ x + 2x 2 ; a 2 ƒ D + C

x2a + bx dx =

2 15b 2

13bx - 2a21a + bx23>2 + C

x 2 2a + bx dx =

25.

dx =

2 105b 3

115b 2x 2 - 12abx + 8a 221a + bx23>2 + C

x dx L 2a + bx

=

2 3b 2

1bx - 2a22a + bx + C

x 2 dx L 2a + bx 2 = 13b 2x 2 - 4abx + 8a 222a + bx + C 15b 3